R Cookbook R Cookbook Paul Teetor Beijing • Cambridge • Farnham • Köln • Sebastopol • Tokyo R Cookbook by Paul Teetor Copyright © 2011 Paul Teetor. All rights reserved. Printed in the United States of America. Published by O’Reilly Media, Inc., 1005 Gravenstein Highway North, Sebastopol, CA 95472. O’Reilly books may be purchased for educational, business, or sales promotional use. Online editions are also available for most titles (http://my.safaribooksonline.com). For more information, contact our corporate/institutional sales department: (800) 998-9938 or corporate@oreilly.com. Editor: Mike Loukides Production Editor: Adam Zaremba Copyeditor: Matt Darnell Proofreader: Jennifer Knight Indexer: Jay Marchand Cover Designer: Karen Montgomery Interior Designer: David Futato Illustrator: Robert Romano Printing History: March 2011: First Edition. Nutshell Handbook, the Nutshell Handbook logo, and the O’Reilly logo are registered trademarks of O’Reilly Media, Inc. R Cookbook, the image of a harpy eagle, and related trade dress are trademarks of O’Reilly Media, Inc. Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book, and O’Reilly Media, Inc. was aware of a trademark claim, the designations have been printed in caps or initial caps. While every precaution has been taken in the preparation of this book, the publisher and authors assume no responsibility for errors or omissions, or for damages resulting from the use of the information con- tained herein. ISBN: 978-0-596-80915-7 [LSI] 1299102737 Table of Contents Preface .................................................................... xiii 1. Getting Started and Getting Help .......................................... 1 1.1 Downloading and Installing R 2 1.2 Starting R 4 1.3 Entering Commands 7 1.4 Exiting from R 8 1.5 Interrupting R 9 1.6 Viewing the Supplied Documentation 10 1.7 Getting Help on a Function 11 1.8 Searching the Supplied Documentation 13 1.9 Getting Help on a Package 14 1.10 Searching the Web for Help 16 1.11 Finding Relevant Functions and Packages 18 1.12 Searching the Mailing Lists 19 1.13 Submitting Questions to the Mailing Lists 20 2. Some Basics ........................................................... 23 2.1 Printing Something 23 2.2 Setting Variables 25 2.3 Listing Variables 26 2.4 Deleting Variables 27 2.5 Creating a Vector 28 2.6 Computing Basic Statistics 30 2.7 Creating Sequences 32 2.8 Comparing Vectors 34 2.9 Selecting Vector Elements 35 2.10 Performing Vector Arithmetic 38 2.11 Getting Operator Precedence Right 40 2.12 Defining a Function 41 2.13 Typing Less and Accomplishing More 43 v 2.14 Avoiding Some Common Mistakes 46 3. Navigating the Software ................................................ 51 3.1 Getting and Setting the Working Directory 51 3.2 Saving Your Workspace 52 3.3 Viewing Your Command History 53 3.4 Saving the Result of the Previous Command 53 3.5 Displaying the Search Path 54 3.6 Accessing the Functions in a Package 55 3.7 Accessing Built-in Datasets 57 3.8 Viewing the List of Installed Packages 58 3.9 Installing Packages from CRAN 59 3.10 Setting a Default CRAN Mirror 61 3.11 Suppressing the Startup Message 62 3.12 Running a Script 62 3.13 Running a Batch Script 63 3.14 Getting and Setting Environment Variables 66 3.15 Locating the R Home Directory 67 3.16 Customizing R 68 4. Input and Output ...................................................... 71 4.1 Entering Data from the Keyboard 72 4.2 Printing Fewer Digits (or More Digits) 73 4.3 Redirecting Output to a File 74 4.4 Listing Files 75 4.5 Dealing with “Cannot Open File” in Windows 76 4.6 Reading Fixed-Width Records 77 4.7 Reading Tabular Data Files 78 4.8 Reading from CSV Files 80 4.9 Writing to CSV Files 82 4.10 Reading Tabular or CSV Data from the Web 83 4.11 Reading Data from HTML Tables 84 4.12 Reading Files with a Complex Structure 86 4.13 Reading from MySQL Databases 89 4.14 Saving and Transporting Objects 92 5. Data Structures ........................................................ 95 5.1 Appending Data to a Vector 101 5.2 Inserting Data into a Vector 103 5.3 Understanding the Recycling Rule 103 5.4 Creating a Factor (Categorical Variable) 105 5.5 Combining Multiple Vectors into One Vector and a Factor 107 5.6 Creating a List 108 vi | Table of Contents 5.7 Selecting List Elements by Position 109 5.8 Selecting List Elements by Name 111 5.9 Building a Name/Value Association List 112 5.10 Removing an Element from a List 114 5.11 Flatten a List into a Vector 115 5.12 Removing NULL Elements from a List 116 5.13 Removing List Elements Using a Condition 117 5.14 Initializing a Matrix 118 5.15 Performing Matrix Operations 119 5.16 Giving Descriptive Names to the Rows and Columns of a Matrix 120 5.17 Selecting One Row or Column from a Matrix 121 5.18 Initializing a Data Frame from Column Data 122 5.19 Initializing a Data Frame from Row Data 123 5.20 Appending Rows to a Data Frame 125 5.21 Preallocating a Data Frame 126 5.22 Selecting Data Frame Columns by Position 127 5.23 Selecting Data Frame Columns by Name 131 5.24 Selecting Rows and Columns More Easily 132 5.25 Changing the Names of Data Frame Columns 133 5.26 Editing a Data Frame 135 5.27 Removing NAs from a Data Frame 136 5.28 Excluding Columns by Name 137 5.29 Combining Two Data Frames 138 5.30 Merging Data Frames by Common Column 140 5.31 Accessing Data Frame Contents More Easily 141 5.32 Converting One Atomic Value into Another 143 5.33 Converting One Structured Data Type into Another 144 6. Data Transformations ................................................. 147 6.1 Splitting a Vector into Groups 148 6.2 Applying a Function to Each List Element 149 6.3 Applying a Function to Every Row 151 6.4 Applying a Function to Every Column 152 6.5 Applying a Function to Groups of Data 154 6.6 Applying a Function to Groups of Rows 156 6.7 Applying a Function to Parallel Vectors or Lists 158 7. Strings and Dates ..................................................... 161 7.1 Getting the Length of a String 163 7.2 Concatenating Strings 163 7.3 Extracting Substrings 164 7.4 Splitting a String According to a Delimiter 165 7.5 Replacing Substrings 166 Table of Contents | vii 7.6 Seeing the Special Characters in a String 167 7.7 Generating All Pairwise Combinations of Strings 168 7.8 Getting the Current Date 169 7.9 Converting a String into a Date 170 7.10 Converting a Date into a String 171 7.11 Converting Year, Month, and Day into a Date 172 7.12 Getting the Julian Date 173 7.13 Extracting the Parts of a Date 174 7.14 Creating a Sequence of Dates 175 8. Probability .......................................................... 177 8.1 Counting the Number of Combinations 179 8.2 Generating Combinations 180 8.3 Generating Random Numbers 180 8.4 Generating Reproducible Random Numbers 182 8.5 Generating a Random Sample 183 8.6 Generating Random Sequences 184 8.7 Randomly Permuting a Vector 185 8.8 Calculating Probabilities for Discrete Distributions 186 8.9 Calculating Probabilities for Continuous Distributions 188 8.10 Converting Probabilities to Quantiles 189 8.11 Plotting a Density Function 190 9. General Statistics ..................................................... 195 9.1 Summarizing Your Data 197 9.2 Calculating Relative Frequencies 199 9.3 Tabulating Factors and Creating Contingency Tables 200 9.4 Testing Categorical Variables for Independence 201 9.5 Calculating Quantiles (and Quartiles) of a Dataset 201 9.6 Inverting a Quantile 202 9.7 Converting Data to Z-Scores 203 9.8 Testing the Mean of a Sample (t Test) 203 9.9 Forming a Confidence Interval for a Mean 205 9.10 Forming a Confidence Interval for a Median 206 9.11 Testing a Sample Proportion 207 9.12 Forming a Confidence Interval for a Proportion 208 9.13 Testing for Normality 209 9.14 Testing for Runs 210 9.15 Comparing the Means of Two Samples 212 9.16 Comparing the Locations of Two Samples Nonparametrically 213 9.17 Testing a Correlation for Significance 215 9.18 Testing Groups for Equal Proportions 216 9.19 Performing Pairwise Comparisons Between Group Means 218 viii | Table of Contents 9.20 Testing Two Samples for the Same Distribution 219 10. Graphics ............................................................. 221 10.1 Creating a Scatter Plot 223 10.2 Adding a Title and Labels 225 10.3 Adding a Grid 226 10.4 Creating a Scatter Plot of Multiple Groups 227 10.5 Adding a Legend 229 10.6 Plotting the Regression Line of a Scatter Plot 231 10.7 Plotting All Variables Against All Other Variables 233 10.8 Creating One Scatter Plot for Each Factor Level 233 10.9 Creating a Bar Chart 236 10.10 Adding Confidence Intervals to a Bar Chart 237 10.11 Coloring a Bar Chart 239 10.12 Plotting a Line from x and y Points 241 10.13 Changing the Type, Width, or Color of a Line 242 10.14 Plotting Multiple Datasets 243 10.15 Adding Vertical or Horizontal Lines 245 10.16 Creating a Box Plot 246 10.17 Creating One Box Plot for Each Factor Level 247 10.18 Creating a Histogram 248 10.19 Adding a Density Estimate to a Histogram 250 10.20 Creating a Discrete Histogram 252 10.21 Creating a Normal Quantile-Quantile (Q-Q) Plot 252 10.22 Creating Other Quantile-Quantile Plots 254 10.23 Plotting a Variable in Multiple Colors 256 10.24 Graphing a Function 258 10.25 Pausing Between Plots 259 10.26 Displaying Several Figures on One Page 260 10.27 Opening Additional Graphics Windows 262 10.28 Writing Your Plot to a File 263 10.29 Changing Graphical Parameters 264 11. Linear Regression and ANOVA ........................................... 267 11.1 Performing Simple Linear Regression 269 11.2 Performing Multiple Linear Regression 270 11.3 Getting Regression Statistics 272 11.4 Understanding the Regression Summary 275 11.5 Performing Linear Regression Without an Intercept 278 11.6 Performing Linear Regression with Interaction Terms 279 11.7 Selecting the Best Regression Variables 281 11.8 Regressing on a Subset of Your Data 284 11.9 Using an Expression Inside a Regression Formula 285 Table of Contents | ix 11.10 Regressing on a Polynomial 286 11.11 Regressing on Transformed Data 287 11.12 Finding the Best Power Transformation (Box–Cox Procedure) 289 11.13 Forming Confidence Intervals for Regression Coefficients 292 11.14 Plotting Regression Residuals 293 11.15 Diagnosing a Linear Regression 293 11.16 Identifying Influential Observations 296 11.17 Testing Residuals for Autocorrelation (Durbin–Watson Test) 298 11.18 Predicting New Values 300 11.19 Forming Prediction Intervals 301 11.20 Performing One-Way ANOVA 302 11.21 Creating an Interaction Plot 303 11.22 Finding Differences Between Means of Groups 304 11.23 Performing Robust ANOVA (Kruskal–Wallis Test) 308 11.24 Comparing Models by Using ANOVA 309 12. Useful Tricks ......................................................... 313 12.1 Peeking at Your Data 313 12.2 Widen Your Output 314 12.3 Printing the Result of an Assignment 315 12.4 Summing Rows and Columns 315 12.5 Printing Data in Columns 316 12.6 Binning Your Data 317 12.7 Finding the Position of a Particular Value 318 12.8 Selecting Every nth Element of a Vector 319 12.9 Finding Pairwise Minimums or Maximums 320 12.10 Generating All Combinations of Several Factors 321 12.11 Flatten a Data Frame 322 12.12 Sorting a Data Frame 323 12.13 Sorting by Two Columns 324 12.14 Stripping Attributes from a Variable 325 12.15 Revealing the Structure of an Object 326 12.16 Timing Your Code 329 12.17 Suppressing Warnings and Error Messages 329 12.18 Taking Function Arguments from a List 331 12.19 Defining Your Own Binary Operators 332 13. Beyond Basic Numerics and Statistics .................................... 335 13.1 Minimizing or Maximizing a Single-Parameter Function 335 13.2 Minimizing or Maximizing a Multiparameter Function 336 13.3 Calculating Eigenvalues and Eigenvectors 338 13.4 Performing Principal Component Analysis 338 13.5 Performing Simple Orthogonal Regression 340 x | Table of Contents 13.6 Finding Clusters in Your Data 342 13.7 Predicting a Binary-Valued Variable (Logistic Regression) 345 13.8 Bootstrapping a Statistic 346 13.9 Factor Analysis 349 14. Time Series Analysis ................................................... 355 14.1 Representing Time Series Data 356 14.2 Plotting Time Series Data 359 14.3 Extracting the Oldest or Newest Observations 361 14.4 Subsetting a Time Series 363 14.5 Merging Several Time Series 364 14.6 Filling or Padding a Time Series 366 14.7 Lagging a Time Series 368 14.8 Computing Successive Differences 369 14.9 Performing Calculations on Time Series 370 14.10 Computing a Moving Average 372 14.11 Applying a Function by Calendar Period 373 14.12 Applying a Rolling Function 375 14.13 Plotting the Autocorrelation Function 376 14.14 Testing a Time Series for Autocorrelation 377 14.15 Plotting the Partial Autocorrelation Function 378 14.16 Finding Lagged Correlations Between Two Time Series 379 14.17 Detrending a Time Series 382 14.18 Fitting an ARIMA Model 383 14.19 Removing Insignificant ARIMA Coefficients 386 14.20 Running Diagnostics on an ARIMA Model 387 14.21 Making Forecasts from an ARIMA Model 389 14.22 Testing for Mean Reversion 391 14.23 Smoothing a Time Series 393 Index ..................................................................... 397 Table of Contents | xi Preface R is a powerful tool for statistics, graphics, and statistical programming. It is used by tens of thousands of people daily to perform serious statistical analyses. It is a free, open source system whose implementation is the collective accomplishment of many intel- ligent, hard-working people. There are more than 2,000 available add-ons, and R is a serious rival to all commercial statistical packages. But R can be frustrating. It’s not obvious how to accomplish many tasks, even simple ones. The simple tasks are easy once you know how, yet figuring out that “how” can be maddening. This book is full of how-to recipes, each of which solves a specific problem. The recipe includes a quick introduction to the solution followed by a discussion that aims to unpack the solution and give you some insight into how it works. I know these recipes are useful and I know they work, because I use them myself. The range of recipes is broad. It starts with basic tasks before moving on to input and output, general statistics, graphics, and linear regression. Any significant work with R will involve most or all of these areas. If you are a beginner then this book will get you started faster. If you are an intermediate user, this book is useful for expanding your horizons and jogging your memory (“How do I do that Kolmogorov–Smirnov test again?”). The book is not a tutorial on R, although you will learn something by studying the recipes. It is not a reference manual, but it does contain a lot of useful information. It is not a book on programming in R, although many recipes are useful inside R scripts. Finally, this book is not an introduction to statistics. Many recipes assume that you are familiar with the underlying statistical procedure, if any, and just want to know how it’s done in R. xiii The Recipes Most recipes use one or two R functions to solve a specific problem. It’s important to remember that I do not describe the functions in detail; rather, I describe just enough to solve the immediate problem. Nearly every such function has additional capabilities beyond those described here, and some have amazing capabilities. I strongly urge you to read the function’s help page. You will likely learn something valuable. Each recipe presents one way to solve a particular problem. Of course, there are likely several reasonable solutions to each problem. When I knew of multiple solutions, I generally selected the simplest one. For any given task, you can probably discover sev- eral alternative solutions yourself. This is a cookbook, not a bible. In particular, R has literally thousands of downloadable add-on packages, many of which implement alternative algorithms and statistical methods. This book concen- trates on the core functionality available through the basic distribution, so your best source of alternative solutions may be searching for an add-on package (Recipe 1.11). A Note on Terminology The goal of every recipe is to solve a problem and solve it quickly. Rather than laboring in tedious prose, I occasionally streamline the description with terminology that is correct but not precise. A good example is the term “generic function”. I refer to print(x) and plot(x) as generic functions because they work for many kinds of x, handling each kind appropriately. A computer scientist would wince at my terminology because, strictly speaking, these are not simply “functions”; they are polymorphic methods with dynamic dispatching. But if I carefully unpacked every such technical detail, the essential solution would be buried in the technicalities. So I just call them functions, which I think is more readable. Another example, taken from statistics, is the complexity surrounding the semantics of statistical hypothesis testing. Using the strict language of probability theory would obscure the practical application of some tests, so I use more colloquial language when describing each statistical test. See the “Introduction” to Chapter 9 for more about how hypothesis tests are presented in the recipes. My goal is to make the power of R available to a wide audience by writing readably, not formally. I hope that experts in their respective fields will understand if my termi- nology is occasionally informal. Software and Platform Notes The base distribution of R has frequent and planned releases, but the language defini- tion and core implementation are stable. The recipes in this book should work with any recent release of the base distribution. xiv | Preface Some recipes have platform-specific considerations, and I have carefully noted them. Those recipes mostly deal with software issues, such as installation and configuration. As far as I know, all other recipes will work on all three major platforms for R: Windows, OS X, and Linux/Unix. Other Resources On the Web The mother ship for all things R is the R project site. From there you can download binaries, add-on packages, documentation, and source code as well as many other resources. Beyond the R project site, I recommend using an R-specific search engine—such as Rseek, created by Sasha Goodman. You can use a generic search engine, such as Google, but the “R” search term brings up too much extraneous stuff. See Recipe 1.10 for more about searching the Web. Reading blogs is a great way to learn about R and stay abreast of leading-edge developments. There are surprisingly many such blogs, so I recommend following two blog-of-blogs: R-bloggers, created by Tal Galili; and PlanetR. By subscribing to their RSS feeds, you will be notified of interesting and useful articles from dozens of websites. R books There are many, many books about learning and using R; listed here are a few that I have found useful. Note that the R project site contains an extensive bibliography of books related to R. I recommend An Introduction to R, by William Venables et al. (Network Theory Limited). It covers many topics and is useful for beginners. You can download the PDF for free from CRAN; or, better yet, buy the printed copy because the profits are donated to the R project. R in a Nutshell, by Joseph Adler (O’Reilly), is the quick tutorial and reference you’ll keep by your side. It covers many more topics than this Cookbook. Anyone doing serious graphics work in R will want R Graphics by Paul Murrell (Chapman & Hall/CRC). Depending on which graphics package you use, you may also want Lattice: Multivariate Data Visualization with R by Deepayan Sarkar (Springer) and ggplot2: Elegant Graphics for Data Analysis by Hadley Wickham (Springer). Modern Applied Statistics with S (4th ed.), by William Venables and Brian Ripley (Springer), uses R to illustrate many advanced statistical techniques. The book’s functions and datasets are available in the MASS package, which is included in the standard distribution of R. Preface | xv I’m not wild about any book on programming in R, although new books appear regularly. For programming, I suggest using R in a Nutshell together with S Pro- gramming by William Venables and Brian Ripley (Springer). I also suggest down- loading the R Language Definition. The Definition is a work in progress, but it can answer many of your detailed questions regarding R as a programming language. Statistics books You will need a good statistics textbook or reference book to accurately interpret the statistical tests performed in R. There are many such fine books—far too many for me to recommend any one above the others. For learning statistics, a great choice is Using R for Introductory Statistics by John Verzani (Chapman & Hall/CRC). It teaches statistics and R together, giving you the necessary computer skills to apply the statistical methods. Increasingly, statistics authors are using R to illustrate their methods. If you work in a specialized field, then you will likely find a useful and relevant book in the R project bibliography. Conventions Used in This Book The following typographical conventions are used in this book: Italic Indicates new terms, URLs, email addresses, filenames, and file extensions. Constant width Used for program listings, as well as within paragraphs to refer to program elements such as variable or function names, databases, data types, environment variables, statements, and keywords. Constant width bold Shows commands or other text that should be typed literally by the user. Constant width italic Shows text that should be replaced with user-supplied values or by values deter- mined by context. This icon signifies a tip, suggestion, or general note. This icon indicates a warning or caution. xvi | Preface Using Code Examples This book is here to help you get your job done. In general, you may use the code in this book in your programs and documentation. You do not need to contact us for permission unless you’re reproducing a significant portion of the code. For example, writing a program that uses several chunks of code from this book does not require permission. Selling or distributing a CD-ROM of examples from O’Reilly books does require permission. Answering a question by citing this book and quoting example code does not require permission. Incorporating a significant amount of example code from this book into your product’s documentation does require permission. We appreciate, but do not require, attribution. An attribution usually includes the title, author, publisher, and ISBN. For example: “R Cookbook by Paul Teetor. Copyright 2011 Paul Teetor, 978-0-596-80915-7.” If you feel your use of code examples falls outside fair use or the permission just de- scribed, feel free to contact us at permissions@oreilly.com. 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How to Contact Us Please address comments and questions concerning this book to the publisher: O’Reilly Media, Inc. 1005 Gravenstein Highway North Sebastopol, CA 95472 800-998-9938 (in the United States or Canada) 707-829-0515 (international or local) 707-829-0104 (fax) Preface | xvii We have a web page for this book, where we list errata, examples, and any additional information. You can access this page at: http://www.oreilly.com/catalog/9780596809157 To comment or ask technical questions about this book, send email to: bookquestions@oreilly.com For more information about our books, courses, conferences, and news, see our website at http://oreilly.com. Find us on Facebook: http://facebook.com/oreilly Follow us on Twitter: http://twitter.com/oreillymedia Watch us on YouTube: http://www.youtube.com/oreillymedia Acknowledgments With gratitude I thank the R community in general and the R Core Team in particular. Their selfless contributions are enormous. The world of statistics is benefiting tremen- dously from their work. I wish to thank the book’s technical reviewers: James D. Long, Timothy McMurry, David Reiner, Jeffery Ryan, and John Verzani. My thanks, also, to Joe Adler for his comments on the text. Their feedback was critical for improving the quality, accuracy, and usefulness of this book. They saved me numerous times from showing the world how foolish I really am. Mike Loukides has been an excellent editor, and I am deeply grateful for his wisdom and guidance. When I started this little project, someone told me that Mike is the best in the business. I believe it. My greatest gratitude is to my dear wife, Anna. Her support made this book possible. Her partnership made it a joy. xviii | Preface CHAPTER 1 Getting Started and Getting Help Introduction This chapter sets the groundwork for the other chapters. It explains how to download, install, and run R. More importantly, it also explains how to get answers to your questions. The R com- munity provides a wealth of documentation and help. You are not alone. Here are some common sources of help: Local, installed documentation When you install R on your computer, a mass of documentation is also installed. You can browse the local documentation (Recipe 1.6) and search it (Recipe 1.8). I am amazed how often I search the Web for an answer only to discover it was already available in the installed documentation. Task views A task view describes packages that are specific to one area of statistical work, such as econometrics, medical imaging, psychometrics, or spatial statistics. Each task view is written and maintained by an expert in the field. There are 28 such task views, so there is likely to be one or more for your areas of interest. I recommend that every beginner find and read at least one task view in order to gain a sense of R’s possibilities (Recipe 1.11). Package documentation Most packages include useful documentation. Many also include overviews and tutorials, called vignettes in the R community. The documentation is kept with the packages in package repositories, such as CRAN, and it is automatically installed on your machine when you install a package. Mailing lists Volunteers have generously donated many hours of time to answer beginners’ questions that are posted to the R mailing lists. The lists are archived, so you can search the archives for answers to your questions (Recipe 1.12). 1 Question and answer (Q&A) websites On a Q&A site, anyone can post a question, and knowledgeable people can re- spond. Readers vote on the answers, so the best answers tend to emerge over time. All this information is tagged and archived for searching. These sites are a cross between a mailing list and a social network; the Stack Overflow site is a good example. The Web The Web is loaded with information about R, and there are R-specific tools for searching it (Recipe 1.10). The Web is a moving target, so be on the lookout for new, improved ways to organize and search information regarding R. 1.1 Downloading and Installing R Problem You want to install R on your computer. Solution Windows and OS X users can download R from CRAN, the Comprehensive R Archive Network. Linux and Unix users can install R packages using their package management tool: Windows 1. Open http://www.r-project.org/ in your browser. 2. Click on “CRAN”. You’ll see a list of mirror sites, organized by country. 3. Select a site near you. 4. Click on “Windows” under “Download and Install R”. 5. Click on “base”. 6. Click on the link for downloading the latest version of R (an .exe file). 7. When the download completes, double-click on the .exe file and answer the usual questions. OS X 1. Open http://www.r-project.org/ in your browser. 2. Click on “CRAN”. You’ll see a list of mirror sites, organized by country. 3. Select a site near you. 4. Click on “MacOS X”. 5. Click on the .pkg file for the latest version of R, under “Files:”, to download it. 6. When the download completes, double-click on the .pkg file and answer the usual questions. 2 | Chapter 1: Getting Started and Getting Help Linux or Unix The major Linux distributions have packages for installing R. Here are some examples: Distribution Package name Ubuntu or Debian r-base Red Hat or Fedora R.i386 Suse R-base Use the system’s package manager to download and install the package. Normally, you will need the root password or sudo privileges; otherwise, ask a system ad- ministrator to perform the installation. Discussion Installing R on Windows or OS X is straightforward because there are prebuilt binaries for those platforms. You need only follow the preceding instructions. The CRAN Web pages also contain links to installation-related resources, such as frequently asked questions (FAQs) and tips for special situations (“How do I install R when using Win- dows Vista?”) that you may find useful. Theoretically, you can install R on Linux or Unix in one of two ways: by installing a distribution package or by building it from scratch. In practice, installing a package is the preferred route. The distribution packages greatly streamline both the initial in- stallation and subsequent updates. On Ubuntu or Debian, use apt-get to download and install R. Run under sudo to have the necessary privileges: $ sudo apt-get install r-base On Red Hat or Fedora, use yum: $ sudo yum install R.i386 Most platforms also have graphical package managers, which you might find more convenient. Beyond the base packages, I recommend installing the documentation packages, too. On my Ubuntu machine, for example, I installed r-base-html (because I like browsing the hyperlinked documentation) as well as r-doc-html, which installs the important R manuals locally: $ sudo apt-get install r-base-html r-doc-html Some Linux repositories also include prebuilt copies of R packages available on CRAN. I don’t use them because I’d rather get my software directly from CRAN itself, which usually has the freshest versions. 1.1 Downloading and Installing R | 3 In rare cases, you may need to build R from scratch. You might have an obscure, un- supported version of Unix; or you might have special considerations regarding per- formance or configuration. The build procedure on Linux or Unix is quite standard. Download the tarball from the home page of your CRAN mirror; it’s called something like R-2.12.1.tar.gz, except the “2.12.1” will be replaced by the latest version. Unpack the tarball, look for a file called INSTALL, and follow the directions. See Also R in a Nutshell (O’Reilly) contains more details of downloading and installing R, in- cluding instructions for building the Windows and OS X versions. Perhaps the ultimate guide is the one entitled R Installation and Administration, available on CRAN, which describes building and installing R on a variety of platforms. This recipe is about installing the base package. See Recipe 3.9 for installing add-on packages from CRAN. 1.2 Starting R Problem You want to run R on your computer. Solution Windows Click on Start → All Programs → R; or double-click on the R icon on your desktop (assuming the installer created an icon for you). OS X Either click on the icon in the Applications directory or put the R icon on the dock and click on the icon there. Alternatively, you can just type R on a Unix command line in a shell. Linux or Unix Start the R program from the shell prompt using the R command (uppercase R). Discussion How you start R depends upon your platform. Starting on Windows When you start R, it opens a new window. The window includes a text pane, called the R Console, where you enter R expressions (see Figure 1-1). 4 | Chapter 1: Getting Started and Getting Help There is an odd thing about the Windows Start menu for R. Every time you upgrade to a new version of R, the Start menu expands to contain the new version while keeping all the previously installed versions. So if you’ve upgraded, you may face several choices such as “R 2.8.1”, “R 2.9.1”, “R 2.10.1”, and so forth. Pick the newest one. (You might also consider uninstalling the older versions to reduce the clutter.) Using the Start menu is cumbersome, so I suggest starting R in one of two other ways: by creating a desktop shortcut or by double-clicking on your .RData file. The installer may have created a desktop icon. If not, creating a shortcut is easy: follow the Start menu to the R program, but instead of left-clicking to run R, press and hold your mouse’s right button on the program name, drag the program name to your desk- top, and release the mouse button. Windows will ask if you want to Copy Here or Move Here. Select Copy Here, and the shortcut will appear on your desktop. Another way to start R is by double-clicking on a .RData file in your working directory. This is the file that R creates to save your workspace. The first time you create a direc- tory, start R and change to that directory. Save your workspace there, either by exiting or using the save.image function. That will create the .RData file. Thereafter, you can simply open the directory in Windows Explorer and then double-click on the .RData file to start R. Perhaps the most baffling aspect of starting R on Windows is embodied in a simple question: When R starts, what is the working directory? The answer, of course, is that “it depends”: Figure 1-1. R on Windows 1.2 Starting R | 5 • If you start R from the Start menu, the working directory is normally either C:\Documents and Settings\\My Documents (Windows XP) or C:\Users \\Documents (Windows Vista, Windows 7). You can override this de- fault by setting the R_USER environment variable to an alternative directory path. • If you start R from a desktop shortcut, you can specify an alternative startup directory that becomes the working directory when R is started. To specify the alternative directory, right-click on the shortcut, select Properties, enter the direc- tory path in the box labeled “Start in”, and click OK. • Starting R by double-clicking on your .RData file is the most straightforward solution to this little problem. R will automatically change its working directory to be the file’s directory, which is usually what you want. In any event, you can always use the getwd function to discover your current working directory (Recipe 3.1). Just for the record, Windows also has a console version of R called Rterm.exe. You’ll find it in the bin subdirectory of your R installation. It is much less convenient than the graphic user interface (GUI) version, and I never use it. I recommend it only for batch (noninteractive) usage such as running jobs from the Windows scheduler. In this book, I assume you are running the GUI version of R, not the console version. Starting on OS X Run R by clicking the R icon in the Applications folder. (If you use R frequently, you can drag it from the folder to the dock.) That will run the GUI version, which is some- what more convenient than the console version. The GUI version displays your working directory, which is initially your home directory. OS X also lets you run the console version of R by typing R at the shell prompt. Starting on Linux and Unix Start the console version of R from the Unix shell prompt simply by typing R, the name of the program. Be careful to type an uppercase R, not a lowercase r. The R program has a bewildering number of command line options. Use the --help option to see the complete list. See Also See Recipe 1.4 for exiting from R, Recipe 3.1 for more about the current working directory, Recipe 3.2 for more about saving your workspace, and Recipe 3.11 for sup- pressing the start-up message. See Chapter 2 of R in a Nutshell. 6 | Chapter 1: Getting Started and Getting Help 1.3 Entering Commands Problem You’ve started R, and you’ve got a command prompt. Now what? Solution Simply enter expressions at the command prompt. R will evaluate them and print (dis- play) the result. You can use command-line editing to facilitate typing. Discussion R prompts you with “>”. To get started, just treat R like a big calculator: enter an expression, and R will evaluate the expression and print the result: > 1+1 [1] 2 The computer adds one and one, giving two, and displays the result. The [1] before the 2 might be confusing. To R, the result is a vector, even though it has only one element. R labels the value with [1] to signify that this is the first element of the vector...which is not surprising, since it’s the only element of the vector. R will prompt you for input until you type a complete expression. The expression max(1,3,5) is a complete expression, so R stops reading input and evaluates what it’s got: > max(1,3,5) [1] 5 In contrast, “max(1,3,” is an incomplete expression, so R prompts you for more input. The prompt changes from greater-than (>) to plus (+), letting you know that R expects more: > max(1,3, + 5) [1] 5 It’s easy to mistype commands, and retyping them is tedious and frustrating. So R includes command-line editing to make life easier. It defines single keystrokes that let you easily recall, correct, and reexecute your commands. My own typical command- line interaction goes like this: 1. I enter an R expression with a typo. 2. R complains about my mistake. 3. I press the up-arrow key to recall my mistaken line. 4. I use the left and right arrow keys to move the cursor back to the error. 5. I use the Delete key to delete the offending characters. 1.3 Entering Commands | 7 6. I type the corrected characters, which inserts them into the command line. 7. I press Enter to reexecute the corrected command. That’s just the basics. R supports the usual keystrokes for recalling and editing com- mand lines, as listed in Table 1-1. Table 1-1. Keystrokes for command-line editing Labeled key Ctrl-key combination Effect Up arrow Ctrl-P Recall previous command by moving backward through the history of commands. Down arrow Ctrl-N Move forward through the history of commands. Backspace Ctrl-H Delete the character to the left of cursor. Delete (Del) Ctrl-D Delete the character to the right of cursor. Home Ctrl-A Move cursor to the start of the line. End Ctrl-E Move cursor to the end of the line. Right arrow Ctrl-F Move cursor right (forward) one character. Left arrow Ctrl-B Move cursor left (back) one character. Ctrl-K Delete everything from the cursor position to the end of the line. Ctrl-U Clear the whole darn line and start over. Tab Name completion (on some platforms). On Windows and OS X, you can also use the mouse to highlight commands and then use the usual copy and paste commands to paste text into a new command line. See Also See Recipe 2.13. From the Windows main menu, follow Help → Console for a complete list of keystrokes useful for command-line editing. 1.4 Exiting from R Problem You want to exit from R. Solution Windows Select File → Exit from the main menu; or click on the red X in the upper-right corner of the window frame. 8 | Chapter 1: Getting Started and Getting Help OS X Press CMD-q (apple-q); or click on the red X in the upper-left corner of the window frame. Linux or Unix At the command prompt, press Ctrl-D. On all platforms, you can also use the q function (as in quit) to terminate the program. > q() Note the empty parentheses, which are necessary to call the function. Discussion Whenever you exit, R asks if you want to save your workspace. You have three choices: • Save your workspace and exit. • Don’t save your workspace, but exit anyway. • Cancel, returning to the command prompt rather than exiting. If you save your workspace, then R writes it to a file called .RData in the current working directory. This will overwrite the previously saved workspace, if any, so don’t save if you don’t like the changes to your workspace (e.g., if you have accidentally erased critical data). See Also See Recipe 3.1 for more about the current working directory and Recipe 3.2 for more about saving your workspace. See Chapter 2 of R in a Nutshell. 1.5 Interrupting R Problem You want to interrupt a long-running computation and return to the command prompt without exiting R. Solution Windows or OS X Either press the Esc key or click on the Stop-sign icon. Linux or Unix Press Ctrl-C. This will interrupt R without terminating it. 1.5 Interrupting R | 9 Discussion Interrupting R can leave your variables in an indeterminate state, depending upon how far the computation had progressed. Check your workspace after interrupting. See Also See Recipe 1.4. 1.6 Viewing the Supplied Documentation Problem You want to read the documentation supplied with R. Solution Use the help.start function to see the documentation’s table of contents: > help.start() From there, links are available to all the installed documentation. Discussion The base distribution of R includes a wealth of documentation—literally thousands of pages. When you install additional packages, those packages contain documentation that is also installed on your machine. It is easy to browse this documentation via the help.start function, which opens a window on the top-level table of contents; see Figure 1-2. The two links in the Reference section are especially useful: Packages Click here to see a list of all the installed packages, both in the base packages and the additional, installed packages. Click on a package name to see a list of its func- tions and datasets. Search Engine & Keywords Click here to access a simple search engine, which allows you to search the docu- mentation by keyword or phrase. There is also a list of common keywords, organized by topic; click one to see the associated pages. See Also The local documentation is copied from the R Project website, which may have updated documents. 10 | Chapter 1: Getting Started and Getting Help 1.7 Getting Help on a Function Problem You want to know more about a function that is installed on your machine. Solution Use help to display the documentation for the function: > help(functionname) Use args for a quick reminder of the function arguments: > args(functionname) Use example to see examples of using the function: > example(functionname) Discussion I present many R functions in this book. Every R function has more bells and whistles than I can possibly describe. If a function catches your interest, I strongly suggest read- Figure 1-2. Documentation table of contents 1.7 Getting Help on a Function | 11 ing the help page for that function. One of its bells or whistles might be very useful to you. Suppose you want to know more about the mean function. Use the help function like this: > help(mean) This will either open a window with function documentation or display the documen- tation on your console, depending upon your platform. A shortcut for the help com- mand is to simply type ? followed by the function name: > ?mean Sometimes you just want a quick reminder of the arguments to a function: What are they, and in what order do they occur? Use the args function: > args(mean) function (x, ...) NULL > args(sd) function (x, na.rm = FALSE) NULL The first line of output from args is a synopsis of the function call. For mean, the synopsis shows one argument, x, which is a vector of numbers. For sd, the synopsis shows the same vector, x, and an optional argument called na.rm. (You can ignore the second line of output, which is often just NULL.) Most documentation for functions includes examples near the end. A cool feature of R is that you can request that it execute the examples, giving you a little demonstration of the function’s capabilities. The documentation for the mean function, for instance, contains examples, but you don’t need to type them yourself. Just use the example function to watch them run: > example(mean) mean> x <- c(0:10, 50) mean> xm <- mean(x) mean> c(xm, mean(x, trim = 0.1)) [1] 8.75 5.50 mean> mean(USArrests, trim = 0.2) Murder Assault UrbanPop Rape 7.42 167.60 66.20 20.16 The user typed example(mean). Everything else was produced by R, which executed the examples from the help page and displayed the results. 12 | Chapter 1: Getting Started and Getting Help See Also See Recipe 1.8 for searching for functions and Recipe 3.5 for more about the search path. 1.8 Searching the Supplied Documentation Problem You want to know more about a function that is installed on your machine, but the help function reports that it cannot find documentation for any such function. Alternatively, you want to search the installed documentation for a keyword. Solution Use help.search to search the R documentation on your computer: > help.search("pattern") A typical pattern is a function name or keyword. Notice that it must be enclosed in quotation marks. For your convenience, you can also invoke a search by using two question marks (in which case the quotes are not required): > ??pattern Discussion You may occasionally request help on a function only to be told R knows nothing about it: > help(adf.test) No documentation for 'adf.test' in specified packages and libraries: you could try 'help.search("adf.test")' This can be frustrating if you know the function is installed on your machine. Here the problem is that the function’s package is not currently loaded, and you don’t know which package contains the function. It’s a kind of catch-22 (the error message indicates the package is not currently in your search path, so R cannot find the help file; see Recipe 3.5 for more details). The solution is to search all your installed packages for the function. Just use the help.search function, as suggested in the error message: > help.search("adf.test") 1.8 Searching the Supplied Documentation | 13 The search will produce a listing of all packages that contain the function: Help files with alias or concept or title matching 'adf.test' using regular expression matching: tseries::adf.test Augmented Dickey-Fuller Test Type '?PKG::FOO' to inspect entry 'PKG::FOO TITLE'. The following output, for example, indicates that the tseries package contains the adf.test function. You can see its documentation by explicitly telling help which pack- age contains the function: > help(adf.test, package="tseries") Alternatively, you can insert the tseries package into your search list and repeat the original help command, which will then find the function and display the documentation. You can broaden your search by using keywords. R will then find any installed docu- mentation that contains the keywords. Suppose you want to find all functions that mention the Augmented Dickey–Fuller (ADF) test. You could search on a likely pattern: > help.search("dickey-fuller") On my machine, the result looks like this because I’ve installed two additional packages (fUnitRoots and urca) that implement the ADF test: Help files with alias or concept or title matching 'dickey-fuller' using fuzzy matching: fUnitRoots::DickeyFullerPValues Dickey-Fuller p Values tseries::adf.test Augmented Dickey-Fuller Test urca::ur.df Augmented-Dickey-Fuller Unit Root Test Type '?PKG::FOO' to inspect entry 'PKG::FOO TITLE'. See Also You can also access the local search engine through the documentation browser; see Recipe 1.6 for how this is done. See Recipe 3.5 for more about the search path and Recipe 4.4 for getting help on functions. 1.9 Getting Help on a Package Problem You want to learn more about a package installed on your computer. 14 | Chapter 1: Getting Started and Getting Help Solution Use the help function and specify a package name (without a function name): > help(package="packagename") Discussion Sometimes you want to know the contents of a package (the functions and datasets). This is especially true after you download and install a new package, for example. The help function can provide the contents plus other information once you specify the package name. This call to help will display the information for the tseries package, a standard pack- age in the base distribution: > help(package="tseries") The information begins with a description and continues with an index of functions and datasets. On my machine, the first few lines look like this: Information on package 'tseries' Description: Package: tseries Version: 0.10-22 Date: 2009-11-22 Title: Time series analysis and computational finance Author: Compiled by Adrian Trapletti Maintainer: Kurt Hornik Description: Package for time series analysis and computational finance Depends: R (>= 2.4.0), quadprog, stats, zoo Suggests: its Imports: graphics, stats, utils License: GPL-2 Packaged: 2009-11-22 19:03:45 UTC; hornik Repository: CRAN Date/Publication: 2009-11-22 19:06:50 Built: R 2.10.0; i386-pc-mingw32; 2009-12-01 19:32:47 UTC; windows Index: NelPlo Nelson-Plosser Macroeconomic Time Series USeconomic U.S. Economic Variables adf.test Augmented Dickey-Fuller Test arma Fit ARMA Models to Time Series . . (etc.) . 1.9 Getting Help on a Package | 15 Some packages also include vignettes, which are additional documents such as intro- ductions, tutorials, or reference cards. They are installed on your computer as part of the package documentation when you install the package. The help page for a package includes a list of its vignettes near the bottom. You can see a list of all vignettes on your computer by using the vignette function: > vignette() You can see the vignettes for a particular package by including its name: > vignette(package="packagename") Each vignette has a name, which you use to view the vignette: > vignette("vignettename") See Also See Recipe 1.7 for getting help on a particular function in a package. 1.10 Searching the Web for Help Problem You want to search the Web for information and answers regarding R. Solution Inside R, use the RSiteSearch function to search by keyword or phrase: > RSiteSearch("key phrase") Inside your browser, try using these sites for searching: http://rseek.org This is a Google custom search that is focused on R-specific websites. http://stackoverflow.com/ Stack Overflow is a searchable Q&A site oriented toward programming issues such as data structures, coding, and graphics. http://stats.stackexchange.com/ The Statistical Analysis area on Stack Exchange is also a searchable Q&A site, but it is oriented more toward statistics than programming. Discussion The RSiteSearch function will open a browser window and direct it to the search engine on the R Project website. There you will see an initial search that you can refine. For example, this call would start a search for “canonical correlation”: > RSiteSearch("canonical correlation") 16 | Chapter 1: Getting Started and Getting Help This is quite handy for doing quick web searches without leaving R. However, the search scope is limited to R documentation and the mailing-list archives. The rseek.org site provides a wider search. Its virtue is that it harnesses the power of the Google search engine while focusing on sites relevant to R. That eliminates the extraneous results of a generic Google search. The beauty of rseek.org is that it organizes the results in a useful way. Figure 1-3 shows the results of visiting rseek.org and searching for “canonical correla- tion”. The left side of the page shows general results for search R sites. The right side is a tabbed display that organizes the search results into several categories: • Introductions • Task Views • Support Lists • Functions • Books • Blogs • Related Tools Figure 1-3. Search results from rseek.org 1.10 Searching the Web for Help | 17 If you click on the Introductions tab, for example, you’ll find tutorial material. The Task Views tab will show any Task View that mentions your search term. Likewise, clicking on Functions will show links to relevant R functions. This is a good way to zero in on search results. Stack Overflow is a so-called Q&A site, which means that anyone can submit a question and experienced users will supply answers—often there are multiple answers to each question. Readers vote on the answers, so good answers tend to rise to the top. This creates a rich database of Q&A dialogs, which you can search. Stack Overflow is strongly problem oriented, and the topics lean toward the programming side of R. Stack Overflow hosts questions for many programming languages; therefore, when entering a term into their search box, prefix it with “[r]” to focus the search on questions tagged for R. For example, searching via “[r] standard error” will select only the ques- tions tagged for R and will avoid the Python and C++ questions. Stack Exchange (not Overflow) has a Q&A area for Statistical Analysis. The area is more focused on statistics than programming, so use this site when seeking answers that are more concerned with statistics in general and less with R in particular. See Also If your search reveals a useful package, use Recipe 3.9 to install it on your machine. 1.11 Finding Relevant Functions and Packages Problem Of the 2,000+ packages for R, you have no idea which ones would be useful to you. Solution • Visit the list of task views at http://cran.r-project.org/web/views/. Find and read the task view for your area, which will give you links to and descriptions of relevant packages. Or visit http://rseek.org, search by keyword, click on the Task Views tab, and select an applicable task view. • Visit crantastic and search for packages by keyword. • To find relevant functions, visit http://rseek.org, search by name or keyword, and click on the Functions tab. Discussion This problem is especially vexing for beginners. You think R can solve your problems, but you have no idea which packages and functions would be useful. A common question on the mailing lists is: “Is there a package to solve problem X?” That is the silent scream of someone drowning in R. 18 | Chapter 1: Getting Started and Getting Help As of this writing, there are more than 2,000 packages available for free download from CRAN. Each package has a summary page with a short description and links to the package documentation. Once you’ve located a potentially interesting package, you would typically click on the “Reference manual” link to view the PDF documentation with full details. (The summary page also contains download links for installing the package, but you’ll rarely install the package that way; see Recipe 3.9.) Sometimes you simply have a generic interest—such as Bayesian analysis, economet- rics, optimization, or graphics. CRAN contains a set of task view pages describing packages that may be useful. A task view is a great place to start since you get an overview of what’s available. You can see the list of task view pages at http://cran.r -project.org/web/views/ or search for them as described in the Solution. Suppose you happen to know the name of a useful package—say, by seeing it men- tioned online. A complete, alphabetical list of packages is available at http://cran.r -project.org/web/packages/ with links to the package summary pages. See Also You can download and install an R package called sos that provides powerful other ways to search for packages; see the vignette at http://cran.r-project.org/web/packages/ sos/vignettes/sos.pdf. 1.12 Searching the Mailing Lists Problem You have a question, and you want to search the archives of the mailing lists to see whether your question was answered previously. Solution • Open http://rseek.org in your browser. Search for a keyword or other search term from your question. When the search results appear, click on the “Support Lists” tab. • You can perform a search within R itself. Use the RSiteSearch function to initiate a search: > RSiteSearch("keyphrase") The initial search results will appear in a browser. Under “Target”, select the R-help sources, clear the other sources, and resubmit your query. 1.12 Searching the Mailing Lists | 19 Discussion This recipe is really just an application of Recipe 1.10. But it’s an important application because you should search the mailing list archives before submitting a new question to the list. Your question has probably been answered before. See Also CRAN has a list of additional resources for searching the Web; see http://cran.r-project .org/search.html. 1.13 Submitting Questions to the Mailing Lists Problem You want to submit a question to the R community via the R-help mailing list. Solution The Mailing Lists page contains general information and instructions for using the R- help mailing list. Here is the general process: 1. Subscribe to the R-help list at the Main R Mailing List. 2. Read the Posting Guide for instructions on writing an effective submission. 3. Write your question carefully and correctly. If appropriate, include a minimal self- reproducing example so that others can reproduce your error or problem. 4. Mail your question to r-help@r-project.org. Discussion The R mailing list is a powerful resource, but please treat it as a last resort. Read the help pages, read the documentation, search the help list archives, and search the Web. It is most likely that your question has already been answered. Don’t kid yourself: very few questions are unique. After writing your question, submitting it is easy. Just mail it to r-help@r-project.org. You must be a list subscriber, however; otherwise your email submission may be rejected. Your question might arise because your R code is causing an error or giving unexpected results. In that case, a critical element of your question is the minimal self-contained example: Minimal Construct the smallest snippet of R code that displays your problem. Remove everything that is irrelevant. 20 | Chapter 1: Getting Started and Getting Help Self-contained Include the data necessary to exactly reproduce the error. If the list readers can’t reproduce it, they can’t diagnose it. For complicated data structures, use the dump function to create an ASCII representation of your data and include it in your message. Including an example clarifies your question and greatly increases the probability of getting a useful answer. There are actually several mailing lists. R-help is the main list for general questions. There are also many special interest group (SIG) mailing lists dedicated to particular domains such as genetics, finance, R development, and even R jobs. You can see the full list at https://stat.ethz.ch/mailman/listinfo. If your question is specific to one such domain, you’ll get a better answer by selecting the appropriate list. As with R-help, however, carefully search the SIG list archives before submitting your question. See Also An excellent essay by Eric Raymond and Rick Moen is entitled “How to Ask Questions the Smart Way”. I suggest that you read it before submitting any question. 1.13 Submitting Questions to the Mailing Lists | 21 CHAPTER 2 Some Basics Introduction The recipes in this chapter lie somewhere between problem-solving ideas and tutorials. Yes, they solve common problems, but the Solutions showcase common techniques and idioms used in most R code, including the code in this Cookbook. If you are new to R, I suggest skimming this chapter to acquaint yourself with these idioms. 2.1 Printing Something Problem You want to display the value of a variable or expression. Solution If you simply enter the variable name or expression at the command prompt, R will print its value. Use the print function for generic printing of any object. Use the cat function for producing custom formatted output. Discussion It’s very easy to ask R to print something: just enter it at the command prompt: > pi [1] 3.141593 > sqrt(2) [1] 1.414214 When you enter expressions like that, R evaluates the expression and then implicitly calls the print function. So the previous example is identical to this: 23 > print(pi) [1] 3.141593 > print(sqrt(2)) [1] 1.414214 The beauty of print is that it knows how to format any R value for printing, including structured values such as matrices and lists: > print(matrix(c(1,2,3,4), 2, 2)) [,1] [,2] [1,] 1 3 [2,] 2 4 > print(list("a","b","c")) [[1]] [1] "a" [[2]] [1] "b" [[3]] [1] "c" This is useful because you can always view your data: just print it. You needn’t write special printing logic, even for complicated data structures. The print function has a significant limitation, however: it prints only one object at a time. Trying to print multiple items gives this mind-numbing error message: > print("The zero occurs at", 2*pi, "radians.") Error in print.default("The zero occurs at", 2 * pi, "radians.") : unimplemented type 'character' in 'asLogical' The only way to print multiple items is to print them one at a time, which probably isn’t what you want: > print("The zero occurs at"); print(2*pi); print("radians") [1] "The zero occurs at" [1] 6.283185 [1] "radians" The cat function is an alternative to print that lets you combine multiple items into a continuous output: > cat("The zero occurs at", 2*pi, "radians.", "\n") The zero occurs at 6.283185 radians. Notice that cat puts a space between each item by default. You must provide a newline character (\n) to terminate the line. The cat function can print simple vectors, too: > fib <- c(0,1,1,2,3,5,8,13,21,34) > cat("The first few Fibonacci numbers are:", fib, "...\n") The first few Fibonacci numbers are: 0 1 1 2 3 5 8 13 21 34 ... 24 | Chapter 2: Some Basics Using cat gives you more control over your output, which makes it especially useful in R scripts. A serious limitation, however, is that it cannot print compound data struc- tures such as matrices and lists. Trying to cat them only produces another mind- numbing message: > cat(list("a","b","c")) Error in cat(list(...), file, sep, fill, labels, append) : argument 1 (type 'list') cannot be handled by 'cat' See Also See Recipe 4.2 for controlling output format. 2.2 Setting Variables Problem You want to save a value in a variable. Solution Use the assignment operator (<-). There is no need to declare your variable first: > x <- 3 Discussion Using R in “calculator mode” gets old pretty fast. Soon you will want to define variables and save values in them. This reduces typing, saves time, and clarifies your work. There is no need to declare or explicitly create variables in R. Just assign a value to the name and R will create the variable: > x <- 3 > y <- 4 > z <- sqrt(x^2 + y^2) > print(z) [1] 5 Notice that the assignment operator is formed from a less-than character (<) and a hyphen (-) with no space between them. When you define a variable at the command prompt like this, the variable is held in your workspace. The workspace is held in the computer’s main memory but can be saved to disk when you exit from R. The variable definition remains in the workspace until you remove it. R is a dynamically typed language, which means that we can change a variable’s data type at will. We could set x to be numeric, as just shown, and then turn around and immediately overwrite that with (say) a vector of character strings. R will not complain: 2.2 Setting Variables | 25 > x <- 3 > print(x) [1] 3 > x <- c("fee", "fie", "foe", "fum") > print(x) [1] "fee" "fie" "foe" "fum" In some R functions you will see assignment statements that use the strange-looking assignment operator <<-: x <<- 3 That forces the assignment to a global variable rather than a local variable. In the spirit of full disclosure, I will reveal that R also supports two other forms of assignment statements. A single equal sign (=) can be used as an assignment operator at the command prompt. A rightward assignment operator (->) can be used anywhere the leftward assignment operator (<-) can be used: > foo = 3 > print(foo) [1] 3 > 5 -> fum > print(fum) [1] 5 These forms are never used in this book, and I recommend that you avoid them. The equals-sign assignment is easily confused with the test for equality. The rightward as- signment is just too unconventional and, worse, becomes difficult to read when the expression is long. See Also See Recipes 2.4, 2.14, and 3.2. See also the help page for the assign function. 2.3 Listing Variables Problem You want to know what variables and functions are defined in your workspace. Solution Use the ls function. Use ls.str for more details about each variable. Discussion The ls function displays the names of objects in your workspace: > x <- 10 > y <- 50 > z <- c("three", "blind", "mice") 26 | Chapter 2: Some Basics > f <- function(n,p) sqrt(p*(1-p)/n) > ls() [1] "f" "x" "y" "z" Notice that ls returns a vector of character strings in which each string is the name of one variable or function. When your workspace is empty, ls returns an empty vector, which produces this puzzling output: > ls() character(0) That is R’s quaint way of saying that ls returned a zero-length vector of strings; that is, it returned an empty vector because nothing is defined in your workspace. If you want more than just a list of names, try ls.str; this will also tell you something about each variable: > ls.str() f : function (n, p) x : num 10 y : num 50 z : chr [1:3] "three" "blind" "mice" The function is called ls.str because it is both listing your variables and applying the str function to them, showing their structure (Recipe 12.15). Ordinarily, ls does not return any name that begins with a dot (.). Such names are considered hidden and are not normally of interest to users. (This mirrors the Unix convention of not listing files whose names begin with dot.) You can force ls to list everything by setting the all.names argument to TRUE: > .hidvar <- 10 > ls() [1] "f" "x" "y" "z" > ls(all.names=TRUE) [1] ".hidvar" "f" "x" "y" "z" See Also See Recipe 2.4 for deleting variables and Recipe 12.15 for inspecting your variables. 2.4 Deleting Variables Problem You want to remove unneeded variables or functions from your workspace or to erase its contents completely. Solution Use the rm function. 2.4 Deleting Variables | 27 Discussion Your workspace can get cluttered quickly. The rm function removes, permanently, one or more objects from the workspace: > x <- 2*pi > x [1] 6.283185 > rm(x) > x Error: object "x" not found There is no “undo”; once the variable is gone, it’s gone. You can remove several variables at once: > rm(x,y,z) You can even erase your entire workspace at once. The rm function has a list argument consisting of a vector of names of variables to remove. Recall that the ls function returns a vector of variables names; hence you can combine rm and ls to erase everything: > ls() [1] "f" "x" "y" "z" > rm(list=ls()) > ls() character(0) Be Polite Never put rm(list=ls()) into code you share with others, such as a library function or sample code sent to a mailing list. Deleting all the variables in someone else’s workspace is worse than rude and will make you extremely unpopular. See Also See Recipe 2.3. 2.5 Creating a Vector Problem You want to create a vector. Solution Use the c(...) operator to construct a vector from given values. 28 | Chapter 2: Some Basics Discussion Vectors are a central component of R, not just another data structure. A vector can contain either numbers, strings, or logical values but not a mixture. The c(...) operator can construct a vector from simple elements: > c(1,1,2,3,5,8,13,21) [1] 1 1 2 3 5 8 13 21 > c(1*pi, 2*pi, 3*pi, 4*pi) [1] 3.141593 6.283185 9.424778 12.566371 > c("Everyone", "loves", "stats.") [1] "Everyone" "loves" "stats." > c(TRUE,TRUE,FALSE,TRUE) [1] TRUE TRUE FALSE TRUE If the arguments to c(...) are themselves vectors, it flattens them and combines them into one single vector: > v1 <- c(1,2,3) > v2 <- c(4,5,6) > c(v1,v2) [1] 1 2 3 4 5 6 Vectors cannot contain a mix of data types, such as numbers and strings. If you create a vector from mixed elements, R will try to accommodate you by converting one of them: > v1 <- c(1,2,3) > v3 <- c("A","B","C") > c(v1,v3) [1] "1" "2" "3" "A" "B" "C" Here, the user tried to create a vector from both numbers and strings. R converted all the numbers to strings before creating the vector, thereby making the data elements compatible. Technically speaking, two data elements can coexist in a vector only if they have the same mode. The modes of 3.1415 and "foo" are numeric and character, respectively: > mode(3.1415) [1] "numeric" > mode("foo") [1] "character" Those modes are incompatible. To make a vector from them, R converts 3.1415 to character mode so it will be compatible with "foo": > c(3.1415, "foo") [1] "3.1415" "foo" > mode(c(3.1415, "foo")) [1] "character" 2.5 Creating a Vector | 29 c is a generic operator, which means that it works with many datatypes and not just vectors. However, it might not do exactly what you expect, so check its behavior before applying it to other datatypes and objects. See Also See the “Introduction” to Chapter 5 for more about vectors and other data structures. 2.6 Computing Basic Statistics Problem You want to calculate basic statistics: mean, median, standard deviation, variance, correlation, or covariance. Solution Use one of these functions as applies, assuming that x and y are vectors: • mean(x) • median(x) • sd(x) • var(x) • cor(x, y) • cov(x, y) Discussion When I first opened the documentation for R, I begin searching for material entitled “Procedures for Calculating Standard Deviation.” I figured that such an important topic would likely require a whole chapter. It’s not that complicated. Standard deviation and other basic statistics are calculated by simple functions. Ordi- narily, the function argument is a vector of numbers and the function returns the cal- culated statistic: > x <- c(0,1,1,2,3,5,8,13,21,34) > mean(x) [1] 8.8 > median(x) [1] 4 > sd(x) [1] 11.03328 > var(x) [1] 121.7333 30 | Chapter 2: Some Basics The sd function calculates the sample standard deviation, and var calculates the sample variance. The cor and cov functions can calculate the correlation and covariance, respectively, between two vectors: > x <- c(0,1,1,2,3,5,8,13,21,34) > y <- log(x+1) > cor(x,y) [1] 0.9068053 > cov(x,y) [1] 11.49988 All these functions are picky about values that are not available (NA). Even one NA value in the vector argument causes any of these functions to return NA or even halt altogether with a cryptic error: > x <- c(0,1,1,2,3,NA) > mean(x) [1] NA > sd(x) [1] NA It’s annoying when R is that cautious, but it is the right thing to do. You must think carefully about your situation. Does an NA in your data invalidate the statistic? If yes, then R is doing the right thing. If not, you can override this behavior by setting na.rm=TRUE, which tells R to ignore the NA values: > x <- c(0,1,1,2,3,NA) > mean(x, na.rm=TRUE) [1] 1.4 > sd(x, na.rm=TRUE) [1] 1.140175 A beautiful aspect of mean and sd is that they are smart about data frames. They un- derstand that each column of the data frame is a different variable, so they calculate their statistic for each column individually. This example calculates those basic statis- tics for a data frame with three columns: > print(dframe) small medium big 1 0.6739635 10.526448 99.83624 2 1.5524619 9.205156 100.70852 3 0.3250562 11.427756 99.73202 4 1.2143595 8.533180 98.53608 5 1.3107692 9.763317 100.74444 6 2.1739663 9.806662 98.58961 7 1.6187899 9.150245 100.46707 8 0.8872657 10.058465 99.88068 9 1.9170283 9.182330 100.46724 10 0.7767406 7.949692 100.49814 > mean(dframe) small medium big 1.245040 9.560325 99.946003 > sd(dframe) 2.6 Computing Basic Statistics | 31 small medium big 0.5844025 0.9920281 0.8135498 Notice that mean and sd both return three values, one for each column defined by the data frame. (Technically, they return a three-element vector whose names attribute is taken from the columns of the data frame.) The var function understands data frames, too, but it behaves quite differently than do mean and sd. It calculates the covariance between the columns of the data frame and returns the covariance matrix: > var(dframe) small medium big small 0.34152627 -0.21516416 -0.04005275 medium -0.21516416 0.98411974 -0.09253855 big -0.04005275 -0.09253855 0.66186326 Likewise, if x is either a data frame or a matrix, then cor(x) returns the correlation matrix and cov(x) returns the covariance matrix: > cor(dframe) small medium big small 1.00000000 -0.3711367 -0.08424345 medium -0.37113670 1.0000000 -0.11466070 big -0.08424345 -0.1146607 1.00000000 > cov(dframe) small medium big small 0.34152627 -0.21516416 -0.04005275 medium -0.21516416 0.98411974 -0.09253855 big -0.04005275 -0.09253855 0.66186326 The median function does not understand data frames, alas. To calculate the medians of data frame columns, use Recipe 6.4 and apply the median function to each column separately. See Also See Recipes 2.14, 6.4, and 9.17. 2.7 Creating Sequences Problem You want to create a sequence of numbers. Solution Use an n:m expression to create the simple sequence n, n+1, n+2, ..., m: > 1:5 [1] 1 2 3 4 5 32 | Chapter 2: Some Basics Use the seq function for sequences with an increment other than 1: > seq(from=1, to=5, by=2) [1] 1 3 5 Use the rep function to create a series of repeated values: > rep(1, times=5) [1] 1 1 1 1 1 Discussion The colon operator (n:m) creates a vector containing the sequence n, n+1, n+2, ..., m: > 0:9 [1] 0 1 2 3 4 5 6 7 8 9 > 10:19 [1] 10 11 12 13 14 15 16 17 18 19 > 9:0 [1] 9 8 7 6 5 4 3 2 1 0 Observe that R was clever with the last expression (9:0). Because 9 is larger than 0, it counts backward from the starting to ending value. The colon operator works for sequences that grow by 1 only. The seq function also builds sequences but supports an optional third argument, which is the increment: > seq(from=0, to=20) [1] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 > seq(from=0, to=20, by=2) [1] 0 2 4 6 8 10 12 14 16 18 20 > seq(from=0, to=20, by=5) [1] 0 5 10 15 20 Alternatively, you can specify a length for the output sequence and then R will calculate the necessary increment: > seq(from=0, to=20, length.out=5) [1] 0 5 10 15 20 > seq(from=0, to=100, length.out=5) [1] 0 25 50 75 100 The increment need not be an integer. R can create sequences with fractional incre- ments, too: > seq(from=1.0, to=2.0, length.out=5) [1] 1.00 1.25 1.50 1.75 2.00 For the special case of a “sequence” that is simply a repeated value you should use the rep function, which repeats its first argument: > rep(pi, times=5) [1] 3.141593 3.141593 3.141593 3.141593 3.141593 See Also See Recipe 7.14 for creating a sequence of Date objects. 2.7 Creating Sequences | 33 2.8 Comparing Vectors Problem You want to compare two vectors or you want to compare an entire vector against a scalar. Solution The comparison operators (==, !=, <, >, <=, >=) can perform an element-by-element comparison of two vectors. They can also compare a vector’s element against a scalar. The result is a vector of logical values in which each value is the result of one element- wise comparison. Discussion R has two logical values, TRUE and FALSE. These are often called Boolean values in other programming languages. The comparison operators compare two values and return TRUE or FALSE, depending upon the result of the comparison: > a <- 3 > a == pi # Test for equality [1] FALSE > a != pi # Test for inequality [1] TRUE > a < pi [1] TRUE > a > pi [1] FALSE > a <= pi [1] TRUE > a >= pi [1] FALSE You can experience the power of R by comparing entire vectors at once. R will perform an element-by-element comparison and return a vector of logical values, one for each comparison: > v <- c( 3, pi, 4) > w <- c(pi, pi, pi) > v == w # Compare two 3-element vectors [1] FALSE TRUE FALSE # Result is a 3-element vector > v != w [1] TRUE FALSE TRUE > v < w [1] TRUE FALSE FALSE > v <= w [1] TRUE TRUE FALSE > v > w [1] FALSE FALSE TRUE 34 | Chapter 2: Some Basics D o wnload from Wow! eBook > v >= w [1] FALSE TRUE TRUE You can also compare a vector against a single scalar, in which case R will expand the scalar to the vector’s length and then perform the element-wise comparison. The pre- vious example can be simplified in this way: > v <- c(3, pi, 4) > v == pi # Compare a 3-element vector against one number [1] FALSE TRUE FALSE > v != pi [1] TRUE FALSE TRUE (This is an application of the Recycling Rule, Recipe 5.3.) After comparing two vectors, you often want to know whether any of the comparisons were true or whether all the comparisons were true. The any and all functions handle those tests. They both test a logical vector. The any function returns TRUE if any element of the vector is TRUE. The all function returns TRUE if all elements of the vector are TRUE: > v <- c(3, pi, 4) > any(v == pi) # Return TRUE if any element of v equals pi [1] TRUE > all(v == 0) # Return TRUE if all elements of v are zero [1] FALSE See Also See Recipe 2.9. 2.9 Selecting Vector Elements Problem You want to extract one or more elements from a vector. Solution Select the indexing technique appropriate for your problem: • Use square brackets to select vector elements by their position, such as v[3] for the third element of v. • Use negative indexes to exclude elements. • Use a vector of indexes to select multiple values. • Use a logical vector to select elements based on a condition. • Use names to access named elements. 2.9 Selecting Vector Elements | 35 Discussion Selecting elements from vectors is another powerful feature of R. Basic selection is handled just as in many other programming languages—use square brackets and a simple index: > fib <- c(0,1,1,2,3,5,8,13,21,34) > fib [1] 0 1 1 2 3 5 8 13 21 34 > fib[1] [1] 0 > fib[2] [1] 1 > fib[3] [1] 1 > fib[4] [1] 2 > fib[5] [1] 3 Notice that the first element has an index of 1, not 0 as in some other programming languages. A cool feature of vector indexing is that you can select multiple elements at once. The index itself can be a vector, and each element of that indexing vector selects an element from the data vector: > fib[1:3] # Select elements 1 through 3 [1] 0 1 1 > fib[4:9] # Select elements 4 through 9 [1] 2 3 5 8 13 21 An index of 1:3 means select elements 1, 2, and 3, as just shown. The indexing vector needn’t be a simple sequence, however. You can select elements anywhere within the data vector—as in this example, which selects elements 1, 2, 4, and 8: > fib[c(1,2,4,8)] [1] 0 1 2 13 R interprets negative indexes to mean exclude a value. An index of −1, for instance, means exclude the first value and return all other values: > fib[-1] # Ignore first element [1] 1 1 2 3 5 8 13 21 34 This method can be extended to exclude whole slices by using an indexing vector of negative indexes: > fib[1:3] # As before [1] 0 1 1 > fib[-(1:3)] # Invert sign of index to exclude instead of select [1] 2 3 5 8 13 21 34 Another indexing technique uses a logical vector to select elements from the data vector. Everywhere that the logical vector is TRUE, an element is selected: 36 | Chapter 2: Some Basics > fib < 10 # This vector is TRUE wherever fib is less than 10 [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE FALSE FALSE > fib[fib < 10] # Use that vector to select elements less than 10 [1] 0 1 1 2 3 5 8 > fib %% 2 == 0 # This vector is TRUE wherever fib is even [1] TRUE FALSE FALSE TRUE FALSE FALSE TRUE FALSE FALSE TRUE > fib[fib %% 2 == 0] # Use that vector to select the even elements [1] 0 2 8 34 Ordinarily, the logical vector should be the same length as the data vector so you are clearly either including or excluding each element. (If the lengths differ then you need to understand the Recycling Rule, Recipe 5.3.) By combining vector comparisons, logical operators, and vector indexing, you can per- form powerful selections with very little R code: Select all elements greater than the median v[ v > median(v) ] Select all elements in the lower and upper 5% v[ (v < quantile(v,0.05)) | (v > quantile(v,0.95)) ] Select all elements that exceed ±2 standard deviations from the mean v[ abs(v-mean(v)) > 2*sd(v) ] Select all elements that are neither NA nor NULL v[ !is.na(v) & !is.null(v) ] One final indexing feature lets you select elements by name. It assumes that the vector has a names attribute, defining a name for each element. This can be done by assigning a vector of character strings to the attribute: > years <- c(1960, 1964, 1976, 1994) > names(years) <- c("Kennedy", "Johnson", "Carter", "Clinton") > years Kennedy Johnson Carter Clinton 1960 1964 1976 1994 Once the names are defined, you can refer to individual elements by name: > years["Carter"] Carter 1976 > years["Clinton"] Clinton 1994 This generalizes to allow indexing by vectors of names: R returns every element named in the index: > years[c("Carter","Clinton")] Carter Clinton 1976 1994 2.9 Selecting Vector Elements | 37 See Also See Recipe 5.3 for more about the Recycling Rule. 2.10 Performing Vector Arithmetic Problem You want to operate on an entire vector at once. Solution The usual arithmetic operators can perform element-wise operations on entire vectors. Many functions operate on entire vectors, too, and return a vector result. Discussion Vector operations are one of R’s great strengths. All the basic arithmetic operators can be applied to pairs of vectors. They operate in an element-wise manner; that is, the operator is applied to corresponding elements from both vectors: > v <- c(11,12,13,14,15) > w <- c(1,2,3,4,5) > v + w [1] 12 14 16 18 20 > v - w [1] 10 10 10 10 10 > v * w [1] 11 24 39 56 75 > v / w [1] 11.000000 6.000000 4.333333 3.500000 3.000000 > w ^ v [1] 1 4096 1594323 268435456 30517578125 Observe that the length of the result here is equal to the length of the original vectors. The reason is that each element comes from a pair of corresponding values in the input vectors. If one operand is a vector and the other is a scalar, then the operation is performed between every vector element and the scalar: > w [1] 1 2 3 4 5 > w + 2 [1] 3 4 5 6 7 > w - 2 [1] -1 0 1 2 3 > w * 2 [1] 2 4 6 8 10 > w / 2 [1] 0.5 1.0 1.5 2.0 2.5 38 | Chapter 2: Some Basics > w ^ 2 [1] 1 4 9 16 25 > 2 ^ w [1] 2 4 8 16 32 For example, you can recenter an entire vector in one expression simply by subtracting the mean of its contents: > w [1] 1 2 3 4 5 > mean(w) [1] 3 > w - mean(w) [1] -2 -1 0 1 2 Likewise, you can calculate the z-score of a vector in one expression: subtract the mean and divide by the standard deviation: > w [1] 1 2 3 4 5 > sd(w) [1] 1.581139 > (w - mean(w)) / sd(w) [1] -1.2649111 -0.6324555 0.0000000 0.6324555 1.2649111 Yet the implementation of vector-level operations goes far beyond elementary arith- metic. It pervades the language, and many functions operate on entire vectors. The functions sqrt and log, for example, apply themselves to every element of a vector and return a vector of results: > w [1] 1 2 3 4 5 > sqrt(w) [1] 1.000000 1.414214 1.732051 2.000000 2.236068 > log(w) [1] 0.0000000 0.6931472 1.0986123 1.3862944 1.6094379 > sin(w) [1] 0.8414710 0.9092974 0.1411200 -0.7568025 -0.9589243 There are two great advantages to vector operations. The first and most obvious is convenience. Operations that require looping in other languages are one-liners in R. The second is speed. Most vectorized operations are implemented directly in C code, so they are substantially faster than the equivalent R code you could write. See Also Performing an operation between a vector and a scalar is actually a special case of the Recycling Rule; see Recipe 5.3. 2.10 Performing Vector Arithmetic | 39 2.11 Getting Operator Precedence Right Problem Your R expression is producing a curious result, and you wonder if operator precedence is causing problems. Solution The full list of operators is shown in Table 2-1, listed in order of precedence from highest to lowest. Operators of equal precedence are evaluated from left to right except where indicated. Table 2-1. Operator precedence Operator Meaning See also [ [[ Indexing Recipe 2.9 :: ::: Access variables in a name space $ @ Component extraction, slot extraction ^ Exponentiation (right to left) - + Unary minus and plus : Sequence creation Recipe 2.7, Recipe 7.14 %any% Special operators Discussion * / Multiplication, division Discussion + - Addition, subtraction == != < > <= >= Comparison Recipe 2.8 ! Logical negation & && Logical “and”, short-circuit “and” | || Logical “or”, short-circuit “or” ~ Formula Recipe 11.1 -> ->> Rightward assignment Recipe 2.2 = Assignment (right to left) Recipe 2.2 <- <<- Assignment (right to left) Recipe 2.2 ? Help Recipe 1.7 Discussion Getting your operator precedence wrong in R is a common problem. It certainly hap- pens to me a lot. I unthinkingly expect that the expression 0:n−1 will create a sequence of integers from 0 to n − 1 but it does not: 40 | Chapter 2: Some Basics > n <− 10 > 0:n−1 [1] −1 0 1 2 3 4 5 6 7 8 9 It creates the sequence from −1 to n − 1 because R interprets it as (0:n)−1. You might not recognize the notation %any% in the table. R interprets any text between percent signs (%...%) as a binary operator. Several such operators have predefined meanings: %% Modulo operator %/% Integer division %*% Matrix multiplication %in% Returns TRUE if the left operand occurs in its right operand; FALSE otherwise You can also define new binary operators using the %...% notation; see Recipe 12.19. The point here is that all such operators have the same precedence. See Also See Recipe 2.10 for more about vector operations, Recipe 5.15 for more about matrix operations, and Recipe 12.19 to define your own operators. See the Arithmetic and Syntax topics in the R help pages as well as Chapters 5 and 6 of R in a Nutshell (O’Reilly). 2.12 Defining a Function Problem You want to define an R function. Solution Create a function by using the function keyword followed by a list of parameters and the function body. A one-liner looks like this: function(param1, ...., paramN) expr The function body can be a series of expressions, in which case curly braces should be used around the function body: 2.12 Defining a Function | 41 function(param1, ..., paramN) { expr1 . . . exprM } Discussion Function definitions are how you tell R, “Here’s how to calculate this.” For example, R does not have a built-in function for calculating the coefficient of variation. You can define the calculation in this way: > cv <- function(x) sd(x)/mean(x) > cv(1:10) [1] 0.5504819 The first line creates a function and assigns it to cv. The second line invokes the function, using 1:10 for the value of parameter x. The function returns the value of its single- expression body, sd(x)/mean(x). After defining a function we can use it anywhere a function is expected, such as the second argument of lapply (Recipe 6.2): > cv <- function(x) sd(x)/mean(x) > lapply(lst, cv) A multiline function uses curly braces to delimit the start and end of the function body. Here is a function that implements Euclid’s algorithm for computing the greatest com- mon divisor of two integers: > gcd <- function(a,b) { + if (b == 0) return(a) + else return(gcd(b, a %% b)) + } R also allows anonymous functions; these are functions with no name that are useful for one-liners. The preceding example using cv and lapply can be shrunk to one line by using an anonymous function that is passed directly into lapply: > lapply(lst, function(x) sd(x)/mean(x)) This is not a book about programming R, so I cannot fully cover the subtleties of coding an R function. But here are a few useful points: Return value All functions return a value. Normally, a function returns the value of the last expression in its body. You can also use return(expr). Call by value Function parameters are “call by value”—in other words, if you change a parameter then the change is local and does not affect the caller’s value. 42 | Chapter 2: Some Basics Local variables You create a local variable simply by assigning a value to it. When the function exits, local variables are lost. Conditional execution The R syntax includes an if statement. See help(Control) for details. Loops The R syntax also includes for loops, while loops, and repeat loops. For details, see help(Control). Global variables Within a function you can change a global variable by using the <<- assignment operator, but this is not encouraged. See Also For more about defining functions, see An Introduction to R and R in a Nutshell. 2.13 Typing Less and Accomplishing More Problem You are getting tired of typing long sequences of commands and especially tired of typing the same ones over and over. Solution Open an editor window and accumulate your reusable blocks of R commands there. Then, execute those blocks directly from that window. Reserve the command line for typing brief or one-off commands. When you are done, you can save the accumulated code blocks in a script file for later use. Discussion The typical beginner to R types an expression and sees what happens. As he gets more comfortable, he types increasingly complicated expressions. Then he begins typing multiline expressions. Soon, he is typing the same multiline expressions over and over, perhaps with small variations, in order to perform his increasingly complicated calculations. 2.13 Typing Less and Accomplishing More | 43 The experienced user does not often retype a complex expression. She may type the same expression once or twice, but when she realizes it is useful and reusable she will cut-and-paste it into an editor window of the GUI. To execute the snippet thereafter, she selects the snippet in the editor window and tells R to execute it, rather than retyping it. This technique is especially powerful as her snippets evolve into long blocks of code. On Windows and OS X, a few features of the GUI facilitate this workstyle: To open an editor window From the main menu, select File → New script. To execute one line of the editor window Position the cursor on the line and then press Ctrl-R to execute it. To execute several lines of the editor window Highlight the lines using your mouse; then press Ctrl-R to execute them. To execute the entire contents of the editor window Press Ctrl-A to select the entire window contents and then Ctrl-R to execute them; or, from the main menu, select Edit → Run all. Copying lines from the console window to the editor window is simply a matter of copy and paste. When you exit R, it will ask if you want to save the new script. You can either save it for future reuse or discard it. I recently used this technique to create a small illustration of the Central Limit Theorem for some students. I wanted a probability density of the population overlaid with the density of the sample mean. I opened R and then began defining variables and executing functions to create the plot of the population density. It dawned on me that I would likely reexecute those few commands, so I opened a script editor window in R and pasted the commands into it as shown in Figure 2-1. I resumed typing commands in the console window, thinking I would overlay the ex- isting graph with the density for the sample mean. However, the second density plot did not fit inside the graph. I needed to fix my function calls and reexecute everything. Here came the big pay-off: I retyped little or nothing. Rather, I pasted a little more code into the editor window, tweaked the call to the plot function, and then reexecuted the whole window contents. Voilà! The scaling problem was fixed. I tweaked the code again (to add titles and labels) and once again reexecuted the whole window contents; this yielded the final result shown in Figure 2-2. I saved the editor contents in a script file when I was done, enabling me to re-create this graph at will. 44 | Chapter 2: Some Basics Figure 2-1. Capturing the commands in a new edit window Figure 2-2. Evolved state of the code 2.13 Typing Less and Accomplishing More | 45 2.14 Avoiding Some Common Mistakes Problem You want to avoid some of the common mistakes made by beginning users—and also by experienced users, for that matter. Discussion Here are some easy ways to make trouble for yourself: Forgetting the parentheses after a function invocation You call an R function by putting parentheses after the name. For instance, this line invokes the ls function: > ls() [1] "x" "y" "z" However, if you omit the parentheses then R does not execute the function. Instead, it shows the function definition, which is almost never what you want: > ls function (name, pos = -1, envir = as.environment(pos), all.names = FALSE, pattern) { if (!missing(name)) { nameValue <- try(name) if (identical(class(nameValue), "try-error")) { name <- substitute(name) . . (etc.) . Forgetting to double up backslashes in Windows file paths This function call appears to read a Windows file called F:\research\bio\assay.csv, but it does not: > tbl <- read.csv("F:\research\bio\assay.csv") Backslashes (\) inside character strings have a special meaning and therefore need to be doubled up. R will interpret this file name as F:researchbioassay.csv, for ex- ample, which is not what the user wanted. See Recipe 4.5 for possible solutions. Mistyping “<-” as “< (blank) -” The assignment operator is <-, with no space between the < and the -: > x <- pi # Set x to 3.1415926... If you accidentally insert a space between < and -, the meaning changes completely: > x < - pi # Oops! We are comparing x instead of setting it! 46 | Chapter 2: Some Basics This is now a comparison (<) between x and negative π (- pi). It does not change x. If you are lucky, x is undefined and R will complain, alerting you that something is fishy: > x < - pi Error: object "x" not found If x is defined, R will perform the comparison and print a logical value, TRUE or FALSE. That should alert you that something is wrong: an assignment does not normally print anything: > x <- 0 # Initialize x to zero > x < - pi # Oops! [1] FALSE Incorrectly continuing an expression across lines R reads your typing until you finish a complete expression, no matter how many lines of input that requires. It prompts you for additional input using the + prompt until it is satisfied. This example splits an expression across two lines: > total <- 1 + 2 + 3 + # Continued on the next line + 4 + 5 > print(total) [1] 15 Problems begin when you accidentally finish the expression prematurely, which can easily happen: > total <- 1 + 2 + 3 # Oops! R sees a complete expression > + 4 + 5 # This is a new expression; R prints its value [1] 9 > print(total) [1] 6 There are two clues that something is amiss: R prompted you with a normal prompt (>), not the continuation prompt (+); and it printed the value of 4 + 5. This common mistake is a headache for the casual user. It is a nightmare for pro- grammers, however, because it can introduce hard-to-find bugs into R scripts. Using = instead of == Use the double-equal operator (==) for comparisons. If you accidentally use the single-equal operator (=), you will irreversibly overwrite your variable: > v == 0 # Compare v against zero > v = 0 # Assign 0 to v, overwriting previous contents Writing 1:n+1 when you mean 1:(n+1) You might think that 1:n+1 is the sequence of numbers 1, 2, ..., n, n + 1. It’s not. It is the sequence 1, 2, ..., n with 1 added to every element, giving 2, 3, ..., n, n + 1. This happens because R interprets 1:n+1 as (1:n)+1. Use parentheses to get ex- actly what you want: 2.14 Avoiding Some Common Mistakes | 47 > n <- 5 > 1:n+1 [1] 2 3 4 5 6 > 1:(n+1) [1] 1 2 3 4 5 6 Getting bitten by the Recycling Rule Vector arithmetic and vector comparisons work well when both vectors have the same length. However, the results can be baffling when the operands are vectors of differing lengths. Guard against this possibility by understanding and remem- bering the Recycling Rule, Recipe 5.3. Installing a package but not loading it with library() or require() Installing a package is the first step toward using it, but one more step is required. Use library or require to load the package into your search path. Until you do so, R will not recognize the functions or datasets in the package. See Recipe 3.6: > truehist(x,n) Error: could not find function "truehist" > library(MASS) # Load the MASS package into R > truehist(x,n) > Writing aList[i] when you mean aList[[i]], or vice versa If the variable lst contains a list, it can be indexed in two ways: lst[[n]] is the nth element of the list; whereas lst[n] is a list whose only element is the nth element of lst. That’s a big difference. See Recipe 5.7. Using & instead of &&, or vice versa; same for | and || Use & and | in logical expressions involving the logical values TRUE and FALSE. See Recipe 2.9. Use && and || for the flow-of-control expressions inside if and while statements. Programmers accustomed to other programming languages may reflexively use && and || everywhere because “they are faster.” But those operators give peculiar results when applied to vectors of logical values, so avoid them unless that’s really what you want. Passing multiple arguments to a single-argument function What do you think is the value of mean(9,10,11)? No, it’s not 10. It’s 9. The mean function computes the mean of the first argument. The second and third arguments are being interpreted as other, positional arguments. Some functions, such as mean, take one argument. Other arguments, such as max and min, take multiple arguments and apply themselves across all arguments. Be sure you know which is which. Thinking that max behaves like pmax, or that min behaves like pmin The max and min functions have multiple arguments and return one value: the maximum or minimum of all their arguments. 48 | Chapter 2: Some Basics The pmax and pmin functions have multiple arguments but return a vector with values taken element-wise from the arguments. See Recipe 12.9. Misusing a function that does not understand data frames Some functions are quite clever regarding data frames. They apply themselves to the individual columns of the data frame, computing their result for each individual column. The mean and sd functions are good examples. These functions can com- pute the mean or standard deviation of each column because they understand that each column is a separate variable and that mixing their data is not sensible. Sadly, not all functions are that clever. This includes the median, max, and min func- tions. They will lump together every value from every column and compute their result from the lump, which might not be what you want. Be aware of which func- tions are savvy to data frames and which are not. Posting a question to the mailing list before searching for the answer Don’t waste your time. Don’t waste other people’s time. Before you post a question to a mailing list or to Stack Overflow, do your homework and search the archives. Odds are, someone has already answered your question. If so, you’ll see the answer in the discussion thread for the question. See Recipe 1.12. See Also See Recipes 1.12, 2.9, 5.3, and 5.7. 2.14 Avoiding Some Common Mistakes | 49 CHAPTER 3 Navigating the Software Introduction R is a big chunk of software, first and foremost. You will inevitably spend time doing what one does with any big piece of software: configuring it, customizing it, updating it, and fitting it into your computing environment. This chapter will help you perform those tasks. There is nothing here about numerics, statistics, or graphics. This is all about dealing with R as software. 3.1 Getting and Setting the Working Directory Problem You want to change your working directory. Or you just want to know what it is. Solution Command line Use getwd to report the working directory, and use setwd to change it: > getwd() [1] "/home/paul/research" > setwd("Bayes") > getwd() [1] "/home/paul/research/Bayes" Windows From the main menu, select File → Change dir... . OS X From the main menu, select Misc → Change Working Directory. For both Windows and OS X, the menu selection opens the current working directory in a file browser. From there, you can navigate to a new working directory if desired. 51 Discussion Your working directory is important because it is the default location for all file input and output—including reading and writing data files, opening and saving script files, and saving your workspace image. When you open a file and do not specify an absolute path, R will assume that the file is in your working directory. The initial working directory depends upon how you started R. See Recipe 1.2. See Also See Recipe 4.5 for dealing with filenames in Windows. 3.2 Saving Your Workspace Problem You want to save your workspace without exiting from R. Solution Call the save.image function: > save.image() Discussion Your workspace holds your R variables and functions, and it is created when R starts. The workspace is held in your computer’s main memory and lasts until you exit from R, at which time you can save it. However, you may want to save your workspace without exiting R. You might go to lunch, for example, and want to protect your work against an unexpected power outage or machine crash. Use the save.image function. The workspace is written to a file called .RData in the working directory. When R starts, it looks for that file and, if found, initializes the workspace from it. A sad fact is that the workspace does not include your open graphs: that cool graph on your screen disappears when you exit R, and there is no simple way to save and restore it. So before you exit, save the data and the R code that will re-create your graphs. See Also See Recipe 1.2 for how to save your workspace when exiting R and Recipe 3.1 for setting the working directory. 52 | Chapter 3: Navigating the Software 3.3 Viewing Your Command History Problem You want to see your recent sequence of commands. Solution Scroll backward by pressing the up arrow or Ctrl-P. Or use the history function to view your most recent input: > history() Discussion The history function will display your most recent commands. By default it shows the most recent 25 lines, but you can request more: > history(100) # Show 100 most recent lines of history > history(Inf) # Show entire saved history For very recent commands, simply scroll backward through your input using the command-line editing feature: pressing either the up arrow or Ctrl-P will cause your previous typing to reappear, one line at a time. If you’ve exited from R then you can still see your command history. It saves the history in a file called .Rhistory in the working directory, if requested. Open the file with a text editor and then scroll to the bottom; you will see your most recent typing. 3.4 Saving the Result of the Previous Command Problem You typed an expression into R that calculated the value, but you forgot to save the result in a variable. Solution A special variable called .Last.value saves the value of the most recently evaluated expression. Save it to a variable before you type anything else. Discussion It is frustrating to type a long expression or call a long-running function but then forget to save the result. Fortunately, you needn’t retype the expression nor invoke the func- tion again—the result was saved in the .Last.value variable: 3.4 Saving the Result of the Previous Command | 53 > aVeryLongRunningFunction() # Oops! Forgot to save the result! [1] 147.6549 > x <- .Last.value # Capture the result now > x [1] 147.6549 A word of caution: the contents of .Last.value are overwritten every time you type another expression, so capture the value immediately. If you don’t remember until another expression has been evaluated, it’s too late. See Also See Recipe 3.3 to recall your command history. 3.5 Displaying the Search Path Problem You want to see the list of packages currently loaded into R. Solution Use the search function with no arguments: > search() Discussion The search path is a list of packages that are currently loaded into memory and available for use. Although many packages may be installed on your computer, only a few of them are actually loaded into the R interpreter at any given moment. You might be wondering which packages are loaded right now. With no arguments, the search function returns the list of loaded packages. It produces an output like this: > search() [1] ".GlobalEnv" "package:stats" "package:graphics" [4] "package:grDevices" "package:utils" "package:datasets" [7] "package:methods" "Autoloads" "package:base" Your machine may return a different result, depending on what’s installed there. The return value of search is a vector of strings. The first string is ".GlobalEnv", which refers to your workspace. Most strings have the form "package:packagename", which indicates that the package called packagename is currently loaded into R. In this example, the loaded packages include stats, graphics, grDevices, utils, and so forth. R uses the search path to find functions. When you type a function name, R searches the path—in the order shown—until it finds the function in a loaded package. If the function is found, R executes it. Otherwise, it prints an error message and stops. (There 54 | Chapter 3: Navigating the Software is actually a bit more to it: the search path can contain environments, not just packages, and the search algorithm is different when initiated by an object within a package; see the R Language Definition for details.) Since your workspace (.GlobalEnv) is first in the list, R looks for functions in your workspace before searching any packages. If your workspace and a package both con- tain a function with the same name, your workspace will “mask” the function; this means that R stops searching after it finds your function and so never sees the package function. This is a blessing if you want to override the package function...and a curse if you still want access to the package function. R also uses the search path to find R datasets (not files) via a similar procedure. Unix users: don’t confuse the R search path with the Unix search path (the PATH envi- ronment variable). They are conceptually similar but two distinct things. The R search path is internal to R and is used by R only to locate functions and datasets, whereas the Unix search path is used by Unix to locate executable programs. See Also See Recipe 3.6 for loading packages into R, Recipe 3.8 for the list of installed packages (not just loaded packages), and Recipe 5.31 for inserting data frames into the search path. 3.6 Accessing the Functions in a Package Problem A package installed on your computer is either a standard package or a package down- loaded by you. When you try using functions in the package, however, R cannot find them. Solution Use either the library function or the require function to load the package into R: > library(packagename) Discussion R comes with several standard packages, but not all of them are automatically loaded when you start R. Likewise, you can download and install many useful packages from CRAN, but they are not automatically loaded when you run R. The MASS package comes standard with R, for example, but you could get this message when using the lda func- tion in that package: > lda(x) Error: could not find function "lda" 3.6 Accessing the Functions in a Package | 55 R is complaining that it cannot find the lda function among the packages currently loaded into memory. When you use the library function or the require function, R loads the package into memory and its contents become immediately available to you: > lda(f ~ x + y) Error: could not find function "lda" > library(MASS) # Load the MASS library into memory > lda(f ~ x + y) # Now R can find the function Call: lda(f ~ x + y) Prior probabilities of groups: . . (etc.) . Before calling library, R does not recognize the function name. Afterward, the package contents are available and calling the lda function works. Notice that you needn’t enclose the package name in quotes. The require function is nearly identical to library. It has two features that are useful for writing scripts. It returns TRUE if the package was successfully loaded and FALSE otherwise. It also generates a mere warning if the load fails—unlike library, which generates an error. Both functions have a key feature: they do not reload packages that are already loaded, so calling twice for the same package is harmless. This is especially nice for writing scripts. The script can load needed packages while knowing that loaded packages will not be reloaded. The detach function will unload a package that is currently loaded: > detach(package:MASS) Observe that the package name must be qualified, as in package:MASS. One reason to unload a package is that it contains a function whose name conflicts with a same-named function lower on the search list. When such a conflict occurs, we say the higher function masks the lower function. You no longer “see” the lower func- tion because R stops searching when it finds the higher function. Hence unloading the higher package unmasks the lower name. See Also See Recipe 3.5. 56 | Chapter 3: Navigating the Software 3.7 Accessing Built-in Datasets Problem You want to use one of R’s built-in datasets. Solution The standard datasets distributed with R are already available to you, since the datasets package is in your search path. To access datasets in other packages, use the data function while giving the dataset name and package name: > data(dsname, package="pkgname") Discussion R comes with many built-in datasets. These datasets are useful when you are learning about R, since they provide data with which to experiment. Many datasets are kept in a package called (naturally enough) datasets, which is dis- tributed with R. That package is in your search path, so you have instant access to its contents. For example, you can use the built-in dataset called pressure: > head(pressure) temperature pressure 1 0 0.0002 2 20 0.0012 3 40 0.0060 4 60 0.0300 5 80 0.0900 6 100 0.2700 If you want to know more about pressure, use the help function to learn about it and other datasets: > help(pressure) # Bring up help page for pressure dataset You can see a table of contents for datasets by calling the data function with no arguments: > data() # Bring up a list of datasets Any R package can elect to include datasets that supplement those supplied in datasets. The MASS package, for example, includes many interesting datasets. Use the data function to access a dataset in a specific package by using the package argument. MASS includes a dataset called Cars93, which you can access in this way: > data(Cars93, package="MASS") After this call to data, the Cars93 dataset is available to you; then you can execute summary(Cars93), head(Cars93), and so forth. 3.7 Accessing Built-in Datasets | 57 When attaching a package to your search list (e.g., via library(MASS)), you don’t need to call data. Its datasets become available automatically when you attach it. You can see a list of available datasets in MASS, or any other package, by using the data function with a package argument and no dataset name: > data(package="pkgname") See Also See Recipe 3.5 for more about the search path and Recipe 3.6 for more about packages and the library function. 3.8 Viewing the List of Installed Packages Problem You want to know what packages are installed on your machine. Solution Use the library function with no arguments for a basic list. Use installed.packages to see more detailed information about the packages. Discussion The library function with no arguments prints a list of installed packages. The list can be quite long. On a Linux computer, these might be the first few lines of output: > library() Packages in library '/usr/local/lib/R/site-library': boot Bootstrap R (S-Plus) Functions (Canty) CGIwithR CGI Programming in R class Functions for Classification cluster Cluster Analysis Extended Rousseeuw et al. DBI R Database Interface expsmooth Data sets for "Forecasting with exponential smoothing" . . (etc.) . On Windows and OS X, the list is displayed in a pop-up window. You can get more details via the installed.packages function, which returns a matrix of information regarding the packages on your machine. Each matrix row corresponds to one installed package. The columns contain the information such as package name, library path, and version. The information is taken from R’s internal database of in- stalled packages. 58 | Chapter 3: Navigating the Software To extract useful information from this matrix, use normal indexing methods. This Windows snippet calls installed.packages and extracts both the Package and Version columns, letting you see what version of each package is installed: > installed.packages()[,c("Package","Version")] Package Version acepack "acepack" "1.3-2.2" alr3 "alr3" "1.0.9" base "base" "2.4.1" boot "boot" "1.2-27" bootstrap "bootstrap" "1.0-20" calibrate "calibrate" "0.0" car "car" "1.2-1" chron "chron" "2.3-12" class "class" "7.2-30" cluster "cluster" "1.11.4" . . (etc.) . See Also See Recipe 3.6 for loading a package into memory. 3.9 Installing Packages from CRAN Problem You found a package on CRAN, and now you want to install it on your computer. Solution Command line Use the install.packages function, putting the name of the package in quotes: > install.packages("packagename") Windows You can also download and install via Packages → Install package(s)... from the main menu. OS X You can also download and install via Packages & Data → Package Installer from the main menu. On all platforms, you will be asked to select a CRAN mirror. On Windows and OS X, you will also be asked to select the packages for download. 3.9 Installing Packages from CRAN | 59 D o wnload from Wow! eBook System-wide Installation on Linux or Unix On Linux or Unix systems, root privileges are required to install packages into the system-wide libraries because those directories are not usually writable by mere mor- tals. For that reason, installations are usually performed by a system administrator. If the administrator is unwilling or unable to do a system-wide installation, you can still install packages in your personal library. If you have root privileges: 1. Run su or sudo to start a root shell. 2. Start an R session in the root shell. 3. From there, execute the install.packages function. If you don’t have root privileges, you’ll know very quickly. The install.packages func- tion will stop, warning you that it cannot write into the necessary directories. Then it will ask if you want to create a personal library instead. If you do, it will create the necessary directories in your home directory and install the package there. Discussion Installing a package locally is the first step toward using it. The installer will prompt you for a mirror site from which it can download the package files: --- Please select a CRAN mirror for use in this session --- It will then display a list of CRAN mirror sites. Select one close to you. The official CRAN server is a relatively modest machine generously hosted by the Department of Statistics and Mathematics at WU Wien, Vienna, Austria. If every R user downloaded from the official server, it would buckle under the load, so there are numerous mirror sites around the globe. You are strongly encouraged to find and use a nearby mirror. If the new package depends upon other packages that are not already installed locally, then the R installer will automatically download and install those required packages. This is a huge benefit that frees you from the tedious task of identifying and resolving those dependencies. There is a special consideration when installing on Linux or Unix. You can install the package either in the system-wide library or in your personal library. Packages in the system-wide library are available to everyone; packages in your personal library are (normally) used only by you. So a popular, well-tested package would likely go in the system-wide library whereas an obscure or untested package would go into your per- sonal library. By default, install.packages assumes you are performing a system-wide install. To install into your personal library, first create a directory for the library—for example, ~/lib/R: 60 | Chapter 3: Navigating the Software $ mkdir ~/lib $ mkdir ~/lib/R Set the R_LIBS environment variable before starting R. Otherwise, it will not know your library’s location: $ export R_LIBS=~/lib/R # bash syntax $ setenv R_LIBS ~/lib/R # csh syntax Then, call install.packages with the lib argument set to that directory: > install.packages("packagename", lib="~/lib/R") See Also See Recipe 1.11 for ways to find relevant packages and Recipe 3.6 for using a package after installing it. 3.10 Setting a Default CRAN Mirror Problem You are downloading packages. You want to set a default CRAN mirror so R does not prompt for one each time. Solution This solution assumes you have an .Rprofile, as described in Recipe 3.16: 1. Call the chooseCRANmirror function: > chooseCRANmirror() R will present a list of CRAN mirrors. 2. Select a CRAN mirror from the list and press OK. 3. To get the URL of the mirror, look at the first element of the repos option: > options("repos")[[1]][1] 4. Add this line to your .Rprofile file: options(repos="URL") where URL is the URL of the mirror. Discussion When you install packages, you probably use the same CRAN mirror each time (namely, the mirror closest to you). You probably get tired of R repeatedly asking you to select a mirror. If you follow the Solution, R will stop prompting for a mirror because it will have a default mirror. 3.10 Setting a Default CRAN Mirror | 61 The repos option is the name of your default mirror. The chooseCRANmirror function has the important side effect of setting the repos option according to your selection. The problem is that R forgets the setting when it exits, leaving no permanent default. By setting repos in your .Rprofile, you restore the setting every time R starts. See Also See Recipe 3.16 for more about the .Rprofile file and the options function. 3.11 Suppressing the Startup Message Problem You are tired of seeing R’s verbose startup message. Solution Use the --quiet command-line option when you start R. Discussion The startup message from R is handy for beginners because it contains useful infor- mation about the R project and getting help. But the novelty wears off pretty quickly. If you start R from the shell prompt, use the --quiet option to hide the startup message: $ R --quiet > On my Linux box, I aliased R like this so I never see the startup message: $ alias R="/usr/bin/R --quiet" If you are starting R in Windows using a shortcut, you can embed the --quiet option inside the shortcut. Right-click on the shortcut icon; select Properties; select the Short- cut tab; and, at the end of the Target string, add the --quiet option. Be careful to leave a space between the end of the program path and --quiet. See Also See Recipe 1.2. 3.12 Running a Script Problem You captured a series of R commands in a text file. Now you want to execute them. 62 | Chapter 3: Navigating the Software Solution The source function instructs R to read the text file and execute its contents: > source("myScript.R") Discussion When you have a long or frequently used piece of R code, capture it inside a text file. That lets you easily rerun the code without having to retype it. Use the source function to read and execute the code, just as if you had typed it into the R console. Suppose the file hello.R contains this one, familiar greeting: print("Hello, World!") Then sourcing the file will execute the file contents: > source("hello.R") [1] "Hello, World!" Setting echo=TRUE will echo the script lines before they are executed, with the R prompt shown before each line: > source("hello.R", echo=TRUE) > print("Hello, World!") [1] "Hello, World!" See Also See Recipe 2.13 for running blocks of R code inside the GUI. 3.13 Running a Batch Script Problem You are writing a command script, such as a shell script in Unix or OS X or a BAT script in Windows. Inside your script, you want to execute an R script. Solution Run the R program with the CMD BATCH subcommand, giving the script name and the output file name: $ R CMD BATCH scriptfile outputfile If you want the output sent to stdout or if you need to pass command-line arguments to the script, consider the Rscript command instead: $ Rscript scriptfile arg1 arg2 arg3 3.13 Running a Batch Script | 63 Discussion R is normally an interactive program, one that prompts the user for input and then displays the results. Sometimes you want to run R in batch mode, reading commands from a script. This is especially useful inside shell scripts, such as scripts that include a statistical analysis. The CMD BATCH subcommand puts R into batch mode, reading from scriptfile and writing to outputfile. It does not interact with a user. You will likely use command-line options to adjust R’s batch behavior to your circum- stances. For example, using --quiet silences the startup messages that would otherwise clutter the output: $ R CMD BATCH --quiet myScript.R results.out Other useful options in batch mode include the following: --slave Like --quiet, but it makes R even more silent by inhibiting echo of the input. --no-restore At startup, do not restore the R workspace. This is important if your script expects R to begin with an empty workspace. --no-save At exit, do not save the R workspace. Otherwise, R will save its workspace and overwrite the .RData file in the working directory. --no-init-file Do not read either the .Rprofile or ~/.Rprofile files. The CMD BATCH subcommand normally calls proc.time when your script completes, showing the execution time. If this annoys you then end your script by calling the q function with runLast=FALSE, which will prevent the call to proc.time. The CMD BATCH subcommand has two limitations: the output always goes to a file, and you cannot easily pass command-line arguments to your script. If either limitation is a problem, consider using the Rscript program that comes with R. The first command- line argument is the script name, and the remaining arguments are given to the script: $ Rscript myScript.R arg1 arg2 arg3 Inside the script, the command-line arguments can be accessed by calling commandArgs, which returns the arguments as a vector of strings: argv <- commandArgs(TRUE) The Rscript program takes the same command-line options as CMD BATCH, which were just described. 64 | Chapter 3: Navigating the Software Output is written to stdout, which R inherits from the calling shell script, of course. You can redirect the output to a file by using the normal redirection: $ Rscript --slave myScript.R arg1 arg2 arg3 >results.out Here is a small R script, arith.R, that takes two command-line arguments and performs four arithmetic operations on them: argv <- commandArgs(TRUE) x <- as.numeric(argv[1]) y <- as.numeric(argv[2]) cat("x =", x, "\n") cat("y =", y, "\n") cat("x + y = ", x + y, "\n") cat("x - y = ", x - y, "\n") cat("x * y = ", x * y, "\n") cat("x / y = ", x / y, "\n") The script is invoked like this: $ Rscript arith.R 2 3.1415 which produces the following output: x = 2 y = 3.1415 x + y = 5.1415 x - y = -1.1415 x * y = 6.283 x / y = 0.6366385 On Linux or Unix, you can make the script fully self-contained by placing a #! line at the head with the path to the Rscript program. Suppose that Rscript is installed in /usr/bin/Rscript on your system. Then adding this line to arith.R makes it a self- contained script: #!/usr/bin/Rscript --slave argv <- commandArgs(TRUE) x <- as.numeric(argv[1]) . . (etc.) . At the shell prompt, we mark the script as executable: $ chmod +x arith.R Now we can invoke the script directly without the Rscript prefix: $ arith.R 2 3.1415 See Also See Recipe 3.12 for running a script from within R. 3.13 Running a Batch Script | 65 3.14 Getting and Setting Environment Variables Problem You want to see the value of an environment variable, or you want to change its value. Solution Use the Sys.getenv function to see values. Use Sys.putenv to change them: > Sys.getenv("SHELL") SHELL "/bin/bash" > Sys.setenv(SHELL="/bin/ksh") Discussion Environment variables are often used on Linux and Unix to configure and control the software. Each process has its own set of environment variables, which are inherited from its parent process. You sometimes need to see the environment variable settings for your R process in order to understand its behavior. Likewise, you sometimes need to change those settings to modify that behavior. Sometimes I start R in one place but actually want the graphics to appear in a different place. For example, I might run R on one Linux terminal but want the graphics to appear on a larger display that an audience can see more easily. Or I might want to run R on my Linux workstation and have the graphics appear on a colleague’s workstation. On Linux, R uses the X Windows system for its graphics displays. X Windows chooses its display device according to an environment variable called DISPLAY. We can use Sys.getenv to see its value: > Sys.getenv("DISPLAY") DISPLAY ":0.0" All environment variables are string-valued, and here the value is ":0.0"—an obscure way of saying that the R session is connected to display 0, screen number 0 on my workstation. To redirect graphics to the local display at 10.0, we can change DISPLAY in this way: > Sys.putenv(DISPLAY="localhost:10.0") Similarly, we could redirect graphics to display 0, screen 0 on the workstation called zeus: > Sys.putenv(DISPLAY="zeus:0.0") In either event, you must change the DISPLAY variable before drawing your graphic. 66 | Chapter 3: Navigating the Software 3.15 Locating the R Home Directory Problem You need to know the R home directory, which is where the configuration and instal- lation files are kept. Solution R creates an environment variable called R_HOME that you can access by using the Sys.getenv function: > Sys.getenv("R_HOME") Discussion Most users will never need to know the R home directory. But system administrators or sophisticated users must know in order to check or change the R installation files. When R starts, it defines an environment variable (not an R variable) called R_HOME, which is the path to the R home directory. The Sys.getenv function can retrieve its value. Here are examples by platform. The exact value reported will almost certainly be different on your own computer: On Windows > Sys.getenv("R_HOME") R_HOME "C:\\PROGRA~1\\R\\R-21~1.1" On OS X > Sys.getenv("R_HOME") "/Library/Frameworks/R.framework/Resources" On Linux or Unix > Sys.getenv("R_HOME") R_HOME "/usr/lib/R" The Windows result looks funky because R reports the old, DOS-style compressed path name. The full, user-friendly path would be C:\Program Files\R\R-2.10.1 in this case. On Unix and OS X, you can also run the R program from the shell and use the RHOME subcommand to display the home directory: $ R RHOME /usr/lib/R Note that the R home directory on Unix and OS X contains the installation files but not necessarily the R executable file. The executable could be in /usr/bin while the R home directory is, for example, /usr/lib/R. 3.15 Locating the R Home Directory | 67 3.16 Customizing R Problem You want to customize your R sessions by, for instance, changing configuration options or preloading packages. Solution Create a script called .Rprofile that customizes your R session. R will execute the .Rprofile script when it starts. The placement of .Rprofile depends upon your platform: OS X, Linux, or Unix Save the file in your home directory (~/.Rprofile). Windows Save the file in your My Documents directory. Discussion R executes profile scripts when it starts, freeing you from repeatedly loading often-used packages or tweaking the R configuration options. You can create a profile script called .Rprofile and place it in your home directory (OS X, Linux, Unix) or your My Documents directory (Windows XP) or Documents direc- tory (Windows Vista, Windows 7). The script can call functions to customize your sessions, such as this simple script that loads the MASS package and sets the prompt to R>: require(MASS) options(prompt="R> ") The profile script executes in a bare-bones environment, so there are limits on what it can do. Trying to open a graphics window will fail, for example, because the graphics package is not yet loaded. Also, you should not attempt long-running computations. You can customize a particular project by putting an .Rprofile file in the directory that contains the project files. When R starts in that directory, it reads the local .Rprofile file; this allows you to do project-specific customizations (e.g., loading packages needed only by the project). However, if R finds a local profile then it does not read the global profile. That can be annoying, but it’s easily fixed: simply source the global profile from the local profile. On Unix, for instance, this local profile would execute the global profile first and then execute its local material: source("~/.Rprofile") # # ... remainder of local .Rprofile... # 68 | Chapter 3: Navigating the Software Setting Options Some customizations are handled via calls to the options function, which sets the R configuration options. There are many such options, and the R help page for options lists them all: > help(options) Here are some examples: browser="path" Path of default HTML browser digits=n Suggested number of digits to print when printing numeric values editor="path" Default text editor prompt="string" Input prompt repos="url" URL for default repository for packages warn=n Controls display of warning messages Loading Packages Another common customization is the preloading of packages. You may want certain packages loaded every time you run R because you use them so often. A simple way to accomplish this is by calling the require function inside your .Rprofile script, like this call (which loads the tseries package): require(tseries) If require generates warning messages that are polluting your output when R starts, surround it with a suppressMessages function. That’ll muzzle it: suppressMessages(require(tseries)) Instead of calling require explicitly, you can set a configuration parameter called defaultPackages. This is the list of packages loaded by R at startup. It initially contains a system-defined list. If you append package names to the list then R will load them, too, saving you from having to call library or require yourself. Here is a clip from my R profile that adjusts the list of loaded packages. I almost always use the zoo package, so I append it to the defaultPackages list: pkgs <- getOption("defaultPackages") # Get system list of packages to load pkgs <- c(pkgs, "zoo") # Append "zoo" to list options(defaultPackages = pkgs) # Update configuration option rm(pkgs) # Clean up our temp variable 3.16 Customizing R | 69 One advantage of using defaultPackages is that you can control the placement of the packages in the search list. Packages appended to the end are loaded last, so in this example the zoo package will appear at the head of the search list. (Remember that packages are inserted at the head of the search list when they are loaded.) The price for always loading the zoo package is that R starts a little more slowly, even when I don’t really need zoo. In my case, I just was fed up with typing library(zoo) nearly every time I ran R, so I decided the slower startup was a good trade-off for the convenience. Startup Sequence Here is a simplified overview of what happens when R starts (type help(Startup) to see the full details): 1. R executes the Rprofile.site script. This is the site-level script that enables system administrators to override default options with localizations. The script’s full path is R_HOME/etc/Rprofile.site. (R_HOME is the R home directory; see Recipe 3.15.) The R distribution does not include an Rprofile.site file. Rather, the system administrator creates one if it is needed. 2. R executes the .Rprofile script in the working directory; or, if that file does not exist, executes the .Rprofile script in your home directory. This is the user’s opportunity to customize R for his or her purposes. The .Rprofile script in the home directory is used for global customizations. The .Rprofile script in a lower-level directory can perform specific customizations when R is started there; for instance, custom- izing R when started in a project-specific directory. 3. R loads the workspace saved in .RData, if that file exists in the working directory. R saves your workspace in the file called .RData when it exits. It reloads your work- space from that file, restoring access to your local variables and functions. 4. R executes the .First function, if you defined one. The .First function is a useful place for users or projects to define startup initialization code. You can define it in your .Rprofile or in your workspace. 5. R executes the .First.sys function. This step loads the default packages. The func- tion is internal to R and not normally changed by either users or administrators. Observe that R does not load the default packages until the final step, when it executes the .First.sys function. Before that, only the base package has been loaded. This is a key fact because it means the previous steps cannot assume that packages other than the base are available. It also explains why trying to open a graphical window in your .Rprofile script fails: the graphics packages aren’t loaded yet. See Also See Recipe 3.6 for more about loading packages. See the R help page for Startup (help(Startup)) and the R help page for options (help(options)). 70 | Chapter 3: Navigating the Software CHAPTER 4 Input and Output Introduction All statistical work begins with data, and most data is stuck inside files and databases. Dealing with input is probably the first step of implementing any significant statistical project. All statistical work ends with reporting numbers back to a client, even if you are the client. Formatting and producing output is probably the climax of your project. Casual R users can solve their input problems by using basic functions such as read.csv to read CSV files and read.table to read more complicated, tabular data. They can use print, cat, and format to produce simple reports. Users with heavy-duty input/output (I/O) needs are strongly encouraged to read the R Data Import/Export guide, available on CRAN at http://cran.r-project.org/doc/man uals/R-data.pdf. This manual includes important information on reading data from sources such as spreadsheets, binary files, other statistical systems, and relational databases. A Philosophical Note Several of my Statistical Analysis System (SAS) friends are disappointed with the input facilities of R. They point out that SAS has an elaborate set of commands for reading and parsing input files in many formats. R does not, and this leads them to conclude that R is not ready for real work. After all, if it can’t read your data, what good is it? I think they do not understand the design philosophy behind R, which is based on a statistical package called S. The authors of S worked at Bell Labs and were steeped in the Unix design philosophy. A keystone of that philosophy is the idea of modular tools. Programs in Unix are not large, monolithic programs that try to do everything. Instead, they are smaller, specialized tools that each do one thing well. The Unix user joins the programs together like building blocks, creating systems from the components. 71 R does statistics and graphics well. Very well, in fact. It is superior in that way to many commercial packages. R is not a great tool for preprocessing data files, however. The authors of S assumed you would perform that munging with some other tool: perl, awk, sed, cut, paste, whatever floats your boat. Why should they duplicate that capability? If your data is difficult to access or difficult to parse, consider using an outboard tool to preprocess the data before loading it into R. Let R do what R does best. 4.1 Entering Data from the Keyboard Problem You have a small amount of data, too small to justify the overhead of creating an input file. You just want to enter the data directly into your workspace. Solution For very small datasets, enter the data as literals using the c() constructor for vectors: > scores <- c(61, 66, 90, 88, 100) Alternatively, you can create an empty data frame and then invoke the built-in, spreadsheet-like editor to populate it: > scores <- data.frame() # Create empty data frame > scores <- edit(score) # Invoke editor, overwrite with edited data On Windows, you can also invoke the editor by selecting Edit → Data frame... from the menu. Discussion When I am working on a toy problem, I don’t want the hassle of creating and then reading the data file. I just want to enter the data into R. The easiest way is by using the c() constructor for vectors, as shown in the Solution. This approach works for data frames, too, by entering each variable (column) as a vector: > points <- data.frame( + label=c("Low", "Mid", "High"), + lbound=c( 0, 0.67, 1.64), + ubound=c(0.674, 1.64, 2.33) + ) See Also See Recipe 5.26 for more about using the built-in data editor, as suggested in the Solution. 72 | Chapter 4: Input and Output 4.2 Printing Fewer Digits (or More Digits) Problem Your output contains too many digits or too few digits. You want to print fewer or more. Solution For print, the digits parameter can control the number of printed digits. For cat, use the format function (which also has a digits parameter) to alter the for- matting of numbers. Discussion R normally formats floating-point output to have seven digits: > pi [1] 3.141593 > 100*pi [1] 314.1593 This works well most of the time but can become annoying when you have lots of numbers to print in a small space. It gets downright misleading when there are only a few significant digits in your numbers and R still prints seven. The print function lets you vary the number of printed digits using the digits parameter: > print(pi, digits=4) [1] 3.142 > print(100*pi, digits=4) [1] 314.2 The cat function does not give you direct control over formatting. Instead, use the format function to format your numbers before calling cat: > cat(pi, "\n") 3.141593 > cat(format(pi,digits=4), "\n") 3.142 This is R, so both print and format will format entire vectors at once: > pnorm(-3:3) [1] 0.001349898 0.022750132 0.158655254 0.500000000 0.841344746 0.977249868 [7] 0.998650102 > print(pnorm(-3:3), digits=3) [1] 0.00135 0.02275 0.15866 0.50000 0.84134 0.97725 0.9986 Notice that print formats the vector elements consistently: finding the number of digits necessary to format the smallest number and then formatting all numbers to have the same width (though not necessarily the same number of digits). This is extremely useful for formating an entire table: 4.2 Printing Fewer Digits (or More Digits) | 73 > q <- seq(from=0,to=3,by=0.5) > tbl <- data.frame(Quant=q, Lower=pnorm(-q), Upper=pnorm(q)) > tbl # Unformatted print Quant Lower Upper 1 0.0 0.500000000 0.5000000 2 0.5 0.308537539 0.6914625 3 1.0 0.158655254 0.8413447 4 1.5 0.066807201 0.9331928 5 2.0 0.022750132 0.9772499 6 2.5 0.006209665 0.9937903 7 3.0 0.001349898 0.9986501 > print(tbl,digits=2) # Formatted print: fewer digits Quant Lower Upper 1 0.0 0.5000 0.50 2 0.5 0.3085 0.69 3 1.0 0.1587 0.84 4 1.5 0.0668 0.93 5 2.0 0.0228 0.98 6 2.5 0.0062 0.99 7 3.0 0.0013 1.00 You can also alter the format of all output by using the options function to change the default for digits: > pi [1] 3.141593 > options(digits=15) > pi [1] 3.14159265358979 But this is a poor choice in my experience, since it also alters the output from R’s built- in functions, and that alteration will likely be unpleasant. See Also Other functions for formatting numbers include sprintf and formatC; see their help pages for details. 4.3 Redirecting Output to a File Problem You want to redirect the output from R into a file instead of your console. Solution You can redirect the output of the cat function by using its file argument: > cat("The answer is", answer, "\n", file="filename") 74 | Chapter 4: Input and Output Use the sink function to redirect all output from both print and cat. Call sink with a filename argument to begin redirecting console output to that file. When you are done, use sink with no argument to close the file and resume output to the console: > sink("filename") # Begin writing output to file . . . other session work . . . > sink() # Resume writing output to console Discussion The print and cat functions normally write their output to your console. The cat func- tion writes to a file if you supply a file argument, which can be either a filename or a connection. The print function cannot redirect its output, but the sink function can force all output to a file. A common use for sink is to capture the output of an R script: > sink("script_output.txt") # Redirect output to file > source("script.R") # Run the script, capturing its output > sink() # Resume writing output to console If you are repeatedly cating items to one file, be sure to set append=TRUE. Otherwise, each call to cat will simply overwrite the file’s contents: cat(data, file="analysisReport.out") cat(results, file="analysisRepart.out", append=TRUE) cat(conclusion, file="analysisReport.out", append=TRUE) Hard-coding file names like this is a tedious and error-prone process. Did you notice that the filename is misspelled in the second line? Instead of hard-coding the filename repeatedly, I suggest opening a connection to the file and writing your output to the connection: con <- file("analysisReport.out", "w") cat(data, file=con) cat(results, file=con) cat(conclusion, file=con) close(con) (You don’t need append=TRUE when writing to a connection because it’s implied.) This technique is especially valuable inside R scripts because it makes your code more reli- able and more maintainable. 4.4 Listing Files Problem You want to see a listing of your files without the hassle of switching to your file browser. 4.4 Listing Files | 75 Solution The list.files function shows the contents of your working directory: > list.files() Discussion This function is just plain handy. If I can’t remember the name of my data file (was it sample_data.csv or sample-data.csv?), I do a quick list.files to refresh my memory: > list.files() [1] "sample-data.csv" "script.R" To see all the files in your subdirectories, too, use list.files(recursive=T). A possible “gotcha” of list.files is that it ignores hidden files—typically, any file whose name begins with a period. If you don’t see the file you expected to see, try setting all.files=TRUE: > list.files(all.files=TRUE) See Also R has other handy functions for working with files; see help(files). 4.5 Dealing with “Cannot Open File” in Windows Problem You are running R on Windows, and you are using file names such as C:\data\sample.txt. R says it cannot open the file, but you know the file does exist. Solution The backslashes in the file path are causing trouble. You can solve this problem in one of two ways: • Change the backslashes to forward slashes: "C:/data/sample.txt". • Double the backslashes: "C:\\data\\sample.txt". Discussion When you open a file in R, you give the file name as a character string. Problems arise when the name contains backslashes (\) because backslashes have a special meaning inside strings. You’ll probably get something like this: > samp <- read.csv("C:\Data\sample-data.csv") Error in file(file, "rt") : cannot open the connection In addition: Warning messages: 1: '\D' is an unrecognized escape in a character string 76 | Chapter 4: Input and Output 2: '\s' is an unrecognized escape in a character string 3: unrecognized escapes removed from "C:\Data\sample-data.csv" 4: In file(file, "rt") : cannot open file 'C:Datasample-data.csv': No such file or directory R escapes every character that follows a backslash and then removes the backslashes. That leaves a meaningless file path, such as C:Datasample-data.csv in this example. The simple solution is to use forward slashes instead of backslashes. R leaves the for- ward slashes alone, and Windows treats them just like backslashes. Problem solved: > samp <- read.csv("C:/Data/sample-data.csv") > An alternative solution is to double the backslashes, since R replaces two consecutive backslashes with a single backslash: > samp <- read.csv("C:\\Data\\sample-data.csv") > 4.6 Reading Fixed-Width Records Problem You are reading data from a file of fixed-width records: records whose data items occur at fixed boundaries. Solution Read the file using the read.fwf function. The main arguments are the file name and the widths of the fields: > records <- read.fwf("filename", widths=c(w1, w2, ..., wn)) Discussion Suppose we want to read an entire file of fixed-width records, such as fixed-width.txt, shown here: Fisher R.A. 1890 1962 Pearson Karl 1857 1936 Cox Gertrude 1900 1978 Yates Frank 1902 1994 Smith Kirstine 1878 1939 We need to know the column widths. In this case the columns are: last name, 10 char- acters; first name, 10 characters; year of birth, 4 characters; and year of death, 4 char- acters. In between the two last columns is a 1-character space. We can read the file this way: > records <- read.fwf("fixed-width.txt", widths=c(10,10,4,-1,4)) 4.6 Reading Fixed-Width Records | 77 D o wnload from Wow! eBook The -1 in the widths argument says there is a one-character column that should be ignored. The result of read.fwf is a data frame: > records V1 V2 V3 V4 1 Fisher R.A. 1890 1962 2 Pearson Karl 1857 1936 3 Cox Gertrude 1900 1978 4 Yates Frank 1902 1994 5 Smith Kirstine 1878 1939 Note that R supplied some funky, synthetic column names. We can override that de- fault by using a col.names argument: > records <- read.fwf("fixed-width.txt", widths=c(10,10,4,-1,4), + col.names=c("Last","First","Born","Died")) > records Last First Born Died 1 Fisher R.A. 1890 1962 2 Pearson Karl 1857 1936 3 Cox Gertrude 1900 1978 4 Yates Frank 1902 1994 5 Smith Kirstine 1878 1939 read.fwf interprets nonnumeric data as a factor (categorical variable) by default. For instance, the Last and First columns just displayed were interpreted as factors. Set stringsAsFactors=FALSE to have the function read them as character strings. The read.fwf function has many bells and whistles that may be useful for reading your data files. It also shares many bells and whistles with the read.table function. I suggest reading the help pages for both functions. See Also See Recipe 4.7 for more discussion of reading text files. 4.7 Reading Tabular Data Files Problem You want to read a text file that contains a table of data. Solution Use the read.table function, which returns a data frame: > dfrm <- read.table("filename") Discussion Tabular data files are quite common. They are text files with a simple format: 78 | Chapter 4: Input and Output • Each line contains one record. • Within each record, fields (items) are separated by a one-character delimiter, such as a space, tab, colon, or comma. • Each record contains the same number of fields. This format is more free-form than the fixed-width format because fields needn’t be aligned by position. Here is the data file of Recipe 4.6 in tabular format, using a space character between fields: Fisher R.A. 1890 1962 Pearson Karl 1857 1936 Cox Gertrude 1900 1978 Yates Frank 1902 1994 Smith Kirstine 1878 1939 The read.table function is built to read this file. By default, it assumes the data fields are separated by white space (blanks or tabs): > dfrm <- read.table("statisticians.txt") > print(dfrm) V1 V2 V3 V4 1 Fisher R.A. 1890 1962 2 Pearson Karl 1857 1936 3 Cox Gertrude 1900 1978 4 Yates Frank 1902 1994 5 Smith Kirstine 1878 1939 If your file uses a separator other than white space, specify it using the sep parameter. If our file used colon (:) as the field separator, we would read it this way: > dfrm <- read.table("statisticians.txt", sep=":") You cannot tell from the printed output, but read.table interpreted the first and last names as factors, not strings. We see that by checking the class of the resulting column: > class(dfrm$V1) [1] "factor" To prevent read.table from interpreting character strings as factors, set the stringsAsFactors parameter to FALSE: > dfrm <- read.table("statisticians.txt", stringsAsFactor=FALSE) > class(dfrm$V1) [1] "character" Now the class of the first column is character, not factor. If any field contains the string “NA”, then read.table assumes that the value is missing and converts it to NA. Your data file might employ a different string to signal missing values, in which case use the na.strings parameter. The SAS convention, for example, is that missing values are signaled by a single period (.). We can read such data files in this way: > dfrm <- read.table("filename.txt", na.strings=".") 4.7 Reading Tabular Data Files | 79 I am a huge fan of self-describing data: data files which describe their own contents. (A computer scientist would say the file contains its own metadata.) The read.table func- tion has two features that support this characteristic. First, you can include a header line at the top of your file that gives names to the columns. The line contains one name for every column, and it uses the same field separator as the data. Here is our data file with a header line that names the columns: lastname firstname born died Fisher R.A. 1890 1962 Pearson Karl 1857 1936 Cox Gertrude 1900 1978 Yates Frank 1902 1994 Smith Kirstine 1878 1939 Now we can tell read.table that our file contains a header line, and it will use the column names when it builds the data frame: > dfrm <- read.table("statisticians.txt", header=TRUE, stringsAsFactor=FALSE) > print(dfrm) lastname firstname born died 1 Fisher R.A. 1890 1962 2 Pearson Karl 1857 1936 3 Cox Gertrude 1900 1978 4 Yates Frank 1902 1994 5 Smith Kirstine 1878 1939 The second feature of read.table is comment lines. Any line that begins with a pound sign (#) is ignored, so you can put comments on those lines: # This is a data file of famous statisticians. # Last edited on 1994-06-18 lastname firstname born died Fisher R.A. 1890 1962 Pearson Karl 1857 1936 Cox Gertrude 1900 1978 Yates Frank 1902 1994 Smith Kirstine 1878 1939 read.table has many parameters for controlling how it reads and interprets the input file. See the help page for details. See Also If your data items are separated by commas, see Recipe 4.8 for reading a CSV file. 4.8 Reading from CSV Files Problem You want to read data from a comma-separated values (CSV) file. 80 | Chapter 4: Input and Output Solution The read.csv function can read CSV files. If your CSV file has a header line, use this: > tbl <- read.csv("filename") If your CSV file does not contain a header line, set the header option to FALSE: > tbl <- read.csv("filename", header=FALSE) Discussion The CSV file format is popular because many programs can import and export data in that format. Such programs include R, Excel, other spreadsheet programs, many database managers, and most statistical packages. It is a flat file of tabular data, where each line in the file is a row of data, and each row contains data items separated by commas. Here is a very simple CSV file with three rows and three columns (the first line is a header line that contains the column names, also separated by commas): label,lbound,ubound low,0,0.674 mid,0.674,1.64 high,1.64,2.33 The read.csv file reads the data and creates a data frame, which is the usual R repre- sentation for tabular data. The function assumes that your file has a header line unless told otherwise: > tbl <- read.csv("table-data.csv") > tbl label lbound ubound 1 low 0.000 0.674 2 mid 0.674 1.640 3 high 1.640 2.330 Observe that read.csv took the column names from the header line for the data frame. If the file did not contain a header, then we would specify header=FALSE and R would synthesize column names for us (V1, V2, and V3 in this case): > tbl <- read.csv("table-data-with-no-header.csv", header=FALSE) > tbl V1 V2 V3 1 low 0.000 0.674 2 mid 0.674 1.640 3 high 1.640 2.330 A good feature of read.csv is that is automatically interprets nonnumeric data as a factor (categorical variable), which is often what you want since after all this is a statistical package, not Perl. The label variable in the tbl data frame just shown is actually a factor, not a character variable. You see that by inspecting the structure of tbl: 4.8 Reading from CSV Files | 81 > str(tbl) 'data.frame': 3 obs. of 3 variables: $ label : Factor w/ 3 levels "high","low","mid": 2 3 1 $ lbound: num 0 0.674 1.64 $ ubound: num 0.674 1.64 2.33 Sometimes you really want your data interpreted as strings, not as a factor. In that case, set the as.is parameter to TRUE; this indicates that R should not interpret nonnumeric data as a factor: > tbl <- read.csv("table-data.csv", as.is=TRUE) > str(tbl) 'data.frame': 3 obs. of 3 variables: $ label : chr "low" "mid" "high" $ lbound: num 0 0.674 1.64 $ ubound: num 0.674 1.64 2.33 Notice that the label variable now has character-string values and is no longer a factor. Another useful feature is that input lines starting with a pound sign (#) are ignored, which lets you embed comments in your data file. Disable this feature by specifying comment.char="". The read.csv function has many useful bells and whistles. These include the ability to skip leading lines in the input file, control the conversion of individual columns, fill out short rows, limit the number of lines, and control the quoting of strings. See the R help page for details. See Also See Recipe 4.9. See the R help page for read.table, which is the basis for read.csv. 4.9 Writing to CSV Files Problem You want to save a matrix or data frame in a file using the comma-separated values format. Solution The write.csv function can write a CSV file: > write.csv(x, file="filename", row.names=FALSE) Discussion The write.csv function writes tabular data to an ASCII file in CSV format. Each row of data creates one line in the file, with data items separated by commas (,): > print(tbl) label lbound ubound 82 | Chapter 4: Input and Output 1 low 0.000 0.674 2 mid 0.674 1.640 3 high 1.640 2.330 > write.csv(tbl, file="table-data.csv", row.names=T) This example creates a file called table-data.csv in the current working directory. The file looks like this: "label","lbound","ubound" "low",0,0.674 "mid",0.674,1.64 "high",1.64,2.33 Notice that the function writes a column header line by default. Set col.names=FALSE to change that. If we do not specify row.names=FALSE, the function prepends each row with a label taken from the row.names attribute of your data. If your data doesn’t have row names then the function just uses the row numbers, which creates a CSV file like this: "","label","lbound","ubound" "1","low",0,0.674 "2","mid",0.674,1.64 "3","high",1.64,2.33 I rarely want row labels in my CSV files, which is why I recommend setting row.names=FALSE. The function is intentionally inflexible. You cannot easily change the defaults because it really, really wants to write files in a valid CSV format. Use the write.table function to save your tabular data in other formats. A sad limitation of write.csv is that it cannot append lines to a file. Use write.table instead. See Also See Recipe 3.1 for more about the current working directory and Recipe 4.14 for other ways to save data to files. 4.10 Reading Tabular or CSV Data from the Web Problem You want to read data directly from the Web into your R workspace. Solution Use the read.csv, read.table, and scan functions, but substitute a URL for a file name. The functions will read directly from the remote server: > tbl <- read.csv("http://www.example.com/download/data.csv") 4.10 Reading Tabular or CSV Data from the Web | 83 You can also open a connection using the URL and then read from the connection, which may be preferable for complicated files. Discussion The Web is a gold mine of data. You could download the data into a file and then read the file into R, but it’s more convenient to read directly from the Web. Give the URL to read.csv, read.table, or scan (depending upon the format of the data), and the data will be downloaded and parsed for you. No fuss, no muss. Aside from using a URL, this recipe is just like reading from a CSV file (Recipe 4.8) or a complex file (Recipe 4.12), so all the comments in those recipes apply here, too. Remember that URLs work for FTP servers, not just HTTP servers. This means that R can also read data from FTP sites using URLs: > tbl <- read.table("ftp://ftp.example.com/download/data.txt") See Also See Recipes 4.8 and 4.12. 4.11 Reading Data from HTML Tables Problem You want to read data from an HTML table on the Web. Solution Use the readHTMLTable function in the XML package. To read all tables on the page, simply give the URL: > library(XML) > url <- 'http://www.example.com/data/table.html' > tbls <- readHTMLTable(url) To read only specific tables, use the which parameter. This example reads the third table on the page: > tbl <- readHTMLTable(url, which=3) Discussion Web pages can contain several HTML tables. Calling readHTMLTable(url) reads all ta- bles on the page and returns them in a list. This can be useful for exploring a page, but it’s annoying if you want just one specific table. In that case, use which=n to select the desired table. You’ll get only the nth table. 84 | Chapter 4: Input and Output The following example, which is taken from the help page for readHTMLTable, loads all tables from the Wikipedia page entitled “World population”: > library(XML) > url <- 'http://en.wikipedia.org/wiki/World_population' > tbls <- readHTMLTable(url) As it turns out, that page contains 17 tables: > length(tbls) [1] 17 In this example we care only about the third table (which lists the largest populations by country), so we specify which=3: > tbl <- readHTMLTable(url, which=3) In that table, columns 2 and 3 contain the country name and population, respectively: > tbl[,c(2,3)] Country / Territory Population 1 Â People's Republic of China[44] 1,338,460,000 2 Â India 1,182,800,000 3 Â United States 309,659,000 4 Â Indonesia 231,369,500 5 Â Brazil 193,152,000 6 Â Pakistan 169,928,500 7 Â Bangladesh 162,221,000 8 Â Nigeria 154,729,000 9 Â Russia 141,927,297 10 Â Japan 127,530,000 11 Â Mexico 107,550,697 12 Â Philippines 92,226,600 13 Â Vietnam 85,789,573 14 Â Germany 81,882,342 15 Â Ethiopia 79,221,000 16 Â Egypt 78,459,000 Right away, we can see problems with the data: the country names have some funky Unicode character stuck to the front. I don’t know why; it probably has something to do with formatting the Wikipedia page. Also, the name of the People’s Republic of China has “[44]” appended. On the Wikipedia website, that was a footnote reference, but now it’s just a bit of unwanted text. Adding insult to injury, the population numbers have embedded commas, so you cannot easily convert them to raw numbers. All these problems can be solved by some string processing, but each problem adds at least one more step to the process. This illustrates the main obstacle to reading HTML tables. HTML was designed for presenting information to people, not to computers. When you “scrape” information off an HTML page, you get stuff that’s useful to people but annoying to computers. If you ever have a choice, choose instead a computer-oriented data representation such as XML, JSON, or CSV. 4.11 Reading Data from HTML Tables | 85 The readHTMLTable function is part of the XML package, which (by necessity) is large and complex. The XML package depends on a software library called libxml2, which you will need to obtain and install first. On Linux, you will also need the Linux package xml2-config, which is necessary for building the R package. See Also See Recipe 3.9 for downloading and installing packages such as the XML package. 4.12 Reading Files with a Complex Structure Problem You are reading data from a file that has a complex or irregular structure. Solution • Use the readLines function to read individual lines; then process them as strings to extract data items. • Alternatively, use the scan function to read individual tokens and use the argument what to describe the stream of tokens in your file. The function can convert tokens into data and then assemble the data into records. Discussion Life would be simple and beautiful if all our data files were organized into neat tables with cleanly delimited data. We could read those files using read.table and get on with living. Dream on. You will eventually encounter a funky file format, and your job—no matter how painful—is to read the file contents into R. The read.table and read.csv functions are line-oriented and probably won’t help. However, the readLines and scan functions are useful here because they let you process the individual lines and even tokens of the file. The readLines function is pretty simple. It reads lines from a file and returns them as a list of character strings: > lines <- readLines("input.txt") You can limit the number of lines by using the n parameter, which gives the number of maximum number of lines to be read: > lines <- readLines("input.txt", n=10) # Read 10 lines and stop 86 | Chapter 4: Input and Output The scan function is much richer. It reads one token at a time and handles it according to your instructions. The first argument is either a filename or a connection (more on connections later). The second argument is called what, and it describes the tokens that scan should expect in the input file. The description is cryptic but quite clever: what=numeric(0) Interpret the next token as a number. what=integer(0) Interpret the next token as an integer. what=complex(0) Interpret the next token as complex number. what=character(0) Interpret the next token as a character string. what=logical(0) Interpret the next token as a logical value. The scan function will apply the given pattern repeatedly until all data is read. Suppose your file is simply a sequence of numbers, like this: 2355.09 2246.73 1738.74 1841.01 2027.85 Use what=numeric(0) to say, “My file is a sequence of tokens, each of which is a number”: > singles <- scan("singles.txt", what=numeric(0)) Read 5 items > singles [1] 2355.09 2246.73 1738.74 1841.01 2027.85 A key feature of scan is that the what can be a list containing several token types. The scan function will assume your file is a repeating sequence of those types. Suppose your file contains triplets of data, like this: 15-Oct-87 2439.78 2345.63 16-Oct-87 2396.21 2,207.73 19-Oct-87 2164.16 1677.55 20-Oct-87 2067.47 1,616.21 21-Oct-87 2081.07 1951.76 Use a list to tell scan that it should expect a repeating, three-token sequence: > triples <- scan("triples.txt", what=list(character(0),numeric(0),numeric(0))) Give names to the list elements, and scan will assign those names to the data: > triples <- scan("triples.txt", + what=list(date=character(0), high=numeric(0), low=numeric(0))) Read 5 records > triples $date [1] "15-Oct-87" "16-Oct-87" "19-Oct-87" "20-Oct-87" "21-Oct-87" $high [1] 2439.78 2396.21 2164.16 2067.47 2081.07 4.12 Reading Files with a Complex Structure | 87 $low [1] 2345.63 2207.73 1677.55 1616.21 1951.76 The scan function has many bells and whistles, but the following are especially useful: n=number Stop after reading this many tokens. (Default: stop at end of file.) nlines=number Stop after reading this many input lines. (Default: stop at end of file.) skip=number Number of input lines to skip before reading data. na.strings=list A list of strings to be interpreted as NA. An Example Let’s use this recipe to read a dataset from StatLib, the repository of statistical data and software maintained by Carnegie Mellon University. Jeff Witmer contributed a dataset called wseries that shows the pattern of wins and losses for every World Series since 1903. The dataset is stored in an ASCII file with 35 lines of comments followed by 23 lines of data. The data itself looks like this: 1903 LWLlwwwW 1927 wwWW 1950 wwWW 1973 WLwllWW 1905 wLwWW 1928 WWww 1951 LWlwwW 1974 wlWWW 1906 wLwLwW 1929 wwLWW 1952 lwLWLww 1975 lwWLWlw 1907 WWww 1930 WWllwW 1953 WWllwW 1976 WWww 1908 wWLww 1931 LWwlwLW 1954 WWww 1977 WLwwlW . . (etc.) . The data is encoded as follows: L = loss at home, l = loss on the road, W = win at home, w = win on the road. The data appears in column order, not row order, which com- plicates our lives a bit. Here is the R code for reading the raw data: # Read the wseries dataset: # - Skip the first 35 lines # - Then read 23 lines of data # - The data occurs in pairs: a year and a pattern (char string) # world.series <- scan("http://lib.stat.cmu.edu/datasets/wseries", skip = 35, nlines = 23, what = list(year = integer(0), pattern = character(0)), ) 88 | Chapter 4: Input and Output The scan function returns a list, so we get a list with two elements: year and pattern. The function reads from left to right, but the dataset is organized by columns and so the years appear in a strange order: > world.series$year [1] 1903 1927 1950 1973 1905 1928 1951 1974 1906 1929 1952 [12] 1975 1907 1930 1953 1976 1908 1931 1954 1977 1909 1932 . . (etc.) . We can fix that by sorting the list elements according to year: > perm <- order(world.series$year) > world.series <- list(year = world.series$year[perm], + pattern = world.series$pattern[perm]) Now the data appears in chronological order: > world.series$year [1] 1903 1905 1906 1907 1908 1909 1910 1911 1912 1913 1914 [12] 1915 1916 1917 1918 1919 1920 1921 1922 1923 1924 1925 . . (etc.) . > world.series$pattern [1] "LWLlwwwW" "wLwWW" "wLwLwW" "WWww" "wWLww" [6] "WLwlWlw" "WWwlw" "lWwWlW" "wLwWlLW" "wLwWw" [11] "wwWW" "lwWWw" "WWlwW" "WWllWw" "wlwWLW" [16] "WWlwwLLw" "wllWWWW" "LlWwLwWw" "WWwW" "LwLwWw" . . (etc.) . 4.13 Reading from MySQL Databases Problem You want access to data stored in a MySQL database. Solution 1. Install the RMySQL package on your computer. 2. Open a database connection using the dbConnect function. 3. Use dbGetQuery to initiate a SELECT and return the result sets. 4. Use dbDisconnect to terminate the database connection when you are done. 4.13 Reading from MySQL Databases | 89 Discussion This recipe requires that the RMySQL package be installed on your computer. That pack- age requires, in turn, the MySQL client software. If the MySQL client software is not already installed and configured, consult the MySQL documentation or your system administrator. Use the dbConnect function to establish a connection to the MySQL database. It returns a connection object which is used in subsequent calls to RMySQL functions: library(RMySQL) con <- dbConnect(MySQL(), user="userid", password="pswd", host="hostname", client.flag=CLIENT_MULTI_RESULTS) Setting client.flag=CLIENT_MULTI_RESULTS is necessary to correctly handle multiple re- sult sets. Even if your queries return a single result set, you must set client.flag this way because MySQL might include additional status result sets after your data. The username, password, and host parameters are the same parameters used for ac- cessing MySQL through the mysql client program. The example given here shows them hard-coded into the dbConnect call. Actually, that is an ill-advised practice. It puts your password out in the open, creating a security problem. It also creates a major headache whenever your password or host change, requiring you to hunt down the hard-coded values. I strongly recommend using the security mechanism of MySQL instead. Put those three parameters into your MySQL configuration file, which is $HOME/.my.cnf on Unix and C:\my.cnf on Windows. Make sure the file is unreadable by anyone except you. The file is delimited into sections with markers such as [client]. Put the parameters into the [client] section, so that your config file will contain something like this: [client] user = userid password = password host = hostname Once the parameters are defined in the config file, you no longer need to supply them in the dbConnect call, which then becomes much simpler: con <- dbConnect(MySQL(), client.flag=CLIENT_MULTI_RESULTS) Use the dbGetQuery function to submit your SQL to the database and read the result sets. Doing so requires an open database connection: sql <- "SELECT * from SurveyResults WHERE City = 'Chicago'" rows <- dbGetQuery(con, sql) You will need to construct your own SQL query, of course; this is just an example. You are not restricted to SELECT statements. Any SQL that generates a result set is OK. I generally use CALL statements, for example, because all my SQL is encapsulated in stored procedures and those stored procedures contain embedded SELECT statements. 90 | Chapter 4: Input and Output Using dbGetQuery is convenient because it packages the result set into a data frame and returns the data frame. This is the perfect representation of an SQL result set. The result set is a tabular data structure of rows and columns, and so is a data frame. The result set’s columns have names given by the SQL SELECT statement, and R uses them for naming the columns of the data frame. After the first result set of data, MySQL can return a second result set containing status information. You can choose to inspect the status or ignore it, but you must read it. Otherwise, MySQL will complain that there are unprocessed result sets and then halt. So call dbNextResult if necessary: if (dbMoreResults(con)) dbNextResult(con) Call dbGetQuery repeatedly to perform multiple queries, checking for the result status after each call (and reading it, if necessary). When you are done, close the database connection using dbDisconnect: dbDisconnect(con) Here is a complete session that reads and prints three rows from my database of stock prices. The query selects the price of IBM stock for the last three days of 2008. It assumes that the username, password, and host are defined in the my.cnf file: > con <- dbConnect(MySQL(), client.flag=CLIENT_MULTI_RESULTS) > sql <- paste("select * from DailyBar where Symbol = 'IBM'", + "and Day between '2008-12-29' and '2008-12-31'") > rows <- dbGetQuery(con, sql) > if (dbMoreResults(con)) dbNextResults(con) > print(rows) Symbol Day Next OpenPx HighPx LowPx ClosePx AdjClosePx 1 IBM 2008-12-29 2008-12-30 81.72 81.72 79.68 81.25 81.25 2 IBM 2008-12-30 2008-12-31 81.83 83.64 81.52 83.55 83.55 3 IBM 2008-12-31 2009-01-02 83.50 85.00 83.50 84.16 84.16 HistClosePx Volume OpenInt 1 81.25 6062600 NA 2 83.55 5774400 NA 3 84.16 6667700 NA > dbDisconnect(con) [1] TRUE See Also See Recipe 3.9 and the documentation for RMySQL, which contains more details about configuring and using the package. R can read from several other RDBMS systems, including Oracle, Sybase, PostgreSQL, and SQLite. For more information, see the R Data Import/Export guide, which is sup- plied with the base distribution (Recipe 1.6) and is also available on CRAN at http:// cran.r-project.org/doc/manuals/R-data.pdf. 4.13 Reading from MySQL Databases | 91 4.14 Saving and Transporting Objects Problem You want to store one or more R objects in a file for later use, or you want to copy an R object from one machine to another. Solution Write the objects to a file using the save function: > save(myData, file="myData.RData") Read them back using the load function, either on your computer or on any platform that supports R: > load("myData.RData") The save function writes binary data. To save in an ASCII format, use dput or dump instead: > dput(myData, file="myData.txt") > dump("myData", file="myData.txt") # Note quotes around variable name Discussion I normally save my data in my workspace, but sometimes I need to save data outside my workspace. I may have a large, complicated data object that I want to load into other workspaces, or I may want to move R objects between my Linux box and my Windows box. The load and save functions let me do all this: save will store the object in a file that is portable across machines, and load can read those files. When you run load, it does not return your data per se; rather, it creates variables in your workspace, loads your data into those variables, and then returns the names of the variables (in a list). The first time I used load, I did this: > myData <- load("myFile.RData") # Achtung! Might not do what you think I was extremely puzzled because myData did not contain my data at all and because my variables had mysteriously appeared in my workspace. Eventually, I broke down and read the documentation for load, which explained everything. The save function writes in a binary format to keep the file small. Sometimes you want an ASCII format instead. When you submit a question to a mailing list, for example, including an ASCII dump of the data lets others re-create your problem. In such cases use dput or dump, which write an ASCII representation. Be careful when you save and load objects created by a particular R package. When you load the objects, R does not automatically load the required packages, too, so it will not “understand” the object unless you previously loaded the package yourself. For instance, suppose you have an object called z created by the zoo package, and 92 | Chapter 4: Input and Output suppose we save the object in a file called z.RData. The following sequence of functions will create some confusion: > load("z.RData") # Create and populate the z variable > plot(z) # Does not plot what we expected: zoo pkg not loaded We should have loaded the zoo package before printing or plotting any zoo objects, like this: > library(zoo) # Load the zoo package into memory > load("z.RData") # Create and populate the z variable > plot(z) # Ahhh. Now plotting works correctly 4.14 Saving and Transporting Objects | 93 CHAPTER 5 Data Structures Introduction You can get pretty far in R just using vectors. That’s what Chapter 2 is all about. This chapter moves beyond vectors to recipes for matrices, lists, factors, and data frames. If you have preconceptions about data structures, I suggest you put them aside. R does data structures differently. If you want to study the technical aspects of R’s data structures, I suggest reading R in a Nutshell (O’Reilly) and the R Language Definition. My notes here are more informal. These are things I wish I’d known when I started using R. Vectors Here are some key properties of vectors: Vectors are homogeneous All elements of a vector must have the same type or, in R terminology, the same mode. Vectors can be indexed by position So v[2] refers to the second element of v. Vectors can be indexed by multiple positions, returning a subvector So v[c(2,3)] is a subvector of v that consists of the second and third elements. Vector elements can have names Vectors have a names property, the same length as the vector itself, that gives names to the elements: > v <- c(10, 20, 30) > names(v) <- c("Moe", "Larry", "Curly") > print(v) Moe Larry Curly 10 20 30 95 If vector elements have names then you can select them by name Continuing the previous example: > v["Larry"] Larry 20 Lists Lists are heterogeneous Lists can contain elements of different types; in R terminology, list elements may have different modes. Lists can even contain other structured objects, such as lists and data frames; this allows you to create recursive data structures. Lists can be indexed by position So lst[[2]] refers to the second element of lst. Note the double square brackets. Lists let you extract sublists So lst[c(2,3)] is a sublist of lst that consists of the second and third elements. Note the single square brackets. List elements can have names Both lst[["Moe"]] and lst$Moe refer to the element named “Moe”. Since lists are heterogeneous and since their elements can be retrieved by name, a list is like a dictionary or hash or lookup table in other programming languages (Recipe 5.9). What’s surprising (and cool) is that in R, unlike most of those other programming languages, lists can also be indexed by position. Mode: Physical Type In R, every object has a mode, which indicates how it is stored in memory: as a number, as a character string, as a list of pointers to other objects, as a function, and so forth: Object Example Mode Number 3.1415 numeric Vector of numbers c(2.7.182, 3.1415) numeric Character string "Moe" character Vector of character strings c("Moe", "Larry", "Curly") character Factor factor(c("NY", "CA", "IL")) numeric List list("Moe", "Larry", "Curly") list Data frame data.frame(x=1:3, y=c("NY", "CA", "IL")) list Function print function The mode function gives us this information: 96 | Chapter 5: Data Structures > mode(3.1415) # Mode of a number [1] "numeric" > mode(c(2.7182, 3.1415)) # Mode of a vector of numbers [1] "numeric" > mode("Moe") # Mode of a character string [1] "character" > mode(list("Moe","Larry","Curly")) # Mode of a list [1] "list" A critical difference between a vector and a list can be summed up this way: • In a vector, all elements must have the same mode. • In a list, the elements can have different modes. Class: Abstract Type In R, every object also has a class, which defines its abstract type. The terminology is borrowed from object-oriented programming. A single number could represent many different things: a distance, a point in time, a weight. All those objects have a mode of “numeric” because they are stored as a number; but they could have different classes to indicate their interpretation. For example, a Date object consists of a single number: > d <- as.Date("2010-03-15") > mode(d) [1] "numeric" > length(d) [1] 1 But it has a class of Date, telling us how to interpret that number; namely, as the number of days since January 1, 1970: > class(d) [1] "Date" R uses an object’s class to decide how to process the object. For example, the generic function print has specialized versions (called methods) for printing objects according to their class: data.frame, Date, lm, and so forth. When you print an object, R calls the appropriate print function according to the object’s class. Scalars The quirky thing about scalars is their relationship to vectors. In some software, scalars and vectors are two different things. In R, they are the same thing: a scalar is simply a vector that contains exactly one element. In this book I often use the term “scalar”, but that’s just shorthand for “vector with one element.” Consider the built-in constant pi. It is a scalar: > pi [1] 3.141593 Introduction | 97 Since a scalar is a one-element vector, you can use vector functions on pi: > length(pi) [1] 1 You can index it. The first (and only) element is π, of course: > pi[1] [1] 3.141593 If you ask for the second element, there is none: > pi[2] [1] NA Matrices In R, a matrix is just a vector that has dimensions. It may seem strange at first, but you can transform a vector into a matrix simply by giving it dimensions. A vector has an attribute called dim, which is initially NULL, as shown here: > A <- 1:6 > dim(A) NULL > print(A) [1] 1 2 3 4 5 6 We give dimensions to the vector when we set its dim attribute. Watch what happens when we set our vector dimensions to 2 × 3 and print it: > dim(A) <- c(2,3) > print(A) [,1] [,2] [,3] [1,] 1 3 5 [2,] 2 4 6 Voilà! The vector was reshaped into a 2 × 3 matrix. A matrix can be created from a list, too. Like a vector, a list has a dim attribute, which is initially NULL: > B <- list(1,2,3,4,5,6) > dim(B) NULL If we set the dim attribute, it gives the list a shape: > dim(B) <- c(2,3) > print(B) [,1] [,2] [,3] [1,] 1 3 5 [2,] 2 4 6 Voilà! We have turned this list into a 2 × 3 matrix. 98 | Chapter 5: Data Structures Arrays The discussion of matrices can be generalized to 3-dimensional or even n-dimensional structures: just assign more dimensions to the underlying vector (or list). The following example creates a 3-dimensional array with dimensions 2 × 3 × 2: > D <- 1:12 > dim(D) <- c(2,3,2) > print(D) , , 1 [,1] [,2] [,3] [1,] 1 3 5 [2,] 2 4 6 , , 2 [,1] [,2] [,3] [1,] 7 9 11 [2,] 8 10 12 Note that R prints one “slice” of the structure at a time, since it’s not possible to print a 3-dimensional structure on a 2-dimensional medium. Matrices Made from Lists It strikes me as very odd that we can turn a list into a matrix just by giving the list a dim attribute. But wait; it gets stranger. Recall that a list can be heterogeneous (mixed modes). We can start with a heteroge- neous list, give it dimensions, and thus create a heterogeneous matrix. This code snippet creates a matrix that is a mix of numeric and character data: > C <- list(1, 2, 3, "X", "Y", "Z") > dim(C) <- c(2,3) > print(C) [,1] [,2] [,3] [1,] 1 3 "Y" [2,] 2 "X" "Z" To me this is strange because I ordinarily assume a matrix is purely numeric, not mixed. R is not that restrictive. The possibility of a heterogeneous matrix may seem powerful and strangely fascinating. However, it creates problems when you are doing normal, day-to-day stuff with ma- trices. For example, what happens when the matrix C (above) is used in matrix multi- plication? What happens if it is converted to a data frame? The answer is that odd things happen. In this book, I generally ignore the pathological case of a heterogeneous matrix. I as- sume you’ve got simple, vanilla matrices. Some recipes involving matrices may work oddly (or not at all) if your matrix contains mixed data. Converting such a matrix to a vector or data frame, for instance, can be problematic (Recipe 5.33). Introduction | 99 Factors A factor looks like a vector, but it has special properties. R keeps track of the unique values in a vector, and each unique value is called a level of the associated factor. R uses a compact representation for factors, which makes them efficient for storage in data frames. In other programming languages, a factor would be represented by a vector of enumerated values. There are two key uses for factors: Categorical variables A factor can represent a categorical variable. Categorical variables are used in con- tingency tables, linear regression, analysis of variance (ANOVA), logistic regres- sion, and many other areas. Grouping This is a technique for labeling or tagging your data items according to their group. See the “Introduction” to Chapter 6. Data Frames A data frame is powerful and flexible structure. Most serious R applications involve data frames. A data frame is intended to mimic a dataset, such as one you might en- counter in SAS or SPSS. A data frame is a tabular (rectangular) data structure, which means that it has rows and columns. It is not implemented by a matrix, however. Rather, a data frame is a list: • The elements of the list are vectors and/or factors.* • Those vectors and factors are the columns of the data frame. • The vectors and factors must all have the same length; in other words, all columns must have the same height. • The equal-height columns give a rectangular shape to the data frame. • The columns must have names. Because a data frame is both a list and a rectangular structure, R provides two different paradigms for accessing its contents: • You can use list operators to extract columns from a data frame, such as dfrm[i], dfrm[[i]], or dfrm$name. • You can use matrix-like notation, such as dfrm[i,j], dfrm[i,], or dfrm[,j]. Your perception of a data frame likely depends on your background: * A data frame can be built from a mixture of vectors, factors, and matrices. The columns of the matrices become columns in the data frame. The number of rows in each matrix must match the length of the vectors and factors. In other words, all elements of a data frame must have the same height. 100 | Chapter 5: Data Structures To a statistician A data frame is a table of observations. Each row contains one observation. Each observation must contain the same variables. These variables are called columns, and you can refer to them by name. You can also refer to the contents by row number and column number, just as with a matrix. To a SQL programmer A data frame is a table. The table resides entirely in memory, but you can save it to a flat file and restore it later. You needn’t declare the column types because R figures that out for you. To an Excel user A data frame is like a worksheet, or perhaps a range within a worksheet. It is more restrictive, however, in that each column has a type. To an SAS user A data frame is like a SAS dataset for which all the data resides in memory. R can read and write the data frame to disk, but the data frame must be in memory while R is processing it. To an R programmer A data frame is a hybrid data structure, part matrix and part list. A column can contain numbers, character strings, or factors but not a mix of them. You can index the data frame just like you index a matrix. The data frame is also a list, where the list elements are the columns, so you can access columns by using list operators. To a computer scientist A data frame is a rectangular data structure. The columns are strongly typed, and each column must be numeric values, character strings, or a factor. Columns must have labels; rows may have labels. The table can be indexed by position, column name, and/or row name. It can also be accessed by list operators, in which case R treats the data frame as a list whose elements are the columns of the data frame. To an executive You can put names and numbers into a data frame. It's easy! A data frame is like a little database. Your staff will enjoy using data frames. 5.1 Appending Data to a Vector Problem You want to append additional data items to a vector. Solution Use the vector constructor (c) to construct a vector with the additional data items: > v <- c(v,newItems) 5.1 Appending Data to a Vector | 101 D o wnload from Wow! eBook For a single item, you can also assign the new item to the next vector element. R will automatically extend the vector: > v[length(v)+1] <- newItem Discussion If you ask me about appending a data item to a vector, I will suggest that maybe you shouldn’t. R works best when you think about entire vectors, not single data items. Are you repeatedly appending items to a vector? If so, then you are probably working inside a loop. That’s OK for small vectors, but for large vectors your program will run slowly. The memory management in R works poorly when you repeatedly extend a vector by one element. Try to replace that loop with vector-level operations. You’ll write less code, and R will run much faster. Nonetheless, one does occasionally need to append data to vectors. My experiments show that the most efficient way is to create a new vector using the vector constructor (c) to join the old and new data. This works for appending single elements or multiple elements: > v <- c(1,2,3) > v <- c(v,4) # Append a single value to v > v [1] 1 2 3 4 > w <- c(5,6,7,8) > v <- c(v,w) # Append an entire vector to v > v [1] 1 2 3 4 5 6 7 8 You can also append an item by assigning it to the position past the end of the vector, as shown in the Solution. In fact, R is very liberal about extending vectors. You can assign to any element and R will expand the vector to accommodate your request: > v <- c(1,2,3) # Create a vector of three elements > v[10] <- 10 # Assign to the 10th element > v # R extends the vector automatically [1] 1 2 3 NA NA NA NA NA NA 10 Note that R did not complain about the out-of-bounds subscript. It just extended the vector to the needed length, filling with NA. R includes an append function that creates a new vector by appending items to an ex- isting vector. However, my experiments show that this function runs more slowly than both the vector constructor and the element assignment. 102 | Chapter 5: Data Structures 5.2 Inserting Data into a Vector Problem You want to insert one or more data items into a vector. Solution Despite its name, the append function inserts data into a vector by using the after parameter, which gives the insertion point for the new item or items: > append(vec, newvalues, after=n) Discussion The new items will be inserted at the position given by after. This example inserts 99 into the middle of a sequence: > append(1:10, 99, after=5) [1] 1 2 3 4 5 99 6 7 8 9 10 The special value of after=0 means insert the new items at the head of the vector: > append(1:10, 99, after=0) [1] 99 1 2 3 4 5 6 7 8 9 10 The comments in Recipe 5.1 apply here, too. If you are inserting single items into a vector, you might be working at the element level when working at the vector level would be easier to code and faster to run. 5.3 Understanding the Recycling Rule Problem You want to understand the mysterious Recycling Rule that governs how R handles vectors of unequal length. Discussion When you do vector arithmetic, R performs element-by-element operations. That works well when both vectors have the same length: R pairs the elements of the vectors and applies the operation to those pairs. But what happens when the vectors have unequal lengths? In that case, R invokes the Recycling Rule. It processes the vector element in pairs, starting at the first elements of both vectors. At a certain point, the shorter vector is exhausted while the longer vector still has unprocessed elements. R returns to the be- ginning of the shorter vector, “recycling” its elements; continues taking elements from 5.3 Understanding the Recycling Rule | 103 the longer vector; and completes the operation. It will recycle the shorter-vector ele- ments as often as necessary until the operation is complete. It’s useful to visualize the Recycling Rule. Here is a diagram of two vectors, 1:6 and 1:3: 1:6 1:3 1 1 2 2 3 3 4 5 6 Obviously, the 1:6 vector is longer than the 1:3 vector. If we try to add the vectors using (1:6) + (1:3), it appears that 1:3 has too few elements. However, R recycles the elements of 1:3, pairing the two vectors like this and producing a six-element vector: 1:6 1:3 (1:6) + (1:3) 1 1 2 2 2 4 3 3 6 4 1 5 5 2 7 6 3 9 Here is what you see at the command line: > (1:6) + (1:3) [1] 2 4 6 5 7 9 It’s not only vector operations that invoke the Recycling Rule; functions can, too. The cbind function can create column vectors, such as the following column vectors of 1:6 and 1:3. The two column have different heights, of course: > cbind(1:6) [,1] [1,] 1 [2,] 2 [3,] 3 [4,] 4 [5,] 5 [6,] 6 104 | Chapter 5: Data Structures > cbind(1:3) [,1] [1,] 1 [2,] 2 [3,] 3 If we try binding these column vectors together into a two-column matrix, the lengths are mismatched. The 1:3 vector is too short, so cbind invokes the Recycling Rule and recycles the elements of 1:3: > cbind(1:6, 1:3) [,1] [,2] [1,] 1 1 [2,] 2 2 [3,] 3 3 [4,] 4 1 [5,] 5 2 [6,] 6 3 If the longer vector’s length is not a multiple of the shorter vector’s length, R gives a warning. That’s good, since the operation is highly suspect and there is likely a bug in your logic: > (1:6) + (1:5) # Oops! 1:5 is one element too short [1] 2 4 6 8 10 7 Warning message: In (1:6) + (1:5) : longer object length is not a multiple of shorter object length Once you understand the Recycling Rule, you will realize that operations between a vector and a scalar are simply applications of that rule. In this example, the 10 is recycled repeatedly until the vector addition is complete: > (1:6) + 10 [1] 11 12 13 14 15 16 5.4 Creating a Factor (Categorical Variable) Problem You have a vector of character strings or integers. You want R to treat them as a factor, which is R’s term for a categorical variable. Solution The factor function encodes your vector of discrete values into a factor: > f <- factor(v) # v is a vector of strings or integers If your vector contains only a subset of possible values and not the entire universe, then include a second argument that gives the possible levels of the factor: > f <- factor(v, levels) 5.4 Creating a Factor (Categorical Variable) | 105 Discussion In R, each possible value of a categorical variable is called a level. A vector of levels is called a factor. Factors fit very cleanly into the vector orientation of R, and they are used in powerful ways for processing data and building statistical models. Most of the time, converting your categorical data into a factor is a simple matter of calling the factor function, which identifies the distinct levels of the categorical data and packs them into a factor: > f <- factor(c("Win","Win","Lose","Tie","Win","Lose")) > f [1] Win Win Lose Tie Win Lose Levels: Lose Tie Win Notice that when we printed the factor, f, R did not put quotes around the values. They are levels, not strings. Also notice that when we printed the factor, R also displayed the distinct levels below the factor. If your vector contains only a subset of all the possible levels, then R will have an incomplete picture of the possible levels. Suppose you have a string-valued variable wday that gives the day of the week on which your data was observed: > f <- factor(wday) > f [1] Wed Thu Mon Wed Thu Thu Thu Tue Thu Tue Levels: Mon Thu Tue Wed R thinks that Monday, Thursday, Tuesday, and Wednesday are the only possible levels. Friday is not listed. Apparently, the lab staff never made observations on Friday, so R does not know that Friday is a possible value. Hence you need to list the possible levels of wday explicitly: > f <- factor(wday, c("Mon","Tue","Wed","Thu","Fri")) > f [1] Wed Thu Mon Wed Thu Thu Thu Tue Thu Tue Levels: Mon Tue Wed Thu Fri Now R understands that f is a factor with five possible levels. It knows their correct order, too. It originally put Thursday before Tuesday because it assumes alphabetical order by default.† The explicit second argument defines the correct order. In many situations it is not necessary to call factor explicitly. When an R function requires a factor, it usually converts your data to a factor automatically. The table function, for instance, works only on factors, so it routinely converts its inputs to factors without asking. You must explicitly create a factor variable when you want to specify the full set of levels or when you want to control the ordering of levels. † More precisely, it orders the names according to your Locale. 106 | Chapter 5: Data Structures See Also See Recipe 12.6 to create a factor from continuous data. 5.5 Combining Multiple Vectors into One Vector and a Factor Problem You have several groups of data, with one vector for each group. You want to combine the vectors into one large vector and simultaneously create a parallel factor that iden- tifies each value’s original group. Solution Create a list that contains the vectors. Use the stack function to combine the list into a two-column data frame: > comb <- stack(list(v1=v1, v2=v2, v3=v3)) # Combine 3 vectors The data frame’s columns are called values and ind. The first column contains the data, and the second column contains the parallel factor. Discussion Why in the world would you want to mash all your data into one big vector and a parallel factor? The reason is that many important statistical functions require the data in that format. Suppose you survey freshmen, sophomores, and juniors regarding their confidence level (“What percentage of the time do you feel confident in school?”). Now you have three vectors, called freshmen, sophomores, and juniors. You want to perform an ANOVA analysis of the differences between the groups. The ANOVA function, aov, requires one vector with the survey results as well as a parallel factor that identifies the group. You can combine the groups using the stack function: > comb <- stack(list(fresh=freshmen, soph=sophomores, jrs=juniors)) > print(comb) values ind 1 0.60 fresh 2 0.35 fresh 3 0.44 fresh 4 0.62 fresh 5 0.60 fresh 6 0.70 soph 7 0.61 soph 8 0.63 soph 9 0.87 soph 10 0.85 soph 11 0.70 soph 12 0.64 soph 5.5 Combining Multiple Vectors into One Vector and a Factor | 107 13 0.76 jrs 14 0.71 jrs 15 0.92 jrs 16 0.87 jrs Now you can perform the ANOVA analysis on the two columns: > aov(values ~ ind, data=comb) When building the list we must provide tags for the list elements (the tags are fresh, soph, and jrs in this example). Those tags are required because stack uses them as the levels of the parallel factor. 5.6 Creating a List Problem You want to create and populate a list. Solution To create a list from individual data items, use the list function: > lst <- list(x, y, z) Discussion Lists can be quite simple, such as this list of three numbers: > lst <- list(0.5, 0.841, 0.977) > lst [[1]] [1] 0.5 [[2]] [1] 0.841 [[3]] [1] 0.977 When R prints the list, it identifies each list element by its position ([[1]], [[2]], [[3]]) and prints the element’s value (e.g., [1] 0.5) under its position. More usefully, lists can—unlike vectors—contain elements of different modes (types). Here is an extreme example of a mongrel created from a scalar, a character string, a vector, and a function: > lst <- list(3.14, "Moe", c(1,1,2,3), mean) > lst [[1]] [1] 3.14 [[2]] 108 | Chapter 5: Data Structures [1] "Moe" [[3]] [1] 1 1 2 3 [[4]] function (x, ...) UseMethod("mean") You can also build a list by creating an empty list and populating it. Here is our “mon- grel” example built in that way: > lst <- list() > lst[[1]] <- 3.14 > lst[[2]] <- "Moe" > lst[[3]] <- c(1,1,2,3) > lst[[4]] <- mean List elements can be named. The list function lets you supply a name for every element: > lst <- list(mid=0.5, right=0.841, far.right=0.977) > lst $mid [1] 0.5 $right [1] 0.841 $far.right [1] 0.977 See Also See the “Introduction” to this chapter for more about lists; see Recipe 5.9 for more about building and using lists with named elements. 5.7 Selecting List Elements by Position Problem You want to access list elements by position. Solution Use one of these ways. Here, lst is a list variable: lst[[n]] Select the nth element from the list. lst[c(n1, n2, ..., nk)] Returns a list of elements, selected by their positions. 5.7 Selecting List Elements by Position | 109 Note that the first form returns a single element and the second returns a list. Discussion Suppose we have a list of four integers, called years: > years <- list(1960, 1964, 1976, 1994) > years [[1]] [1] 1960 [[2]] [1] 1964 [[3]] [1] 1976 [[4]] [1] 1994 We can access single elements using the double-square-bracket syntax: > years[[1]] [1] 1960 We can extract sublists using the single-square-bracket syntax: > years[c(1,2)] [[1]] [1] 1960 [[2]] [1] 1964 This syntax can be confusing because of a subtlety: there is an important difference between lst[[n]] and lst[n]. They are not the same thing: lst[[n]] This is an element, not a list. It is the nth element of lst. lst[n] This is a list, not an element. The list contains one element, taken from the nth element of lst. This is a special case of lst[c(n1, n2, ..., nk)] in which we elim- inated the c(...) construct because there is only one n. The difference becomes apparent when we inspect the structure of the result—one is a number; the other is a list: > class(years[[1]]) [1] "numeric" > class(years[1]) [1] "list" 110 | Chapter 5: Data Structures The difference becomes annoyingly apparent when we cat the value. Recall that cat can print atomic values or vectors but complains about printing structured objects: > cat(years[[1]], "\n") 1960 > cat(years[1], "\n") Error in cat(list(...), file, sep, fill, labels, append) : argument 1 (type 'list') cannot be handled by 'cat' We got lucky here because R alerted us to the problem. In other contexts, you might work long and hard to figure out that you accessed a sublist when you wanted an element, or vice versa. 5.8 Selecting List Elements by Name Problem You want to access list elements by their names. Solution Use one of these forms. Here, lst is a list variable: lst[["name"]] Selects the element called name. Returns NULL if no element has that name. lst$name Same as previous, just different syntax. lst[c(name1, name2, ..., namek)] Returns a list built from the indicated elements of lst. Note that the first two forms return an element whereas the third form returns a list. Discussion Each element of a list can have a name. If named, the element can be selected by its name. This assignment creates a list of four named integers: > years <- list(Kennedy=1960, Johnson=1964, Carter=1976, Clinton=1994) These next two expressions return the same value—namely, the element that is named “Kennedy”: > years[["Kennedy"]] [1] 1960 > years$Kennedy [1] 1960 The following two expressions return sublists extracted from years: > years[c("Kennedy","Johnson")] $Kennedy 5.8 Selecting List Elements by Name | 111 [1] 1960 $Johnson [1] 1964 > years["Carter"] $Carter [1] 1976 Just as with selecting list elements by position (Recipe 5.7), there is an important dif- ference between lst[["name"]] and lst["name"]. They are not the same: lst[["name"]] This is an element, not a list. lst["name"] This is a list, not an element. This is a special case of lst[c(name1, name2, ..., namek)] in which we don’t need the c(...) construct because there is only one name. See Also See Recipe 5.7 to access elements by position rather than by name. 5.9 Building a Name/Value Association List Problem You want to create a list that associates names and values—as would a dictionary, hash, or lookup table in another programming language. Solution The list function lets you give names to elements, creating an association between each name and its value: > lst <- list(mid=0.5, right=0.841, far.right=0.977) If you have parallel vectors of names and values, you can create an empty list and then populate the list by using a vectorized assignment statement: > lst <- list() > lst[names] <- values Discussion Each element of a list can be named, and you can retrieve list elements by name. This gives you a basic programming tool: the ability to associate names with values. You can assign element names when you build the list. The list function allows ar- guments of the form name=value: 112 | Chapter 5: Data Structures > lst <- list( + far.left=0.023, + left=0.159, + mid=0.500, + right=0.841, + far.right=0.977) > lst $far.left [1] 0.023 $left [1] 0.159 $mid [1] 0.5 $right [1] 0.841 $far.right [1] 0.977 One way to name the elements is to create an empty list and then populate it via assignment statements: > lst <- list() > lst$far.left <- 0.023 > lst$left <- 0.159 > lst$mid <- 0.500 > lst$right <- 0.841 > lst$far.right <- 0.977 Sometimes you have a vector of names and a vector of corresponding values: > values <- pnorm(-2:2) > names <- c("far.left", "left", "mid", "right", "far.right") You can associate the names and the values by creating an empty list and then popu- lating it with a vectorized assignment statement: > lst <- list() > lst[names] <- values > lst $far.left [1] 0.02275013 $left [1] 0.1586553 $mid [1] 0.5 $right [1] 0.8413447 $far.right [1] 0.9772499 5.9 Building a Name/Value Association List | 113 Once the association is made, the list can “translate” names into values through a simple list lookup: > cat("The left limit is", lst[["left"]], "\n") The left limit is 0.1586553 > cat("The right limit is", lst[["right"]], "\n") The right limit is 0.8413447 > for (nm in names(lst)) cat("The", nm, "limit is", lst[[nm]], "\n") The far.left limit is 0.02275013 The left limit is 0.1586553 The mid limit is 0.5 The right limit is 0.8413447 The far.right limit is 0.9772499 5.10 Removing an Element from a List Problem You want to remove an element from a list. Solution Assign NULL to the element. R will remove it from the list. Discussion To remove a list element, select it by position or by name, and then assign NULL to the selected element: > years $Kennedy [1] 1960 $Johnson [1] 1964 $Carter [1] 1976 $Clinton [1] 1994 > years[["Johnson"]] <- NULL # Remove the element labeled "Johnson" > years $Kennedy [1] 1960 $Carter [1] 1976 $Clinton [1] 1994 114 | Chapter 5: Data Structures You can remove multiple elements this way, too: > years[c("Carter","Clinton")] <- NULL # Remove two elements > years $Kennedy [1] 1960 5.11 Flatten a List into a Vector Problem You want to flatten all the elements of a list into a vector. Solution Use the unlist function. Discussion There are many contexts that require a vector. Basic statistical functions work on vec- tors but not on lists, for example. If iq.scores is a list of numbers, then we cannot directly compute their mean: > mean(iq.scores) [1] NA Warning message: In mean.default(iq.scores) : argument is not numeric or logical: returning NA Instead, we must flatten the list into a vector using unlist and then compute the mean of the result: > mean(unlist(iq.scores)) [1] 106.4452 Here is another example. We can cat scalars and vectors, but we cannot cat a list: > cat(iq.scores, "\n") Error in cat(list(...), file, sep, fill, labels, append) : argument 1 (type 'list') cannot be handled by 'cat' One solution is to flatten the list into a vector before printing: > cat("IQ Scores:", unlist(iq.scores), "\n") IQ Scores: 89.73383 116.5565 113.0454 See Also Conversions such as this are discussed more fully in Recipe 5.33. 5.11 Flatten a List into a Vector | 115 5.12 Removing NULL Elements from a List Problem Your list contains NULL values. You want to remove them. Solution Suppose lst is a list some of whose elements are NULL. This expression will remove the NULL elements: > lst[sapply(lst, is.null)] <- NULL Discussion Finding and removing NULL elements from a list is surprisingly tricky. I wrote the fol- lowing expression after trying several other ways, including the obvious ones, and fail- ing. Here’s how it works: 1. R calls sapply to apply the is.null function to every element of the list. 2. sapply returns a vector of logical values that are TRUE wherever the corresponding list element is NULL. 3. R selects values from the list according to that vector. 4. R assigns NULL to the selected items, removing them from the list. The curious reader may be wondering how a list can contain NULL elements, given that we remove elements by setting them to NULL (Recipe 5.10). The answer is that we can create a list containing NULL elements: > lst <- list("Moe", NULL, "Curly") # Create list with NULL element > lst [[1]] [1] "Moe" [[2]] NULL [[3]] [1] "Curly" > lst[sapply(lst, is.null)] <- NULL # Remove NULL element from list > lst [[1]] [1] "Moe" [[2]] [1] "Curly" 116 | Chapter 5: Data Structures See Also See Recipe 5.10 for how to remove list elements. 5.13 Removing List Elements Using a Condition Problem You want to remove elements from a list according to a conditional test, such as re- moving elements that are negative or smaller than some threshold. Solution Build a logical vector based on the condition. Use the vector to select list elements and then assign NULL to those elements. This assignment, for example, removes all negative value from lst: > lst[lst < 0] <- NULL Discussion This recipe is based on two useful features of R. First, a list can be indexed by a logical vector. Wherever the vector element is TRUE, the corresponding list element is selected. Second, you can remove a list element by assigning NULL to it. Suppose we want to remove elements from lst whose value is zero. We construct a logical vector which identifies the unwanted values (lst == 0). Then we select those elements from the list and assign NULL to them: > lst[lst == 0] <- NULL This expression will remove NA values from the list: > lst[is.na(lst)] <- NULL So far, so good. The problems arise when you cannot easily build the logical vector. That often happens when you want to use a function that cannot handle a list. Suppose you want to remove list elements whose absolute value is less than 1. The abs function will not handle a list, unfortunately: > lst[abs(lst) < 1] <- NULL Error in abs(lst) : non-numeric argument to function The simplest solution is flattening the list into a vector by calling unlist and then testing the vector: > lst[abs(unlist(lst)) < 1] <- NULL A more elegant solution uses lapply to apply the function to every element of the list: > lst[lapply(lst,abs) < 1] <- NULL 5.13 Removing List Elements Using a Condition | 117 Lists can hold complex objects, too, not just atomic values. Suppose that mods is a list of linear models created by the lm function. This expression will remove any model whose R2 value is less than 0.30: > mods[sapply(mods, function(m) summary(m)$r.squared < 0.3)] <- NULL See Also See Recipes 5.7, 5.10, 5.11, 6.2, and 11.1. 5.14 Initializing a Matrix Problem You want to create a matrix and initialize it from given values. Solution Capture the data in a vector or list, and then use the matrix function to shape the data into a matrix. This example shapes a vector into a 2 × 3 matrix (i.e., two rows and three columns): > matrix(vec, 2, 3) Discussion Suppose we want to create and initialize a 2 × 3 matrix. We can capture the initial data inside a vector and then shape it using the matrix function: > theData <- c(1.1, 1.2, 2.1, 2.2, 3.1, 3.2) > mat <- matrix(theData, 2, 3) > mat [,1] [,2] [,3] [1,] 1.1 2.1 3.1 [2,] 1.2 2.2 3.2 The first argument of matrix is the data, the second argument is the number of rows, and the third argument is the number of columns. Observe that the matrix was filled column by column, not row by row. It’s common to initialize an entire matrix to one value such as zero or NA. If the first argument of matrix is a single value, then R will apply the Recycling Rule and auto- matically replicate the value to fill the entire matrix: > matrix(0, 2, 3) # Create an all-zeros matrix [,1] [,2] [,3] [1,] 0 0 0 [2,] 0 0 0 > matrix(NA, 2, 3) # Create a matrix populated with NA [,1] [,2] [,3] 118 | Chapter 5: Data Structures [1,] NA NA NA [2,] NA NA NA You can create a matrix with a one-liner, of course, but it becomes difficult to read: > mat <- matrix(c(1.1, 1.2, 1.3, 2.1, 2.2, 2.3), 2, 3) A common idiom in R is typing the data itself in a rectangular shape that reveals the matrix structure: > theData <- c(1.1, 1.2, 1.3, + 2.1, 2.2, 2.3) > mat <- matrix(theData, 2, 3, byrow=TRUE) Setting byrow=TRUE tells matrix that the data is row-by-row and not column-by-column (which is the default). In condensed form, that becomes: > mat <- matrix(c(1.1, 1.2, 1.3, + 2.1, 2.2, 2.3), + 2, 3, byrow=TRUE) Expressed this way, the reader quickly sees the two rows and three columns of data. There is a quick-and-dirty way to turn a vector into a matrix: just assign dimensions to the vector. This was discussed in the “Introduction”. The following example creates a vanilla vector and then shapes it into a 2 × 3 matrix: > v <- c(1.1, 1.2, 1.3, 2.1, 2.2, 2.3) > dim(v) <- c(2,3) > v [,1] [,2] [,3] [1,] 1.1 1.3 2.2 [2,] 1.2 2.1 2.3 Personally, I find this more opaque than using matrix, especially since there is no byrow option here. See Also See Recipe 5.3. 5.15 Performing Matrix Operations Problem You want to perform matrix operations such as transpose, matrix inversion, matrix multiplication, or constructing an identity matrix. Solution t(A) Matrix transposition of A 5.15 Performing Matrix Operations | 119 solve(A) Matrix inverse of A A %*% B Matrix multiplication of A and B diag(n) An n-by-n diagonal (identity) matrix Discussion Recall that A*B is element-wise multiplication whereas A %*% B is matrix multiplication. All these functions return a matrix. Their arguments can be either matrices or data frames. If they are data frames then R will first convert them to matrices (although this is useful only if the data frame contains exclusively numeric values). 5.16 Giving Descriptive Names to the Rows and Columns of a Matrix Problem You want to assign descriptive names to the rows or columns of a matrix. Solution Every matrix has a rownames attribute and a colnames attribute. Assign a vector of char- acter strings to the appropriate attribute: > rownames(mat) <- c("rowname1", "rowname2", ..., "rownamem") > colnames(mat) <- c("colname1", "colname2", ..., "colnamen") Discussion R lets you assign names to the rows and columns of a matrix, which is useful for printing the matrix. R will display the names if they are defined, enhancing the readability of your output. Consider this matrix of correlations between the prices of IBM, Microsoft, and Google stock: > print(tech.corr) [,1] [,2] [,3] [1,] 1.000 0.556 0.390 [2,] 0.556 1.000 0.444 [3,] 0.390 0.444 1.000 In this form, the matrix output’s interpretation is not self-evident. Yet if we define names for the rows and columns, then R will annotate the matrix output with the names: 120 | Chapter 5: Data Structures > colnames(tech.corr) <- c("IBM","MSFT","GOOG") > rownames(tech.corr) <- c("IBM","MSFT","GOOG") > print(tech.corr) IBM MSFT GOOG IBM 1.000 0.556 0.390 MSFT 0.556 1.000 0.444 GOOG 0.390 0.444 1.000 Now the reader knows at a glance which rows and columns apply to which stocks. Another advantage of naming rows and columns is that you can refer to matrix elements by those names: > tech.corr["IBM","GOOG"] # What is the correlation between IBM and GOOG? [1] 0.39 5.17 Selecting One Row or Column from a Matrix Problem You want to select a single row or a single column from a matrix. Solution The solution depends on what you want. If you want the result to be a simple vector, just use normal indexing: > vec <- mat[1,] # First row > vec <- mat[,3] # Third column If you want the result to be a one-row matrix or a one-column matrix, then include the drop=FALSE argument: > row <- mat[1,,drop=FALSE] # First row in a one-row matrix > col <- mat[,3,drop=FALSE] # Third column in a one-column matrix Discussion Normally, when you select one row or column from a matrix, R strips off the dimen- sions. The result is a dimensionless vector: > mat[1,] [1] 1 4 7 10 > mat[,3] [1] 7 8 9 When you include the drop=FALSE argument, however, R retains the dimensions. In that case, selecting a row returns a row vector (a 1 × n matrix): > mat[1,,drop=FALSE] [,1] [,2] [,3] [,4] [1,] 1 4 7 10 5.17 Selecting One Row or Column from a Matrix | 121 Likewise, selecting a column with drop=FALSE returns a column vector (an n × 1 matrix): > mat[,3,drop=FALSE] [,1] [1,] 7 [2,] 8 [3,] 9 5.18 Initializing a Data Frame from Column Data Problem Your data is organized by columns, and you want to assemble it into a data frame. Solution If your data is captured in several vectors and/or factors, use the data.frame function to assemble them into a data frame: > dfrm <- data.frame(v1, v2, v3, f1, f2) If your data is captured in a list that contains vectors and/or factors, use instead as.data.frame: > dfrm <- as.data.frame(list.of.vectors) Discussion A data frame is a collection of columns, each of which corresponds to an observed variable (in the statistical sense, not the programming sense). If your data is already organized into columns, then it’s easy to build a data frame. The data.frame function can construct a data frame from vectors, where each vector is one observed variable. Suppose you have two numeric predictor variables, one cate- gorical predictor variable, and one response variable. The data.frame function can cre- ate a data frame from your vectors: > dfrm <- data.frame(pred1, pred2, pred3, resp) > dfrm pred1 pred2 pred3 resp 1 -2.7528917 -1.40784130 AM 12.57715 2 -0.3626909 0.31286963 AM 21.02418 3 -1.0416039 -0.69685664 PM 18.94694 4 1.2666820 -1.27511434 PM 18.98153 5 0.7806372 -0.27292745 AM 19.59455 6 -1.0832624 0.73383339 AM 20.71605 7 -2.0883305 0.96816822 PM 22.70062 8 -0.7063653 -0.84476203 PM 18.40691 9 -0.8394022 0.31530793 PM 21.00930 10 -0.4966884 -0.08030948 AM 19.31253 122 | Chapter 5: Data Structures Notice that data.frame takes the column names from your program variables. You can override that default by supplying explicit column names: > dfrm <- data.frame(p1=pred1, p2=pred2, p3=pred3, r=resp) > dfrm p1 p2 p3 r 1 -2.7528917 -1.40784130 AM 12.57715 2 -0.3626909 0.31286963 AM 21.02418 3 -1.0416039 -0.69685664 PM 18.94694 . . (etc.) . Alternatively, your data may be organized into vectors but those vectors are held in a list, not individual program variables, like this: > lst <- list(p1=pred1, p2=pred2, p3=pred3, r=resp) No problem. Use the as.data.frame function to create a data frame from the list of vectors: > as.data.frame(lst) p1 p2 p3 r 1 -2.7528917 -1.40784130 AM 12.57715 2 -0.3626909 0.31286963 AM 21.02418 3 -1.0416039 -0.69685664 PM 18.94694 . . (etc.) . 5.19 Initializing a Data Frame from Row Data Problem Your data is organized by rows, and you want to assemble it into a data frame. Solution Store each row in a one-row data frame. Store the one-row data frames in a list. Use rbind and do.call to bind the rows into one, large data frame: > dfrm <- do.call(rbind, obs) Here, obs is a list of one-row data frames. Discussion Data often arrives as a collection of observations. Each observation is a record or tuple that contains several values, one for each observed variable. The lines of a flat file are usually like that: each line is one record, each record contains several columns, and each column is a different variable (see Recipe 4.12). Such data is organized by 5.19 Initializing a Data Frame from Row Data | 123 observation, not by variable. In other words, you are given rows one at a time rather than columns one at a time. Each such row might be stored in several ways. One obvious way is as a vector. If you have purely numerical data, use a vector. However, many datasets are a mixture of numeric, character, and categorical data, in which case a vector won’t work. I recommend storing each such heterogeneous row in a one-row data frame. (You could store each row in a list, but this recipe gets a little more complicated.) For concreteness, let’s assume that you have ten rows with four variables per observa- tion: pred1, pred2, pred2, and resp. Each row is stored in a one-row data frame, so you have ten such data frames. Those data frames are stored in a list called obs. The first element of obs might look like this: > obs[[1]] pred1 pred2 pred3 resp 1 -1.197 0.36 AM 18.701 This recipe works also if your observations are stored in vectors rather than one-row data frames. We need to bind together those rows into a data frame. That’s what the rbind function does. It binds its arguments in such a way that each argument becomes one row in the result. If we rbind the first two observations, for example, we get a two-row data frame: > rbind(obs[[1]], obs[[2]]) pred1 pred2 pred3 resp 1 -1.197 0.36 AM 18.701 2 -0.952 1.23 PM 25.709 We want to bind together every observation, not just the first two, so we tap into the vector processing of R. The do.call function will expand obs into one, long argument list and call rbind with that long argument list: > do.call(rbind,obs) pred1 pred2 pred3 resp 1 -1.197 0.360 AM 18.701 2 -0.952 1.230 PM 25.709 3 0.279 0.423 PM 21.572 4 -1.445 -1.846 AM 14.392 5 0.822 -0.246 AM 19.841 6 1.247 1.254 PM 25.637 7 -0.394 1.563 AM 24.585 8 -1.248 -1.264 PM 16.770 9 -0.652 -2.344 PM 14.915 10 -1.171 -0.776 PM 17.948 The result is a data frame built from our rows of data. Sometimes, for reasons beyond your control, the rows of your data are stored in lists rather than one-row data frames. You may be dealing with rows returned by a database package, for example. In that case, obs will be a list of lists, not a list of data frames. 124 | Chapter 5: Data Structures We first transform the rows into data frames using the Map function and then apply this recipe: > dfrm <- do.call(rbind,Map(as.data.frame,obs)) See Also See Recipe 5.18 if your data is organized by columns, not rows; see Recipe 12.18 to learn more about do.call. 5.20 Appending Rows to a Data Frame Problem You want to append one or more new rows to a data frame. Solution Create a second, temporary data frame containing the new rows. Then use the rbind function to append the temporary data frame to the original data frame. Discussion Suppose we want to append a new row to our data frame of Chicago-area cities. First, we create a one-row data frame with the new data: > newRow <- data.frame(city="West Dundee", county="Kane", state="IL", pop=5428) Next, we use the rbind function to append that one-row data frame to our existing data frame: > suburbs <- rbind(suburbs, newRow) > suburbs city county state pop 1 Chicago Cook IL 2853114 2 Kenosha Kenosha WI 90352 3 Aurora Kane IL 171782 4 Elgin Kane IL 94487 5 Gary Lake(IN) IN 102746 6 Joliet Kendall IL 106221 7 Naperville DuPage IL 147779 8 Arlington Heights Cook IL 76031 9 Bolingbrook Will IL 70834 10 Cicero Cook IL 72616 11 Evanston Cook IL 74239 12 Hammond Lake(IN) IN 83048 13 Palatine Cook IL 67232 14 Schaumburg Cook IL 75386 15 Skokie Cook IL 63348 16 Waukegan Lake(IL) IL 91452 17 West Dundee Kane IL 5428 5.20 Appending Rows to a Data Frame | 125 The rbind function tells R that we are appending a new row to suburbs, not a new column. It may be obvious to you that newRow is a row and not a column, but it is not obvious to R. (Use the cbind function to append a column.) One word of caution. The new row must use the same column names as the data frame. Otherwise, rbind will fail. We can combine these two steps into one, of course: > suburbs <- rbind(suburbs, + data.frame(city="West Dundee", county="Kane", state="IL", pop=5428)) We can even extend this technique to multiple new rows because rbind allows multiple arguments: > suburbs <- rbind(suburbs, + data.frame(city="West Dundee", county="Kane", state="IL", pop=5428), + data.frame(city="East Dundee", county="Kane", state="IL", pop=2955)) Do not use this recipe to append many rows to a large data frame. That would force R to reallocate a large data structure repeatedly, which is a very slow process. Build your data frame using more efficient means, such as those in Recipes 5.19 or 5.21. 5.21 Preallocating a Data Frame Problem You are building a data frame, row by row. You want to preallocate the space instead of appending rows incrementally. Solution Create a data frame from generic vectors and factors using the functions numeric(n), character(n), and factor(n): > dfrm <- data.frame(colname1=numeric(n), colname2=character(n), ... etc. ... ) Here, n is the number of rows needed for the data frame. Discussion Theoretically, you can build a data frame by appending new rows, one by one. That’s OK for small data frames, but building a large data frame in that way can be tortuous. The memory manager in R works poorly when one new row is repeatedly appended to a large data structure. Hence your R code will run very slowly. One solution is to preallocate the data frame—assuming you know the required num- ber of rows. By preallocating the data frame once and for all, you sidestep problems with the memory manager. 126 | Chapter 5: Data Structures Suppose you want to create a data frame with 1,000,000 rows and three columns: two numeric and one character. Use the numeric and character functions to preallocate the columns; then join them together using data.frame: > N <- 1000000 > dfrm <- data.frame(dosage=numeric(N), lab=character(N), response=numeric(N)) Now you have a data frame with the correct dimensions, 1,000,000 × 3, waiting to receive its contents. Data frames can contain factors, but preallocating a factor is a little trickier. You can’t simply call factor(n). You need to specify the factor’s levels because you are creating it. Continuing our example, suppose you want the lab column to be a factor, not a character string, and that the possible levels are NJ, IL, and CA. Include the levels in the column specification, like this: > N <- 1000000 > dfrm <- data.frame(dosage=numeric(N), + lab=factor(N, levels=c("NJ", "IL", "CA")), + response=numeric(N) ) 5.22 Selecting Data Frame Columns by Position Problem You want to select columns from a data frame according to their position. Solution To select a single column, use this list operator: dfrm[[n]] Returns one column—specifically, the nth column of dfrm. To select one or more columns and package them in a data frame, use the following sublist expressions: dfrm[n] Returns a data frame consisting solely of the nth column of dfrm. dfrm[c(n1, n2, ..., nk)] Returns a data frame built from the columns in positions n1, n2, ..., nk of dfrm. You can use matrix-style subscripting to select one or more columns: dfrm[, n] Returns the nth column (assuming that n contains exactly one value). dfrm[, c(n1, n2, ..., nk)] Returns a data frame built from the columns in positions n1, n2, ..., nk. 5.22 Selecting Data Frame Columns by Position | 127 Note that the matrix-style subscripting can return two different data types (either col- umn or data frame) depending upon whether you select one column or multiple columns. Discussion There are a bewildering number of ways to select columns from a data frame. The choices can be confusing until you understand the logic behind the alternatives. As you read this explanation, notice how a slight change in syntax—a comma here, a double- bracket there—changes the meaning of the expression. Let’s play with the population data for the 16 largest cities in the Chicago metropolitan area: > suburbs city county state pop 1 Chicago Cook IL 2853114 2 Kenosha Kenosha WI 90352 3 Aurora Kane IL 171782 4 Elgin Kane IL 94487 5 Gary Lake(IN) IN 102746 6 Joliet Kendall IL 106221 7 Naperville DuPage IL 147779 8 Arlington Heights Cook IL 76031 9 Bolingbrook Will IL 70834 10 Cicero Cook IL 72616 11 Evanston Cook IL 74239 12 Hammond Lake(IN) IN 83048 13 Palatine Cook IL 67232 14 Schaumburg Cook IL 75386 15 Skokie Cook IL 63348 16 Waukegan Lake(IL) IL 91452 Use simple list notation to select exactly one column, such as the first column: > suburbs[[1]] [1] "Chicago" "Kenosha" "Aurora" "Elgin" [5] "Gary" "Joliet" "Naperville" "Arlington Heights" [9] "Bolingbrook" "Cicero" "Evanston" "Hammond" [13] "Palatine" "Schaumburg" "Skokie" "Waukegan" The first column of suburbs is a vector, so that’s what suburbs[[1]] returns: a vector. If the first column were a factor, we’d get a factor. The result differs when you use the single-bracket notation, as in suburbs[1] or suburbs[c(1,3)]. You still get the requested columns, but R wraps them in a data frame. This example returns the first column wrapped in a data frame: > suburbs[1] city 1 Chicago 2 Kenosha 3 Aurora 4 Elgin 128 | Chapter 5: Data Structures 5 Gary 6 Joliet 7 Naperville 8 Arlington Heights 9 Bolingbrook 10 Cicero 11 Evanston 12 Hammond 13 Palatine 14 Schaumburg 15 Skokie 16 Waukegan The next example returns the first and third columns wrapped in a data frame: > suburbs[c(1,3)] city pop 1 Chicago 2853114 2 Kenosha 90352 3 Aurora 171782 4 Elgin 94487 5 Gary 102746 6 Joliet 106221 7 Naperville 147779 8 Arlington Heights 76031 9 Bolingbrook 70834 10 Cicero 72616 11 Evanston 74239 12 Hammond 83048 13 Palatine 67232 14 Schaumburg 75386 15 Skokie 63348 16 Waukegan 91452 A major source of confusion is that suburbs[[1]] and suburbs[1] look similar but pro- duce very different results: suburbs[[1]] This returns one column. suburbs[1] This returns a data frame, and the data frame contains exactly one column. This is a special case of dfrm[c(n1,n2, ..., nk)]. We don’t need the c(...) construct because there is only one n. The point here is that “one column” is different from “a data frame that contains one column.” The first expression returns a column, so it’s a vector or a factor. The second expression returns a data frame, which is different. R lets you use matrix notation to select columns, as shown in the Solution. But an odd quirk can bite you: you might get a column or you might get a data frame, depending upon many subscripts you use. In the simple case of one index you get a column, like this: 5.22 Selecting Data Frame Columns by Position | 129 > suburbs[,1] [1] "Chicago" "Kenosha" "Aurora" "Elgin" [5] "Gary" "Joliet" "Naperville" "Arlington Heights" [9] "Bolingbrook" "Cicero" "Evanston" "Hammond" [13] "Palatine" "Schaumburg" "Skokie" "Waukegan" But using the same matrix-style syntax with multiple indexes returns a data frame: > suburbs[,c(1,4)] city pop 1 Chicago 2853114 2 Kenosha 90352 3 Aurora 171782 4 Elgin 94487 5 Gary 102746 6 Joliet 106221 7 Naperville 147779 8 Arlington Heights 76031 9 Bolingbrook 70834 10 Cicero 72616 11 Evanston 74239 12 Hammond 83048 13 Palatine 67232 14 Schaumburg 75386 15 Skokie 63348 16 Waukegan 91452 This creates a problem. Suppose you see this expression in some old R script: dfrm[,vec] Quick, does that return a column or a data frame? Well, it depends. If vec contains one value then you get a column; otherwise, you get a data frame. You cannot tell from the syntax alone. To avoid this problem, you can include drop=FALSE in the subscripts; this forces R to return a data frame: dfrm[,vec,drop=FALSE] Now there is no ambiguity about the returned data structure. It’s a data frame. When all is said and done, using matrix notation to select columns from data frames is not the best procedure. I recommend that you instead use the list operators described previously. They just seem clearer. See Also See Recipe 5.17 for more about using drop=FALSE. 130 | Chapter 5: Data Structures 5.23 Selecting Data Frame Columns by Name Problem You want to select columns from a data frame according to their name. Solution To select a single column, use one of these list expressions: dfrm[["name"]] Returns one column, the column called name. dfrm$name Same as previous, just different syntax. To select one or more columns and package them in a data frame, use these list expressions: dfrm["name"] Selects one column and packages it inside a data frame object. dfrm[c("name1", "name2", ..., "namek")] Selects several columns and packages them in a data frame. You can use matrix-style subscripting to select one or more columns: dfrm[, "name"] Returns the named column. dfrm[, c("name1", "name2", ..., "namek")] Selects several columns and packages in a data frame. Once again, the matrix-style subscripting can return two different data types (column or data frame) depending upon whether you select one column or multiple columns. Discussion All columns in a data frame must have names. If you know the name, it’s usually more convenient and readable to select by name, not by position. The solutions just described are similar to those for Recipe 5.22, where we selected columns by position. The only difference is that here we use column names instead of column numbers. All the observations made in Recipe 5.22 apply here: • dfrm[["name"]] returns one column, not a data frame. • dfrm[c("name1", "name2", ..., "namek")] returns a data frame, not a column. • dfrm["name"] is a special case of the previous expression and so returns a data frame, not a column. 5.23 Selecting Data Frame Columns by Name | 131 • The matrix-style subscripting can return either a column or a data frame, so be careful how many names you supply. See Recipe 5.22 for a discussion of this “gotcha” and using drop=FALSE. There is one new addition: dfrm$name This is identical in effect to dfrm[["name"]], but it’s easier to type and to read. See Also See Recipe 5.22 to understand these ways to select columns. 5.24 Selecting Rows and Columns More Easily Problem You want an easier way to select rows and columns from a data frame or matrix. Solution Use the subset function. The select argument is a column name, or a vector of column names, to be selected: > subset(dfrm, select=colname) > subset(dfrm, select=c(colname1, ..., colnameN)) Note that you do not quote the column names. The subset argument is a logical expression that selects rows. Inside the expression, you can refer to the column names as part of the logical expression. In this example, response is a column in the data frame, and we are selecting rows with a positive response: > subset(dfrm, subset=(response > 0)) subset is most useful when you combine the select and subset arguments: > subset(dfrm, select=c(predictor,response), subset=(response > 0)) Discussion Indexing is the “official” way to select rows and columns from a data frame, as described in Recipes 5.22 and 5.23. However, indexing is cumbersome when the index expres- sions become complicated. The subset function provides a more convenient and readable way to select rows and columns. It’s beauty is that you can refer to the columns of the data frame right inside the expressions for selecting columns and rows. 132 | Chapter 5: Data Structures D o wnload from Wow! eBook Here are some examples using the Cars93 dataset in the MASS package. Recall that the dataset includes columns for Manufacturer, Model, MPG.city, MPG.highway, Min.Price, and Max.Price: Select the model name for cars that can exceed 30 miles per gallon (MPG) in the city > subset(Cars93, select=Model, subset=(MPG.city > 30)) Model 31 Festiva 39 Metro 42 Civic . . (etc.) . Select the model name and price range for four-cylinder cars made in the United States > subset(Cars93, select=c(Model,Min.Price,Max.Price), + subset=(Cylinders == 4 & Origin == "USA")) Model Min.Price Max.Price 6 Century 14.2 17.3 12 Cavalier 8.5 18.3 13 Corsica 11.4 11.4 . . (etc.) . Select the manufacturer’s name and the model name for all cars whose highway MPG value is above the median > subset(Cars93, select=c(Manufacturer,Model), + subset=c(MPG.highway > median(MPG.highway))) Manufacturer Model 1 Acura Integra 5 BMW 535i 6 Buick Century . . (etc.) . The subset function is actually more powerful than this recipe implies. It can select from lists and vectors, too. See the help page for details. 5.25 Changing the Names of Data Frame Columns Problem You converted a matrix or list into a data frame. R gave names to the columns, but the names are at best uninformative and at worst bizarre. 5.25 Changing the Names of Data Frame Columns | 133 Solution Data frames have a colnames attribute that is a vector of column names. You can update individual names or the entire vector: > colnames(dfrm) <- newnames # newnames is a vector of character strings Discussion The columns of data frames must have names. If you convert a vanilla matrix into a data frame, R will synthesize names that are reasonable but boring—for example, V1, V2, V3, and so forth: > mat [,1] [,2] [,3] [1,] -0.818 -0.667 -0.494 [2,] -0.819 -0.946 -0.205 [3,] 0.385 1.531 -0.611 [4,] -2.155 -0.535 -0.316 > as.data.frame(mat) V1 V2 V3 1 -0.818 -0.667 -0.494 2 -0.819 -0.946 -0.205 3 0.385 1.531 -0.611 4 -2.155 -0.535 -0.316 If the matrix had column names defined, R would have used those names instead of synthesizing new ones. However, converting a list into a data frame produces some strange synthetic names: > lst [[1]] [1] -0.284 -1.114 -1.097 -0.873 [[2]] [1] -1.673 0.929 0.306 0.778 [[3]] [1] 0.323 0.368 0.067 -0.080 > as.data.frame(lst) c..0.284...1.114...1.097...0.873. c..1.673..0.929..0.306..0.778. 1 -0.284 -1.673 2 -1.114 0.929 3 -1.097 0.306 4 -0.873 0.778 c.0.323..0.368..0.067...0.08. 1 0.323 2 0.368 3 0.067 4 -0.080 Again, if the list elements had names then R would have used them. 134 | Chapter 5: Data Structures Fortunately, you can overwrite the synthetic names with names of your own by setting the colnames attribute: > dfrm <- as.data.frame(lst) > colnames(dfrm) <- c("before","treatment","after") > dfrm before treatment after 1 -0.284 -1.673 0.323 2 -1.114 0.929 0.368 3 -1.097 0.306 0.067 4 -0.873 0.778 -0.080 See Also See Recipe 5.33. 5.26 Editing a Data Frame Problem Data in a data frame are incorrect or missing. You want a convenient way to edit the data frame contents. Solution R includes a data editor that displays your data frame in a spreadsheet-like window. Invoke the editor using the edit function: > temp <- edit(dfrm) > dfrm <- temp # Overwrite only if you're happy with the changes! Make your changes and then close the editor window. The edit function will return the updated data, which here are assigned to temp. If you are happy with the changes, overwrite your data frame with the results. If you are feeling brave, the fix function invokes the editor and overwrites your variable with the result. There is no “undo”, however: > fix(dfrm) Discussion Figure 5-1 is a screenshot of the data editor on Windows during the editing of data from Recipe 5.22. As of this writing, the editor is quite primitive. It does not include the common features of a modern editor—not even an “undo” command, for instance. I cannot recommend the data editor for regular use, but it’s OK for emergency touch-ups. 5.26 Editing a Data Frame | 135 Since there is no undo, take note of the solution that assigns the edited result to a temporary, intermediate variable. If you mess up your data, you can just delete the temporary variable without affecting the original data. See Also Several of the add-on GUI frontends provide data editors that are better than the native editor. 5.27 Removing NAs from a Data Frame Problem Your data frame contains NA values, which is creating problems for you. Solution Use na.omit to remove rows that contain any NA values. > clean <- na.omit(dfrm) Figure 5-1. Editing a data frame 136 | Chapter 5: Data Structures Discussion I frequently stumble upon situations where just a few NA values in my data frame cause everything to fall apart. One solution is simply to remove all rows that contain NAs. That’s what na.omit does. Here we can see cumsum fail because the input contains NA values: > dfrm x y 1 -0.9714511 -0.4578746 2 NA 3.1663282 3 0.3367627 NA 4 1.7520504 0.7406335 5 0.4918786 1.4543427 > cumsum(dfrm) x y 1 -0.971451 -0.4578746 2 NA 2.7084536 3 NA NA 4 NA NA 5 NA NA If we remove the NA values, cumsum can complete its summations: > cumsum(na.omit(dfrm)) x y 1 -0.9714511 -0.4578746 4 0.7805993 0.2827589 5 1.2724779 1.7371016 This recipe works for vectors and matrices, too, but not for lists. Will You Still Have Enough Data? The obvious danger here is that simply dropping observations from your data could render the results computationally or statistically meaningless. Make sure that omitting data makes sense in your context. Remember that na.omit will remove entire rows, not just the NA values, which could eliminate a lot of useful information. 5.28 Excluding Columns by Name Problem You want to exclude a column from a data frame using its name. Solution Use the subset function with a negated argument for the select parameter: > subset(dfrm, select = -badboy) # All columns except badboy 5.28 Excluding Columns by Name | 137 Discussion We can exclude a column by position (e.g., dfrm[-1]), but how do we exclude a column by name? The subset function can exclude columns from a data frame. The select parameter is a normally a list of columns to include, but prefixing a minus sign (-) to the name causes the column to be excluded instead. I often encounter this problem when calculating the correlation matrix of a data frame and I want to exclude nondata columns such as labels: > cor(patient.data) patient.id pre dosage post patient.id 1.00000000 0.02286906 0.3643084 -0.13798149 pre 0.02286906 1.00000000 0.2270821 -0.03269263 dosage 0.36430837 0.22708208 1.0000000 -0.42006280 post -0.13798149 -0.03269263 -0.4200628 1.00000000 This correlation matrix includes the meaningless “correlation” between patient ID and other variables, which is annoying. We can exclude the patient ID column to clean up the output: > cor(subset(patient.data, select = -patient.id)) pre dosage post pre 1.00000000 0.2270821 -0.03269264 dosage 0.22708207 1.0000000 -0.42006280 post -0.03269264 -0.4200628 1.00000000 We can exclude multiple columns by giving a vector of negated names: > cor(subset(patient.data, select = c(-patient.id,-dosage))) pre post pre 1.00000000 -0.03269264 post -0.03269264 1.00000000 See Also See Recipe 5.24 for more about the subset function. 5.29 Combining Two Data Frames Problem You want to combine the contents of two data frames into one data frame. Solution To combine the columns of two data frames side by side, use cbind: > all.cols <- cbind(dfrm1,dfrm2) To “stack” the rows of two data frames, use rbind: > all.rows <- rbind(dfrm1,dfrm2) 138 | Chapter 5: Data Structures Discussion You can combine data frames in one of two ways: either by putting the columns side by side to create a wider data frame; or by “stacking” the rows to create a taller data frame. The cbind function will combine data frames side by side, as shown here when combining stooges and birth: > stooges name n.marry n.child 1 Moe 1 2 2 Larry 1 2 3 Curly 4 2 > birth birth.year birth.place 1 1887 Bensonhurst 2 1902 Philadelphia 3 1903 Brooklyn > cbind(stooges,birth) name n.marry n.child birth.year birth.place 1 Moe 1 2 1887 Bensonhurst 2 Larry 1 2 1902 Philadelphia 3 Curly 4 2 1903 Brooklyn You would normally combine columns with the same height (number of rows). Tech- nically speaking, however, cbind does not require matching heights. If one data frame is short, it will invoke the Recycling Rule to extend the short columns as necessary (Recipe 5.3), which may or may not be what you want. The rbind function will “stack” the rows of two data frames, as shown here when combining stooges and guys: > stooges name n.marry n.child 1 Moe 1 2 2 Larry 1 2 3 Curly 4 2 > guys name n.marry n.child 1 Tom 4 2 2 Dick 1 4 3 Harry 1 1 > rbind(stooges,guys) name n.marry n.child 1 Moe 1 2 2 Larry 1 2 3 Curly 4 2 4 Tom 4 2 5 Dick 1 4 6 Harry 1 1 The rbind function requires that the data frames have the same width: same number of columns and same column names. The columns need not be in the same order, however; rbind will sort that out. 5.29 Combining Two Data Frames | 139 Finally, this recipe is slightly more general than the title implies. First, you can combine more than two data frames because both rbind and cbind accept multiple arguments. Second, you can apply this recipe to other data types because rbind and cbind work also with vectors, lists, and matrices. See Also The merge function can combine data frames that are otherwise incompatible owing to missing or different columns. The reshape2 and plyr packages, available on CRAN, include some powerful functions for slicing, dicing, and recombining data frames. 5.30 Merging Data Frames by Common Column Problem You have two data frames that share a common column. You want to merge their rows into one data frame by matching on the common column. Solution Use the merge function to join the data frames into one new data frame based on the common column: > m <- merge(df1, df2, by="name") Here name is the name of the column that is common to data frames df1 and df2. Discussion Suppose you have two data frames, born and died, that each contain a column called name: > born name year.born place.born 1 Moe 1887 Bensonhurst 2 Larry 1902 Philadelphia 3 Curly 1903 Brooklyn 4 Harry 1964 Moscow > died name year.died 1 Curly 1952 2 Moe 1975 3 Larry 1975 We can merge them into one data frame by using name to combine matched rows: > merge(born, died, by="name") name year.born place.born year.died 1 Curly 1903 Brooklyn 1952 2 Larry 1902 Philadelphia 1975 3 Moe 1887 Bensonhurst 1975 140 | Chapter 5: Data Structures Notice that merge does not require the rows to be sorted or even to occur in the same order. It found the matching rows for Curly even though they occur in different posi- tions. It also discards rows that appear in only one data frame or the other. In SQL terms, the merge function essentially performs a join operation on the two data frames. It has many options for controlling that join operation, all of which are described on the help page for merge. See Also See Recipe 5.29 for other ways to combine data frames. 5.31 Accessing Data Frame Contents More Easily Problem Your data is stored in a data frame. You are getting tired of repeatedly typing the data frame name and want to access the columns more easily. Solution For quick, one-off expressions, use the with function to expose the column names: > with(dataframe, expr) Inside expr, you can refer to the columns of dataframe by their names—as if they were simple variables. For repetitive access, use the attach function to insert the data frame into your search list. You can then refer to the data frame columns by name without mentioning the data frame: > attach(dataframe) Use the detach function to remove the data frame from your search list. Discussion A data frame is a great way to store your data, but accessing individual columns can become tedious. For a data frame called suburbs that contains a column called pop, here is the naïve way to calculate the z-scores of pop: > z <- (suburbs$pop - mean(suburbs$pop)) / sd(suburbs$pop) Call me a whiner, but all that typing gets tedious. The with function lets you expose the columns of a data frame as distinct variables. It takes two arguments, a data frame and an expression to be evaluated. Inside the expression, you can refer to the data frame columns by their names: > z <- with(suburbs, (pop - mean(pop)) / sd(pop)) 5.31 Accessing Data Frame Contents More Easily | 141 That is useful for one-liners. If you will be working repeatedly with columns in your data frame, attach the data frame to your search list and the columns will become available as variables: > attach(suburbs) After the attach, the second item in the search list is the suburbs data frame: > search() [1] ".GlobalEnv" "suburbs" "package:stats" [4] "package:graphics" "package:grDevices" "package:utils" [7] "package:datasets" "package:methods" "Autoloads" [10] "package:base" Now we can refer to the columns of the data frame as if they were variables: > z <- (pop - mean(pop)) / sd(pop) When you are done, use a detach (with no arguments) to remove the second location in the search list: > detach() > search() [1] ".GlobalEnv" "package:stats" "package:graphics" [4] "package:grDevices" "package:utils" "package:datasets" [7] "package:methods" "Autoloads" "package:base" Observe that suburbs is no longer in the search list. Attaching a data frame has a big quirk: R attaches a temporary copy of the data frame, which means that changes to the original data frame are hidden. In this session frag- ment, notice how changing the data frame does not change our view of the attached data: > attach(suburbs) > pop [1] 2853114 90352 171782 94487 102746 106221 147779 76031 70834 [10] 72616 74239 83048 67232 75386 63348 91452 > suburbs$pop <- 0 # Overwrite data frame contents > pop # Hey! It seems nothing changed [1] 2853114 90352 171782 94487 102746 106221 147779 76031 70834 [10] 72616 74239 83048 67232 75386 63348 91452 > suburbs$pop # Contents of data frame did indeed change [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Another source of confusion is that assigning values to the exposed names does not work as you might expect. In the following fragment, you might think we are scaling pop by 1,000 but we are actually creating a new local variable: > attach(suburbs) > pop [1] 2853114 90352 171782 94487 102746 106221 147779 76031 70834 [10] 72616 74239 83048 67232 75386 63348 91452 > pop <- pop / 1000 # Achtung! This is creating a local variable called "pop" > ls() # We can see the new variable in our workspace [1] "pop" "suburbs" 142 | Chapter 5: Data Structures > suburbs$pop # Original data is unchanged [1] 2853114 90352 171782 94487 102746 106221 147779 76031 70834 [10] 72616 74239 83048 67232 75386 63348 91452 5.32 Converting One Atomic Value into Another Problem You have a data value which has an atomic data type: character, complex, double, integer, or logical. You want to convert this value into one of the other atomic data types. Solution For each atomic data type, there is a function for converting values to that type. The conversion functions for atomic types include: • as.character(x) • as.complex(x) • as.numeric(x) or as.double(x) • as.integer(x) • as.logical(x) Discussion Converting one atomic type into another is usually pretty simple. If the conversion works, you get what you would expect. If it does not work, you get NA: > as.numeric(" 3.14 ") [1] 3.14 > as.integer(3.14) [1] 3 > as.numeric("foo") [1] NA Warning message: NAs introduced by coercion > as.character(101) [1] "101" If you have a vector of atomic types, these functions apply themselves to every value. So the preceding examples of converting scalars generalize easily to converting entire vectors: > as.numeric(c("1","2.718","7.389","20.086")) [1] 1.000 2.718 7.389 20.086 > as.numeric(c("1","2.718","7.389","20.086", "etc.")) [1] 1.000 2.718 7.389 20.086 NA Warning message: NAs introduced by coercion 5.32 Converting One Atomic Value into Another | 143 > as.character(101:105) [1] "101" "102" "103" "104" "105" When converting logical values into numeric values, R converts FALSE to 0 and TRUE to 1: > as.numeric(FALSE) [1] 0 > as.numeric(TRUE) [1] 1 This behavior is useful when you are counting occurrences of TRUE in vectors of logical values. If logvec is a vector of logical values, then sum(logvec) does an implicit conver- sion from logical to integer and returns the number of TRUEs. 5.33 Converting One Structured Data Type into Another Problem You want to convert a variable from one structured data type to another—for example, converting a vector into a list or a matrix into a data frame. Solution These functions convert their argument into the corresponding structured data type: • as.data.frame(x) • as.list(x) • as.matrix(x) • as.vector(x) Some of these conversions may surprise you, however. I suggest you review Table 5-1. Discussion Converting between structured data types can be tricky. Some conversions behave as you’d expect. If you convert a matrix into a data frame, for instance, the rows and columns of the matrix become the rows and columns of the data frame. No sweat. Table 5-1. Data conversions Conversion How Notes Vector→List as.list(vec) Don’t use list(vec); that creates a 1-element list whose only element is a copy of vec. Vector→Matrix To create a 1-column matrix: cbind(vec) or as.matrix(vec) See Recipe 5.14. To create a 1-row matrix: rbind(vec) 144 | Chapter 5: Data Structures Conversion How Notes To create an n × m matrix: matrix(vec,n,m) Vector→Data frame To create a 1-column data frame: as.data.frame(vec) To create a 1-row data frame: as.data.frame(rbind(vec)) List→Vector unlist(lst) Use unlist rather than as.vector; see Note 1 and Recipe 5.11. List→Matrix To create a 1-column matrix: as.matrix(lst) To create a 1-row matrix: as.matrix(rbind(lst)) To create an n × m matrix: matrix(lst,n,m) List→Data frame If the list elements are columns of data: as.data.frame(lst) If the list elements are rows of data: Recipe 5.19 Matrix→Vector as.vector(mat) Returns all matrix elements in a vector. Matrix→List as.list(mat) Returns all matrix elements in a list. Matrix→Data frame as.data.frame(mat) Data frame→Vector To convert a 1-row data frame: dfrm[1,] See Note 2. To convert a 1-column data frame: dfrm[,1] or dfrm[[1]] Data frame→List as.list(dfrm) See Note 3. Data frame→Matrix as.matrix(dfrm) See Note 4. In other cases, the results might surprise you. Table 5-1 summarizes some noteworthy examples. The following Notes are cited in that table: 1. When you convert a list into a vector, the conversion works cleanly if your list contains atomic values that are all of the same mode. Things become complicated if either (a) your list contains mixed modes (e.g., numeric and character), in which case everything is converted to characters; or (b) your list contains other structured data types, such as sublists or data frames—in which case very odd things happen, so don’t do that. 2. Converting a data frame into a vector makes sense only if the data frame contains one row or one column. To extract all its elements into one, long vector, use as.vector(as.matrix(dfrm)). But even that makes sense only if the data frame is all-numeric or all-character; if not, everything is first converted to character strings. 3. Converting a data frame into a list may seem odd in that a data frame is already a list (i.e., a list of columns). Using as.list essentially removes the class (data.frame) and thereby exposes the underlying list. That is useful when you want R to treat your data structure as a list—say, for printing. 5.33 Converting One Structured Data Type into Another | 145 4. Be careful when converting a data frame into a matrix. If the data frame contains only numeric values then you get a numeric matrix. If it contains only character values, you get a character matrix. But if the data frame is a mix of numbers, char- acters, and/or factors, then all values are first converted to characters. The result is a matrix of character strings. Problems with matrices The matrix conversions detailed here assume that your matrix is homogeneous: all elements have the same mode (e.g, all numeric or all character). A matrix can to be heterogeneous, too, when the matrix is built from a list. If so, conversions become messy. For example, when you convert a mixed-mode matrix to a data frame, the data frame’s columns are actually lists (to accommodate the mixed data). See Also See Recipe 5.32 for converting atomic data types; see the “Introduction” to this chapter for remarks on problematic conversions. 146 | Chapter 5: Data Structures CHAPTER 6 Data Transformations Introduction This chapter is all about the apply functions: apply, lapply, sapply, tapply, mapply; and their cousins, by and split. These functions let you take data in great gulps and process the whole gulp at once. Where traditional programming languages use loops, R uses vectorized operations and the apply functions to crunch data in batches, greatly stream- lining the calculations. Defining Groups Via a Factor An important idiom of R is using a factor to define a group. Suppose we have a vector and a factor, both of the same length, that were created as follows: > v <- c(40,2,83,28,58) > f <- factor(c("A","C","C","B","C")) We can visualize the vector elements and factors levels side by side, like this: Vector Factor 40 A 2 C 83 A 28 B 58 C The factor level identifies the group of each vector element: 40 and 83 are in group A; 28 is in group B; and 2 and 58 are in group C. In this book, I refer to such factors as grouping factors. They effectively slice and dice our data by putting them into groups. This is powerful because processing data in groups occurs often in statistics when comparing group means, comparing group pro- portions, performing ANOVA analysis, and so forth. 147 This chapter has recipes that use grouping factors to split vector elements into their respective groups (Recipe 6.1), apply a function to groups within a vector (Rec- ipe 6.5), and apply a function to groups of rows within a data frame (Recipe 6.6). In other chapters, the same idiom is used to test group means (Recipe 9.19), perform one- way ANOVA analysis (Recipe 11.20), and plot data points by groups (Recipe 10.4), among other uses. 6.1 Splitting a Vector into Groups Problem You have a vector. Each element belongs to a different group, and the groups are iden- tified by a grouping factor. You want to split the elements into the groups. Solution Suppose the vector is x and the factor is f. You can use the split function: > groups <- split(x, f) Alternatively, you can use the unstack function: > groups <- unstack(data.frame(x,f)) Both functions return a list of vectors, where each vector contains the elements for one group. The unstack function goes one step further: if all vectors have the same length, it converts the list into a data frame. Discussion The Cars93 dataset contains a factor called Origin that has two levels, USA and non- USA. It also contains a column called MPG.city. We can split the MPG data according to origin as follows: > library(MASS) > split(Cars93$MPG.city, Cars93$Origin) $USA [1] 22 19 16 19 16 16 25 25 19 21 18 15 17 17 20 23 20 29 23 22 17 21 18 29 20 [26] 31 23 22 22 24 15 21 18 17 18 23 19 24 23 18 19 23 31 23 19 19 19 28 $non-USA [1] 25 18 20 19 22 46 30 24 42 24 29 22 26 20 17 18 18 29 28 26 18 17 20 19 29 [26] 18 29 24 17 21 20 33 25 23 39 32 25 22 18 25 17 21 18 21 20 The usefulness of splitting is that we can analyze the data by group. This example computes the median MPG for each group: 148 | Chapter 6: Data Transformations > g <- split(Cars93$MPG.city, Cars93$Origin) > median(g[[1]]) [1] 20 > median(g[[2]]) [1] 22 See Also See the “Introduction” to this chapter for more about grouping factors. Recipe 6.5 shows another way to apply functions, such as median, to each group. The unstack function can perform other, more powerful transformations beside this; see the help page. 6.2 Applying a Function to Each List Element Problem You have a list, and you want to apply a function to each element of the list. Solution Use either the lapply function or the sapply function, depending upon the desired form of the result. lapply always returns the results in list, whereas sapply returns the results in a vector if that is possible: > lst <- lapply(lst, fun) > vec <- sapply(lst, fun) Discussion These functions will call your function (fun, in the solution example) once for every element on your list. Your function should expect one argument, an element from the list. The lapply and sapply functions will collect the returned values. lapply collects them into a list and returns the list. The “s” in “sapply” stands for “simplify.” The function tries to simplify the results into a vector or matrix. For that to happen, all the returned values must have the same length. If that length is 1 then you get a vector; otherwise, you get a matrix. If the lengths vary, simplification is impossible and you get a list. Let’s say I teach an introductory statistics class four times and administer comparable final exams each time. Here are the exam scores from the four semesters: > scores $S1 [1] 89 85 85 86 88 89 86 82 96 85 93 91 98 87 94 77 87 98 85 89 [21] 95 85 93 93 97 71 97 93 75 68 98 95 79 94 98 95 $S2 [1] 60 98 94 95 99 97 100 73 93 91 98 86 66 83 77 6.2 Applying a Function to Each List Element | 149 [16] 97 91 93 71 91 95 100 72 96 91 76 100 97 99 95 [31] 97 77 94 99 88 100 94 93 86 $S3 [1] 95 86 90 90 75 83 96 85 83 84 81 98 77 94 84 89 93 99 91 77 [21] 95 90 91 87 85 76 99 99 97 97 97 77 93 96 90 87 97 88 $S4 [1] 67 93 63 83 87 97 96 92 93 96 87 90 94 90 82 91 85 93 83 90 [21] 87 99 94 88 90 72 81 93 93 94 97 89 96 95 82 97 Each semester starts with 40 students but, alas, not everyone makes it to the finish line; hence each semester has a different number of scores. We can count this number with the length function: lapply will return a list of lengths, and sapply will return a vector of lengths: > lapply(scores, length) $S1 [1] 36 $S2 [1] 39 $S3 [1] 38 $S4 [1] 36 > sapply(scores, length) S1 S2 S3 S4 36 39 38 36 We can see the mean and standard deviation of the scores just as easily: > sapply(scores, mean) S1 S2 S3 S4 88.77778 89.79487 89.23684 88.86111 > sapply(scores, sd) S1 S2 S3 S4 7.720515 10.543592 7.178926 8.208542 If the called function returns a vector, sapply will form the results into a matrix. The range function, for example, returns a two-element vector: > sapply(scores, range) S1 S2 S3 S4 [1,] 68 60 75 63 [2,] 98 100 99 99 If the called function returns a structured object, such as a list, then you will need to use lapply rather than sapply. Structured objects cannot be put into a vector. Suppose we want to perform a t test on every semester. The t.test function returns a list, so we must use lapply: > tests <- lapply(scores, t.test) 150 | Chapter 6: Data Transformations Now the result tests is a list: it is a list of t.test results. We can use sapply to extract elements from the t.test results, such as the bounds of the confidence interval: > sapply(tests, function(t) t$conf.int) S1 S2 S3 S4 [1,] 86.16553 86.37703 86.87719 86.08374 [2,] 91.39002 93.21271 91.59650 91.63848 See Also See Recipe 2.12. 6.3 Applying a Function to Every Row Problem You have a matrix. You want to apply a function to every row, calculating the function result for each row. Solution Use the apply function. Set the second argument to 1 to indicate row-by-row application of a function: > results <- apply(mat, 1, fun) # mat is a matrix, fun is a function The apply function will call fun once for each row, assemble the returned values into a vector, and then return that vector. Discussion Suppose your matrix long is longitudinal data. Each row contains data for one subject, and the columns contain the repeated observations over time: > long trial1 trial2 trial3 trial4 trial5 Moe -1.8501520 -1.406571 -1.0104817 -3.7170704 -0.2804896 Larry 0.9496313 1.346517 -0.1580926 1.6272786 2.4483321 Curly -0.5407272 -1.708678 -0.3480616 -0.2757667 -1.2177024 You could calculate the average observation for each subject by applying the mean func- tion to the rows. The result is a vector: > apply(long, 1, mean) Moe Larry Curly -1.6529530 1.2427334 -0.8181872 Note that apply uses the rownames from your matrix to identify the elements of the resulting vector, which is handy. 6.3 Applying a Function to Every Row | 151 The function being called (fun, described previously) should expect one argument, a vector, which will be one row from the matrix. The function can return a scalar or a vector. In the vector case, apply assembles the results into a matrix. The range function returns a vector of two elements, the minimum and the maximum, so applying it to long produces a matrix: > apply(long, 1, range) Moe Larry Curly [1,] -3.7170704 -0.1580926 -1.7086779 [2,] -0.2804896 2.4483321 -0.2757667 You can employ this recipe on data frames as well. It works if the data frame is homogeneous—either all numbers or all character strings. When the data frame has columns of different types, extracting vectors from the rows isn’t sensible because vec- tors must be homogeneous. 6.4 Applying a Function to Every Column Problem You have a matrix or data frame, and you want to apply a function to every column. Solution For a matrix, use the apply function. Set the second argument to 2, which indicates column-by-column application of the function: > results <- apply(mat, 2, fun) For a data frame, use the lapply or sapply functions. Either one will apply a function to successive columns of your data frame. Their difference is that lapply assembles the return values into a list whereas sapply assembles them into a vector: > lst <- lapply(dfrm, fun) > vec <- sapply(dfrm, fun) You can use apply on data frames, too, but only if the data frame is homogeneous (i.e., either all numeric values or all character strings). Discussion The apply function is intended for processing a matrix. In Recipe 6.3 we used apply to process the rows of a matrix. This is the same situation, but now we are processing the columns. The second argument of apply determines the direction: • 1 means process row by row. • 2 means process column by column. This is more mnemonic than it looks. We speak of matrices in “rows and columns”, so rows are first and columns second; 1 and 2, respectively. 152 | Chapter 6: Data Transformations A data frame is a more complicated data structure than a matrix, so there are more options. You can simply use apply, in which case R will convert your data frame to a matrix and then apply your function. That will work if your data frame contains only numeric data or character data. It probably will not work if you have mixed data types. In that case, R will force all columns to have identical types, likely performing an un- wanted conversion as a result. Fortunately, there is an alternative. Recall that a data frame is a kind of list: it is a list of the columns of the data frame. You can use lapply and sapply to process the columns, as described in Recipe 6.2: > lst <- lapply(dfrm, fun) # Returns a list > vec <- sapply(dfrm, fun) # Returns a vector The function fun should expect one argument: a column from the data frame. I often use this recipe to check the types of columns in data frames. The batch column of this data frame seems to contain numbers: > head(batches) batch clinic dosage shrinkage 1 1 IL 3 -0.11810714 2 3 IL 4 -0.29932107 3 2 IL 4 -0.27651716 4 1 IL 5 -0.18925825 5 2 IL 2 -0.06804804 6 3 NJ 5 -0.38279193 But printing the classes of the columns reveals it to be a factor instead: > sapply(batches, class) batch clinic dosage shrinkage "factor" "factor" "integer" "numeric" A cool example of this recipe is removing low-correlation variables from a set of pre- dictors. Suppose that resp is a response variable and pred is a data frame of predictor variables. Suppose further that we have too many predictors and therefore want to select the top 10 as measured by correlation with the response. The first step is to calculate the correlation between each variable and resp. In R, that’s a one-liner: > cors <- sapply(pred, cor, y=resp) The sapply function will call the function cor for every column in pred. Note that we gave a third argument, y=resp, to sapply. Any arguments beyond the second one are passed to cor. Every time that sapply calls cor, the first argument will be a column and the second argument will be y=resp. With those arguments, the function call will be cor(column,y=resp), which calculates the correlation between the given column and resp. 6.4 Applying a Function to Every Column | 153 D o wnload from Wow! eBook The result from sapply is a vector of correlations, one for each column. We use the rank function to find the positions of the correlations that have the largest magnitude: > mask <- (rank(-abs(cors)) <= 10) This expression is a comparison (<=), so it returns a vector of logical values. It is cleverly constructed so that the top 10 correlations have corresponding TRUE values and all others are FALSE. Using that vector of logical values, we can select just those columns from the data frame: > best.pred <- pred[,mask] At this point, we can regress resp against best.pred, knowing that we have chosen the predictors with the highest correlations: > lm(resp ~ best.pred) That’s pretty good for four lines of code. See Also See Recipes 5.22, 6.2, and 6.3. 6.5 Applying a Function to Groups of Data Problem Your data elements occur in groups. You want to process the data by groups—for example, summing by group or averaging by group. Solution Create a grouping factor (of the same length as your vector) that identifies the group of each corresponding datum. Then use the tapply function, which will apply a function to each group of data: > tapply(x, f, fun) Here, x is a vector, f is a grouping factor, and fun is a function. The function should expect one argument, which is a vector of elements taken from x according to their group. Discussion Suppose I have a vector with the populations of the 16 largest cities in the greater Chicago metropolitan area, taken from the data frame called suburbs: > attach(suburbs) > pop [1] 2853114 90352 171782 94487 102746 106221 147779 76031 70834 [10] 72616 74239 83048 67232 75386 63348 91452 154 | Chapter 6: Data Transformations We can easily compute sums and averages for all the cities: > sum(pop) [1] 4240667 > mean(pop) [1] 265041.7 What if we want the sum and average broken out by county? We will need a factor, say county, the same length as pop, where each level of the factor gives the corresponding county (there are two Lake counties: one in Illinois and one in Indiana) : > county [1] Cook Kenosha Kane Kane Lake(IN) Kendall DuPage Cook [9] Will Cook Cook Lake(IN) Cook Cook Cook Lake(IL) Levels: Cook DuPage Kane Kendall Kenosha Lake(IL) Lake(IN) Will Now I can use the county factor and tapply function to process items in groups. The tapply function has three main parameters: the vector of data, the factor that defines the groups, and a function. It will extract each group, apply the function to each group, and return a vector with the collected results. This example shows summing the pop- ulations by county: > tapply(pop,county,sum) Cook DuPage Kane Kendall Kenosha Lake(IL) Lake(IN) Will 3281966 147779 266269 106221 90352 91452 185794 70834 The next example computes average populations by county: > tapply(pop,county,mean) Cook DuPage Kane Kendall Kenosha Lake(IL) Lake(IN) Will 468852.3 147779.0 133134.5 106221.0 90352.0 91452.0 92897.0 70834.0 The function given to tapply should expect a single argument: a vector containing all the members of one group. A good example is the length function, which takes a vector parameter and returns the vector’s length. Use it to count the number of data in each group; in this case, the number of cities in each county: > tapply(pop,county,length) Cook DuPage Kane Kendall Kenosha Lake(IL) Lake(IN) Will 7 1 2 1 1 1 2 1 In most cases you will use functions that return a scalar and tapply will collect the returned scalars into a vector. Your function can return complex objects, too, in which case tapply will return them in a list. See the tapply help page for details. See Also See this chapter’s “Introduction” for more about grouping factors. 6.5 Applying a Function to Groups of Data | 155 6.6 Applying a Function to Groups of Rows Problem You want to apply a function to groups of rows within a data frame. Solution Define a grouping factor—that is, a factor with one level (element) for every row in your data frame—that identifies the data groups. For each such group of rows, the by function puts the rows into a temporary data frame and calls your function with that argument. The by function collects the returned values into a list and returns the list: > by(dfrm, fact, fun) Here, dfrm is the data frame, fact is the grouping factor, and fun is a function. The function should expect one argument, a data frame. Discussion The advantage of the by function is that it calls your function with a data frame, which is useful if your function handles data frames in a special way. For instance, the print, summary, and mean functions perform special processing for data frames. Suppose you have a data frame from clinical trials, called trials, where the dosage was randomized to study its effect: > trials sex pre dose1 dose2 post 1 F 5.931640 2 1 3.162600 2 F 4.496187 1 2 3.293989 3 M 6.161944 1 1 4.446643 4 F 4.322465 2 1 3.334748 5 M 4.153510 1 1 4.429382 . . (etc.) . The data includes a factor for the subject’s sex, so by can split the data according to sex and call summary for the two groups. The result is two summaries, one for men and one for women: > by(trials, trials$sex, summary) trials$sex: F sex pre dose1 dose2 post F:7 Min. :4.156 Min. :1.000 Min. :1.000 Min. :2.886 M:0 1st Qu.:4.409 1st Qu.:1.000 1st Qu.:1.000 1st Qu.:3.075 Median :4.895 Median :1.000 Median :2.000 Median :3.163 Mean :5.020 Mean :1.429 Mean :1.571 Mean :3.174 3rd Qu.:5.668 3rd Qu.:2.000 3rd Qu.:2.000 3rd Qu.:3.314 Max. :5.932 Max. :2.000 Max. :2.000 Max. :3.389 156 | Chapter 6: Data Transformations ------------------------------------------------------------------ trials$sex: M sex pre dose1 dose2 post F:0 Min. :3.998 Min. :1.000 Min. :1.000 Min. :3.738 M:9 1st Qu.:4.773 1st Qu.:1.000 1st Qu.:1.000 1st Qu.:3.800 Median :5.110 Median :2.000 Median :1.000 Median :4.194 Mean :5.189 Mean :1.556 Mean :1.444 Mean :4.148 3rd Qu.:5.828 3rd Qu.:2.000 3rd Qu.:2.000 3rd Qu.:4.429 Max. :6.658 Max. :2.000 Max. :2.000 Max. :4.517 We can also build linear models of post as a function of the dosages, with one model for men and one model for women: > models <- by(trials, trials$sex, function(df) lm(post~pre+dose1+dose2, data=df)) Observe that the parameter to our function is a data frame, so we can use it as the data argument of lm. The result is a two-element list of linear models. When we print the list, we see a model for each sex: > print(models) trials$sex: F Call: lm(formula = post ~ pre + dose1 + dose2, data = df) Coefficients: (Intercept) pre dose1 dose2 4.30804 -0.08161 -0.16225 -0.31354 ------------------------------------------------------------ trials$sex: M Call: lm(formula = post ~ pre + dose1 + dose2, data = df) Coefficients: (Intercept) pre dose1 dose2 5.29981 -0.02713 -0.36851 -0.30323 We have a list of models, so we can apply the confint function to each list element and see the confidence intervals for each model’s coefficients: > lapply(models, confint) $F 2.5 % 97.5 % (Intercept) 3.0841733 5.53191431 pre -0.2950747 0.13184560 dose1 -0.4711773 0.14667409 dose2 -0.6044273 -0.02264593 $M 2.5 % 97.5 % (Intercept) 4.8898433 5.70978218 pre -0.1070276 0.05277108 dose1 -0.4905828 -0.24644057 dose2 -0.4460200 -0.16043211 6.6 Applying a Function to Groups of Rows | 157 In this case, we see that pre is not significant for either model because its confidence interval contains zero; in contrast, dose1 and dose2 are significant for both models. The key fact, however, is that the models have significantly different intercepts, alerting us to a potentially different response for men and women. See Also See this chapter’s “Introduction” for more about grouping factors. See Recipe 6.2 for more about lapply. 6.7 Applying a Function to Parallel Vectors or Lists Problem You have a function, say f, that takes multiple arguments. You want to apply the func- tion element-wise to vectors and obtain a vector result. Unfortunately, the function is not vectorized; that is, it works on scalars but not on vectors. Solution Use the mapply function. It will apply the function f to your arguments element-wise: > mapply(f, vec1, vec2, ..., vecN) There must be one vector for each argument expected by f. If the vector arguments are of unequal length, the Recycling Rule is applied. The mapply function also works with list arguments: > mapply(f, list1, list2, ..., listN) Discussion The basic operators of R, such as x + y, are vectorized; this means that they compute their result element-by-element and return a vector of results. Also, many R functions are vectorized. Not all functions are vectorized, however, and those that are not work only on scalars. Using vector arguments produces errors at best and meaningless results at worst. In such cases, the mapply function can effectively vectorize the function for you. Consider the gcd function from Recipe 2.12, which takes two arguments: > gcd <- function(a,b) { + if (b == 0) return(a) + else return(gcd(b, a %% b)) + } 158 | Chapter 6: Data Transformations If we apply gcd to two vectors, the result is wrong answers and a pile of error messages: > gcd(c(1,2,3), c(9,6,3)) [1] 1 2 0 Warning messages: 1: In if (b == 0) return(a) else return(gcd(b, a%%b)) : the condition has length > 1 and only the first element will be used 2: In if (b == 0) return(a) else return(gcd(b, a%%b)) : the condition has length > 1 and only the first element will be used 3: In if (b == 0) return(a) else return(gcd(b, a%%b)) : the condition has length > 1 and only the first element will be used The function is not vectorized, but we can use mapply to vectorize it. This gives the element-wise GCDs between two vectors: > mapply(gcd, c(1,2,3), c(9,6,3)) [1] 1 2 3 6.7 Applying a Function to Parallel Vectors or Lists | 159 CHAPTER 7 Strings and Dates Introduction Strings? Dates? In a statistical programming package? As soon as you read files or print reports, you need strings. When you work with real- world problems, you need dates. R has facilities for both strings and dates. They are clumsy compared to string-oriented languages such as Perl, but then it’s a matter of the right tool for the job. I wouldn’t want to perform logistic regression in Perl. Classes for Dates and Times R has a variety of classes for working with dates and times; which is nice if you prefer having a choice but annoying if you prefer living simply. There is a critical distinction among the classes: some are date-only classes, some are datetime classes. All classes can handle calendar dates (e.g., March 15, 2010), but not all can represent a datetime (11:45 AM on March 1, 2010). The following classes are included in the base distribution of R: Date The Date class can represent a calendar date but not a clock time. It is a solid, general-purpose class for working with dates, including conversions, formatting, basic date arithmetic, and time-zone handling. Most of the date-related recipes in this book are built on the Date class. POSIXct This is a datetime class, and it can represent a moment in time with an accuracy of one second. Internally, the datetime is stored as the number of seconds since January 1, 1970, and so is a very compact representation. This class is recommen- ded for storing datetime information (e.g., in data frames). 161 POSIXlt This is also a datetime class, but the representation is stored in a nine-element list that includes the year, month, day, hour, minute, and second. That representation makes it easy to extract date parts, such as the month or hour. Obviously, this representation is much less compact than the POSIXct class; hence it is normally used for intermediate processing and not for storing data. The base distribution also provides functions for easily converting between represen- tations: as.Date, as.POSIXct, and as.POSIXlt. The following packages are available for downloading from CRAN: chron The chron package can represent both dates and times but without the added complexities of handling time zones and daylight savings time. It’s therefore easier to use than Date but less powerful than POSIXct and POSIXlt. It would be useful for work in econometrics or time series analysis. lubridate This is a relatively new, general-purpose package. It’s designed to make working with dates and times easier while keeping the important bells and whistles such as time zones. It’s especially clever regarding datetime arithmetic. mondate This is a specialized package for handling dates in units of months in addition to days and years. Such needs arise in accounting and actuarial work, for example, where month-by-month calculations are needed. timeDate This is a high-powered package with well-thought-out facilities for handling dates and times, including date arithmetic, business days, holidays, conversions, and generalized handling of time zones. It was originally part of the Rmetrics software for financial modeling, where precision in dates and times is critical. If you have a demanding need for date facilities, consider this package. Which class should you select? The article “Date and Time Classes in R” by Grothen- dieck and Petzoldt offers this general advice: When considering which class to use, always choose the least complex class that will support the application. That is, use Date if possible, otherwise use chron and otherwise use the POSIX classes. Such a strategy will greatly reduce the potential for error and increase the reliability of your application. See Also See help(DateTimeClasses) for more details regarding the built-in facilities. See the June 2004 article “Date and Time Classes in R” by Gabor Grothendieck and Thomas Pet- zoldt for a great introduction to the date and time facilities. The June 2001 article 162 | Chapter 7: Strings and Dates “Date-Time Classes” by Brian Ripley and Kurt Hornik discusses the two POSIX classes in particular. 7.1 Getting the Length of a String Problem You want to know the length of a string. Solution Use the nchar function, not the length function. Discussion The nchar function takes a string and returns the number of characters in the string: > nchar("Moe") [1] 3 > nchar("Curly") [1] 5 If you apply nchar to a vector of strings, it returns the length of each string: > s <- c("Moe", "Larry", "Curly") > nchar(s) [1] 3 5 5 You might think the length function returns the length of a string. Nope—it returns the length of a vector. When you apply the length function to a single string, R returns the value 1 because it views that string as a singleton vector—a vector with one element: > length("Moe") [1] 1 > length(c("Moe","Larry","Curly")) [1] 3 7.2 Concatenating Strings Problem You want to join together two or more strings into one string. Solution Use the paste function. 7.2 Concatenating Strings | 163 Discussion The paste function concatenates several strings together. In other words, it creates a new string by joining the given strings end to end: > paste("Everybody", "loves", "stats.") [1] "Everybody loves stats." By default, paste inserts a single space between pairs of strings, which is handy if that’s what you want and annoying otherwise. The sep argument lets you specify a different separator. Use an empty string ("") to run the strings together without separation: > paste("Everybody", "loves", "stats.", sep="-") [1] "Everybody-loves-stats." > paste("Everybody", "loves", "stats.", sep="") [1] "Everybodylovesstats." The function is very forgiving about nonstring arguments. It tries to convert them to strings using the as.character function: > paste("The square root of twice pi is approximately", sqrt(2*pi)) [1] "The square root of twice pi is approximately 2.506628274631" If one or more arguments are vectors of strings, paste will generate all combinations of the arguments: > stooges <- c("Moe", "Larry", "Curly") > paste(stooges, "loves", "stats.") [1] "Moe loves stats." "Larry loves stats." "Curly loves stats." Sometimes you want to join even those combinations into one, big string. The collapse parameter lets you define a top-level separator and instructs paste to concat- enate the generated strings using that separator: > paste(stooges, "loves", "stats", collapse=", and ") [1] "Moe loves stats, and Larry loves stats, and Curly loves stats" 7.3 Extracting Substrings Problem You want to extract a portion of a string according to position. Solution Use substr(string,start,end) to extract the substring that begins at start and ends at end. Discussion The substr function takes a string, a starting point, and an ending point. It returns the substring between the starting to ending points: 164 | Chapter 7: Strings and Dates > substr("Statistics", 1, 4) # Extract first 4 characters [1] "Stat" > substr("Statistics", 7, 10) # Extract last 4 characters [1] "tics" Just like many R functions, substr lets the first argument be a vector of strings. In that case, it applies itself to every string and returns a vector of substrings: > ss <- c("Moe", "Larry", "Curly") > substr(ss, 1, 3) # Extract first 3 characters of each string [1] "Moe" "Lar" "Cur" In fact, all the arguments can be vectors, in which case substr will treat them as parallel vectors. From each string, it extracts the substring delimited by the corresponding en- tries in the starting and ending points. This can facilitate some useful tricks. For ex- ample, the following code snippet extracts the last two characters from each string; each substring starts on the penultimate character of the original string and ends on the final character: > cities <- c("New York, NY", "Los Angeles, CA", "Peoria, IL") > substr(cities, nchar(cities)-1, nchar(cities)) [1] "NY" "CA" "IL" You can extend this trick into mind-numbing territory by exploiting the Recycling Rule, but I suggest you avoid the temptation. 7.4 Splitting a String According to a Delimiter Problem You want to split a string into substrings. The substrings are separated by a delimiter. Solution Use strsplit, which takes two arguments—the string and the delimiter of the sub- strings: > strsplit(string, delimiter) The delimiter can be either a simple string or a regular expression. Discussion It is common for a string to contain multiple substrings separated by the same delimiter. One example is a file path, whose components are separated by slashes (/): > path <- "/home/mike/data/trials.csv" We can split that path into its components by using strsplit with a delimiter of /: > strsplit(path, "/") [[1]] [1] "" "home" "mike" "data" "trials.csv" 7.4 Splitting a String According to a Delimiter | 165 Notice that the first “component” is actually an empty string because nothing preceded the first slash. Also notice that strsplit returns a list and that each element of the list is a vector of substrings. This two-level structure is necessary because the first argument can be a vector of strings. Each string is split into its substrings (a vector); then those vectors are returned in a list. This example splits three file paths and returns a three-element list: > paths <- c("/home/mike/data/trials.csv", + "/home/mike/data/errors.csv", + "/home/mike/corr/reject.doc") > strsplit(paths, "/") [[1]] [1] "" "home" "mike" "data" "trials.csv" [[2]] [1] "" "home" "mike" "data" "errors.csv" [[3]] [1] "" "home" "mike" "corr" "reject.doc" The second argument of strsplit (the delimiter argument) is actually much more powerful than these examples indicate. It can be a regular expression, letting you match patterns far more complicated than a simple string. In fact, to defeat the regular ex- pression feature (and its interpretation of special characters) you must include the fixed=TRUE argument. See Also To learn more about regular expressions in R, see the help page for regexp. See O’Reilly’s Mastering Regular Expressions, by Jeffrey E.F. Friedl to learn more about regular expressions in general. 7.5 Replacing Substrings Problem Within a string, you want to replace one substring with another. Solution Use sub to replace the first instance of a substring: > sub(old, new, string) Use gsub to replace all instances of a substring: > gsub(old, new, string) 166 | Chapter 7: Strings and Dates Discussion The sub function finds the first instance of the old substring within string and replaces it with the new substring: > s <- "Curly is the smart one. Curly is funny, too." > sub("Curly", "Moe", s) [1] "Moe is the smart one. Curly is funny, too." gsub does the same thing, but it replaces all instances of the substring (a global replace), not just the first: > gsub("Curly", "Moe", s) [1] "Moe is the smart one. Moe is funny, too." To remove a substring altogether, simply set the new substring to be empty: > sub(" and SAS", "", "For really tough problems, you need R and SAS.") [1] "For really tough problems, you need R." The old argument can be regular expression, which allows you to match patterns much more complicated than a simple string. This is actually assumed by default, so you must set the fixed=TRUE argument if you don’t want sub and gsub to interpret old as a regular expression. See Also To learn more about regular expressions in R, see the help page for regexp. See Mas- tering Regular Expressions to learn more about regular expressions in general. 7.6 Seeing the Special Characters in a String Problem Your string contains special characters that are unprintable. You want to know what they are. Solution Use print to see the special characters in a string. The cat function will not reveal them. Discussion I once encountered a bizarre situation in which a string contained, unbeknownst to me, unprintable characters. My R script used cat to display the string, and the script output was garbled until I realized there were unprintable characters in the string. 7.6 Seeing the Special Characters in a String | 167 In this example, the string seems to contain 13 characters but cat shows only a 6-character output: > nchar(s) [1] 13 > cat(s) second The reason is that the string contains special characters, which are characters that do not display when printed. When we use print, it uses escapes (backslashes) to show the special characters: > print(s) [1] "first\rsecond\n" Notice that the string contains a return character (\r) in the middle, so “first” is over- written by “second” when cat displays the string. 7.7 Generating All Pairwise Combinations of Strings Problem You have two sets of strings, and you want to generate all combinations from those two sets (their Cartesian product). Solution Use the outer and paste functions together to generate the matrix of all possible combinations: > m <- outer(strings1, strings2, paste, sep="") Discussion The outer function is intended to form the outer product. However, it allows a third argument to replace simple multiplication with any function. In this recipe we replace multiplication with string concatenation (paste), and the result is all combinations of strings. Suppose you have four test sites and three treatments: > locations <- c("NY", "LA", "CHI", "HOU") > treatments <- c("T1", "T2", "T3") We can apply outer and paste to generate all combinations of test sites and treatments: > outer(locations, treatments, paste, sep="-") [,1] [,2] [,3] [1,] "NY-T1" "NY-T2" "NY-T3" [2,] "LA-T1" "LA-T2" "LA-T3" [3,] "CHI-T1" "CHI-T2" "CHI-T3" [4,] "HOU-T1" "HOU-T2" "HOU-T3" 168 | Chapter 7: Strings and Dates The fourth argument of outer is passed to paste. In this case, we passed sep="-" in order to define a hyphen as the separator between the strings. The result of outer is a matrix. If you want the combinations in a vector instead, flatten the matrix using the as.vector function. In the special case when you are combining a set with itself and order does not matter, the result will be duplicate combinations: > outer(treatments, treatments, paste, sep="-") [,1] [,2] [,3] [1,] "T1-T1" "T1-T2" "T1-T3" [2,] "T2-T1" "T2-T2" "T2-T3" [3,] "T3-T1" "T3-T2" "T3-T3" But suppose we want all unique pairwise combinations of treatments. We can eliminate the duplicates by removing the lower triangle (or upper triangle). The lower.tri func- tion identifies that triangle, so inverting it identifies all elements outside the lower triangle: > m <- outer(treatments, treatments, paste, sep="-") > m[!lower.tri(m)] [1] "T1-T1" "T1-T2" "T2-T2" "T1-T3" "T2-T3" "T3-T3" See Also See Recipe 7.2 for using paste to generate combinations of strings. 7.8 Getting the Current Date Problem You need to know today’s date. Solution The Sys.Date function returns the current date: > Sys.Date() [1] "2010-02-11" Discussion The Sys.Date function returns a Date object. In the preceding example it seems to return a string because the result is printed inside double quotes. What really happened, however, is that Sys.Date returned a Date object and then R converted that object into a string for printing purposes. You can see this by checking the class of the result from Sys.Date: > class(Sys.Date()) [1] "Date" 7.8 Getting the Current Date | 169 See Also See Recipe 7.10. 7.9 Converting a String into a Date Problem You have the string representation of a date, such as “2010-12-31”, and you want to convert that into a Date object. Solution You can use as.Date, but you must know the format of the string. By default, as.Date assumes the string looks like yyyy-mm-dd. To handle other formats, you must specify the format parameter of as.Date. Use format="%m/%d/%Y" if the date is in American style, for instance. Discussion This example shows the default format assumed by as.Date, which is the ISO 8601 standard format of yyyy-mm-dd: > as.Date("2010-12-31") [1] "2010-12-31" The as.Date function returns a Date object that is here being converted back to a string for printing; this explains the double quotes around the output. The string can be in other formats, but you must provide a format argument so that as.Date can interpret your string. See the help page for the stftime function for details about allowed formats. Being a simple American, I often mistakenly try to convert the usual American date format (mm/dd/yyyy) into a Date object, with these unhappy results: > as.Date("12/31/2010") Error in charToDate(x) : character string is not in a standard unambiguous format Here is the correct way to convert an American-style date: > as.Date("12/31/2010", format="%m/%d/%Y") [1] "2010-12-31" Observe that the Y in the format string is capitalized to indicate a 4-digit year. If you’re using 2-digit years, specify a lowercase y. 170 | Chapter 7: Strings and Dates 7.10 Converting a Date into a String Problem You want to convert a Date object into a character string, usually because you want to print the date. Solution Use either format or as.character: > format(Sys.Date()) [1] "2010-04-01" > as.character(Sys.Date()) [1] "2010-04-01" Both functions allow a format argument that controls the formatting. Use format="%m/%d/%Y" to get American-style dates, for example: > format(Sys.Date(), format="%m/%d/%Y") [1] "04/01/2010" Discussion The format argument defines the appearance of the resulting string. Normal characters, such as slash (/) or hyphen (-) are simply copied to the output string. Each two-letter combination of a percent sign (%) followed by another character has special meaning. Some common ones are: %b Abbreviated month name (“Jan”) %B Full month name (“January”) %d Day as a two-digit number %m Month as a two-digit number %y Year without century (00–99) %Y Year with century See the help page for the strftime function for a complete list of formatting codes. 7.10 Converting a Date into a String | 171 7.11 Converting Year, Month, and Day into a Date Problem You have a date represented by its year, month, and day. You want to merge these elements into a single Date object representation. Solution Use the ISOdate function: > ISOdate(year, month, day) The result is a POSIXct object that you can convert into a Date object: > as.Date(ISOdate(year, month, day)) Discussion It is common for input data to contain dates encoded as three numbers: year, month, and day. The ISOdate function can combine them into a POSIXct object: > ISOdate(2012,2,29) [1] "2012-02-29 12:00:00 GMT" You can keep your date in the POSIXct format. However, when working with pure dates (not dates and times), I often convert to a Date object and truncate the unused time information: > as.Date(ISOdate(2012,2,29)) [1] "2012-02-29" Trying to convert an invalid date results in NA: > ISOdate(2013,2,29) # Oops! 2013 is not a leap year [1] NA ISOdate can process entire vectors of years, months, and days, which is quite handy for mass conversion of input data. The following example starts with the year/month/day numbers for the third Wednesday in January of several years and then combines them all into Date objects: > years [1] 2010 2011 2012 2013 2014 > months [1] 1 1 1 1 1 > days [1] 15 21 20 18 17 > ISOdate(years, months, days) [1] "2010-01-15 12:00:00 GMT" "2011-01-21 12:00:00 GMT" [3] "2012-01-20 12:00:00 GMT" "2013-01-18 12:00:00 GMT" [5] "2014-01-17 12:00:00 GMT" > as.Date(ISOdate(years, months, days)) [1] "2010-01-15" "2011-01-21" "2012-01-20" "2013-01-18" "2014-01-17" 172 | Chapter 7: Strings and Dates Purists will note that the vector of months is redundant and that the last expression can therefore be further simplified by invoking the Recycling Rule: > as.Date(ISOdate(years, 1, days)) [1] "2010-01-15" "2011-01-21" "2012-01-20" "2013-01-18" "2014-01-17" This recipe can also be extended to handle year, month, day, hour, minute, and second data by using the ISOdatetime function (see the help page for details): > ISOdatetime(year, month, day, hour, minute, second) 7.12 Getting the Julian Date Problem Given a Date object, you want to extract the Julian date—which is, in R, the number of days since January 1, 1970. Solution Either convert the Date object to an integer or use the julian function: > d <- as.Date("2010-03-15") > as.integer(d) [1] 14683 > julian(d) [1] 14683 attr(,"origin") [1] "1970-01-01" Discussion A Julian “date” is simply the number of days since a more-or-less arbitrary starting point. In the case of R, that starting point is January 1, 1970, the same starting point as Unix systems. So the Julian date for January 1, 1970 is zero, as shown here: > as.integer(as.Date("1970-01-01")) [1] 0 > as.integer(as.Date("1970-01-02")) [1] 1 > as.integer(as.Date("1970-01-03")) [1] 2 . . (etc.) . 7.12 Getting the Julian Date | 173 7.13 Extracting the Parts of a Date Problem Given a Date object, you want to extract a date part such as the day of the week, the day of the year, the calendar day, the calendar month, or the calendar year. Solution Convert the Date object to a POSIXlt object, which is a list of date parts. Then extract the desired part from that list: > d <- as.Date("2010-03-15") > p <- as.POSIXlt(d) > p$mday # Day of the month [1] 15 > p$mon # Month (0 = January) [1] 2 > p$year + 1900 # Year [1] 2010 Discussion The POSIXlt object represents a date as a list of date parts. Convert your Date object to POSIXlt by using the as.POSIXlt function, which will give you a list with these members: sec Seconds (0–61) min Minutes (0–59) hour Hours (0–23) mday Day of the month (1–31) mon Month (0–11) year Years since 1900 wday Day of the week (0–6, 0 = Sunday) yday Day of the year (0–365) isdst Daylight savings time flag 174 | Chapter 7: Strings and Dates Using these date parts, we can learn that April 1, 2010, is a Thursday (wday = 4) and the 91st day of the year (because yday = 0 on January 1): > d <- as.Date("2010-04-01") > as.POSIXlt(d)$wday [1] 4 > as.POSIXlt(d)$yday [1] 90 A common mistake is failing to add 1900 to the year, giving the impression you are living a long, long time ago: > as.POSIXlt(d)$year # Oops! [1] 110 > as.POSIXlt(d)$year + 1900 [1] 2010 7.14 Creating a Sequence of Dates Problem You want to create a sequence of dates, such as a sequence of daily, monthly, or annual dates. Solution The seq function is a generic function that has a version for Date objects. It can create a Date sequence similarly to the way it creates a sequence of numbers. Discussion A typical use of seq specifies a starting date (from), ending date (to), and increment (by). An increment of 1 indicates daily dates: > s <- as.Date("2012-01-01") > e <- as.Date("2012-02-01") > seq(from=s, to=e, by=1) # One month of dates [1] "2012-01-01" "2012-01-02" "2012-01-03" "2012-01-04" "2012-01-05" "2012-01-06" [7] "2012-01-07" "2012-01-08" "2012-01-09" "2012-01-10" "2012-01-11" "2012-01-12" [13] "2012-01-13" "2012-01-14" "2012-01-15" "2012-01-16" "2012-01-17" "2012-01-18" [19] "2012-01-19" "2012-01-20" "2012-01-21" "2012-01-22" "2012-01-23" "2012-01-24" [25] "2012-01-25" "2012-01-26" "2012-01-27" "2012-01-28" "2012-01-29" "2012-01-30" [31] "2012-01-31" "2012-02-01" Another typical use specifies a starting date (from), increment (by), and number of dates (length.out): > seq(from=s, by=1, length.out=7) # Dates, one week apart [1] "2012-01-01" "2012-01-02" "2012-01-03" "2012-01-04" "2012-01-05" "2012-01-06" [7] "2012-01-07" 7.14 Creating a Sequence of Dates | 175 D o wnload from Wow! eBook The increment (by) is flexible and can be specified in days, weeks, months, or years: > seq(from=s, by="month", length.out=12) # First of the month for one year [1] "2012-01-01" "2012-02-01" "2012-03-01" "2012-04-01" "2012-05-01" "2012-06-01" [7] "2012-07-01" "2012-08-01" "2012-09-01" "2012-10-01" "2012-11-01" "2012-12-01" > seq(from=s, by="3 months", length.out=4) # Quarterly dates for one year [1] "2012-01-01" "2012-04-01" "2012-07-01" "2012-10-01" > seq(from=s, by="year", length.out=10) # Year-start dates for one decade [1] "2012-01-01" "2013-01-01" "2014-01-01" "2015-01-01" "2016-01-01" "2017-01-01" [7] "2018-01-01" "2019-01-01" "2020-01-01" "2021-01-01" Be careful with by="month" near month-end. In this example, the end of February over- flows into March, which is probably not what you wanted: > seq(as.Date("2010-01-29"), by="month", len=3) [1] "2010-01-29" "2010-03-01" "2010-03-29" 176 | Chapter 7: Strings and Dates CHAPTER 8 Probability Introduction Probability theory is the foundation of statistics, and R has plenty of machinery for working with probability, probability distributions, and random variables. The recipes in this chapter show you how to calculate probabilities from quantiles, calculate quan- tiles from probabilities, generate random variables drawn from distributions, plot dis- tributions, and so forth. Names of Distributions R has an abbreviated name for every probability distribution. This name is used to identify the functions associated with the distribution. For example, the name of the Normal distribution is “norm”, which is the root of these function names: Function Purpose dnorm Normal density pnorm Normal distribution function qnorm Normal quantile function rnorm Normal random variates Table 8-1 describes some common discrete distributions, and Table 8-2 describes sev- eral common continuous distributions. Table 8-1. Discrete distributions Discrete distribution R name Parameters Binomial binom n = number of trials; p = probability of success for one trial Geometric geom p = probability of success for one trial Hypergeometric hyper m = number of white balls in urn; n = number of black balls in urn; k = number of balls drawn from urn 177 Discrete distribution R name Parameters Negative binomial (NegBinomial) nbinom size = number of successful trials; either prob = probability of successful trial or mu = mean Poisson pois lambda = mean Table 8-2. Continuous distributions Continuous distribution R name Parameters Beta beta shape1; shape2 Cauchy cauchy location; scale Chi-squared (Chisquare) chisq df = degrees of freedom Exponential exp rate F f df1 and df2 = degrees of freedom Gamma gamma rate; either rate or scale Log-normal (Lognormal) lnorm meanlog = mean on logarithmic scale; sdlog = standard deviation on logarithmic scale Logistic logis location; scale Normal norm mean; sd = standard deviation Student’s t (TDist) t df = degrees of freedom Uniform unif min = lower limit; max = upper limit Weibull weibull shape; scale Wilcoxon wilcox m = number of observations in first sample; n = number of observations in second sample All distribution-related functions require distributional parameters, such as size and prob for the binomial or prob for the geometric. The big “gotcha” is that the distributional parameters may not be what you expect. For example, I would expect the parameter of an exponential distribution to be β, the mean. The R convention, however, is for the exponential distribution to be defined by the rate = 1/β, so I often supply the wrong value. The moral is, study the help page before you use a function related to a distribution. Be sure you’ve got the parameters right. Getting Help on Probability Distributions To see the R functions related to a particular probability distribution, use the help command and the full name of the distribution. For example, this will show the func- tions related to the Normal distribution: > ?Normal 178 | Chapter 8: Probability Some distributions have names that don’t work well with the help command, such as “Student’s t”. They have special help names, as noted in Tables 8-1 and 8-2: NegBi- nomial, Chisquare, Lognormal, and TDist. Thus, to get help on the Student’s t distri- bution, use this: > ?TDist See Also There are many other distributions implemented in downloadable packages; see the CRAN task view devoted to probability distributions. The SuppDists package is part of the R base, and it includes ten supplemental distributions. The MASS package, which is also part of the base, provides additional support for distributions, such as maximum- likelihood fitting for some common distributions as well as sampling from a multivari- ate Normal distribution. 8.1 Counting the Number of Combinations Problem You want to calculate the number of combinations of n items taken k at a time. Solution Use the choose function: > choose(n, k) Discussion A common problem in computing probabilities of discrete variables is counting com- binations: the number of distinct subsets of size k that can be created from n items. The number is given by n!/r!(n − r)!, but it’s much more convenient to use the choose function—especially as n and k grow larger: > choose(5,3) # How many ways can we select 3 items from 5 items? [1] 10 > choose(50,3) # How many ways can we select 3 items from 50 items? [1] 19600 > choose(50,30) # How many ways can we select 30 items from 50 items? [1] 4.712921e+13 These numbers are also known as binomial coefficients. See Also This recipe merely counts the combinations; see Recipe 8.2 to actually generate them. 8.1 Counting the Number of Combinations | 179 8.2 Generating Combinations Problem You want to generate all combinations of n items taken k at a time. Solution Use the combn function: > combn(items, k) Discussion We can use combn(1:5,3) to generate all combinations of the numbers 1 through 5 taken three at a time: > combn(1:5,3) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] 1 1 1 1 1 1 2 2 2 3 [2,] 2 2 2 3 3 4 3 3 4 4 [3,] 3 4 5 4 5 5 4 5 5 5 The function is not restricted to numbers. We can generate combinations of strings, too. Here are all combinations of five treatments taken three at a time: > combn(c("T1","T2","T3","T4","T5"), 3) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] "T1" "T1" "T1" "T1" "T1" "T1" "T2" "T2" "T2" "T3" [2,] "T2" "T2" "T2" "T3" "T3" "T4" "T3" "T3" "T4" "T4" [3,] "T3" "T4" "T5" "T4" "T5" "T5" "T4" "T5" "T5" "T5" As the number of items, n, increases, the number of combinations can explode—especially if k is not near to 1 or n. See Also See Recipe 8.1 to count the number of possible combinations before you generate a huge set. 8.3 Generating Random Numbers Problem You want to generate random numbers. 180 | Chapter 8: Probability Solution The simple case of generating a uniform random number between 0 and 1 is handled by the runif function. This example generates one uniform random number: > runif(1) R can generate random variates from other distributions, however. For a given distri- bution, the name of the random number generator is “r” prefixed to the distribution’s abbreviated name (e.g., rnorm for the Normal distribution’s random number generator). This example generates one random value from the standard normal distribution: > rnorm(1) Discussion Most programming languages have a wimpy random number generator that generates one random number, uniformly distributed between 0.0 and 1.0, and that’s all. Not R. R can generate random numbers from many probability distributions other than the uniform distribution. The simple case of generating uniform random numbers between 0 and 1 is handled by the runif function: > runif(1) [1] 0.5119812 The argument of runif is the number of random values to be generated. Generating a vector of 10 such values is as easy as generating one: > runif(10) [1] 0.03475948 0.88950680 0.90005434 0.95689496 0.55829493 0.18407604 [7] 0.87814788 0.71057726 0.11140864 0.66392239 There are random number generators for all built-in distributions. Simply prefix the distribution name with “r” and you have the name of the corresponding random num- ber generator. Here are some common ones: > runif(1, min=-3, max=3) # One uniform variate between -3 and +3 [1] 2.954591 > rnorm(1) # One standard Normal variate [1] 1.048491 > rnorm(1, mean=100, sd=15) # One Normal variate, mean 100 and SD 15 [1] 108.7300 > rbinom(1, size=10, prob=0.5) # One binomial variate [1] 3 > rpois(1, lambda=10) # One Poisson variate [1] 13 > rexp(1, rate=0.1) # One exponential variate [1] 8.430267 > rgamma(1, shape=2, rate=0.1) # One gamma variate [1] 20.47334 8.3 Generating Random Numbers | 181 As with runif, the first argument is the number of random values to be generated. Subsequent arguments are the parameters of the distribution, such as mean and sd for the Normal distribution or size and prob for the binomial. See the function’s R help page for details. The examples given so far use simple scalars for distributional parameters. Yet the parameters can also be vectors, in which case R will cycle through the vector while generating random values. The following example generates three normal random val- ues drawn from distributions with means of −10, 0, and +10, respectively (all distri- butions have a standard deviation of 1.0): > rnorm(3, mean=c(-10,0,+10), sd=1) [1] -11.195667 2.615493 10.294831 That is a powerful capability in such cases as hierarchical models, where the parameters are themselves random. The next example calculates 100 draws of a normal variate whose mean is itself randomly distributed and with hyperparameters of μ = 0 and σ = 0.2: > means <- rnorm(100, mean=0, sd=0.2) > rnorm(100, mean=means, sd=1) [1] -0.410299581 1.030662055 -0.920933054 -0.572994026 0.984743043 [6] 1.251189879 0.873930251 0.509027943 -0.788626886 -0.113224062 [11] -0.035042586 0.150925067 0.634036678 1.627473761 -0.812021925 . . (etc.) . If you are generating many random values and the vector of parameters is too short, R will apply the Recycling Rule to the parameter vector. See Also See the “Introduction” to this chapter. 8.4 Generating Reproducible Random Numbers Problem You want to generate a sequence of random numbers, but you want to reproduce the same sequence every time your program runs. Solution Before running your R code, call the set.seed function to initialize the random number generator to a known state: > set.seed(666) # Or use any other positive integer 182 | Chapter 8: Probability Discussion After generating random numbers, you may often want to reproduce the same sequence of “random” numbers every time your program executes. That way, you get the same results from run to run. I once supported a complicated Monte Carlo analysis of a huge portfolio of securities. The users complained about getting a slightly different result each time the program ran. No kidding, I replied; it’s all based on random numbers, so of course there is randomness in the output. The solution was to set the random number generator to a known state at the beginning of the program. That way, it would generate the same (quasi-)random numbers each time and thus yield consistent, re- producible results. In R, the set.seed function sets the random number generator to a known state. The function takes one argument, an integer. Any positive integer will work, but you must use the same one in order to get the same initial state. The function returns nothing. It works behind the scenes, initializing (or reinitializing) the random number generator. The key here is that using the same seed restarts the random number generator back at the same place: > set.seed(165) # Initialize the random number generator to a known state > runif(10) # Generate ten random numbers [1] 0.1159132 0.4498443 0.9955451 0.6106368 0.6159386 0.4261986 0.6664884 [8] 0.1680676 0.7878783 0.4421021 > set.seed(165) # Reinitialize to the same known state > runif(10) # Generate the same ten "random" numbers [1] 0.1159132 0.4498443 0.9955451 0.6106368 0.6159386 0.4261986 0.6664884 [8] 0.1680676 0.7878783 0.4421021 When you set the seed value and freeze your sequence of random num- bers, you are eliminating a source of randomness that is critical to al- gorithms such as Monte Carlo simulations. Before you call set.seed in your application, ask yourself: Am I undercutting the value of my pro- gram or perhaps even damaging its logic? See Also See Recipe 8.3 for more about generating random numbers. 8.5 Generating a Random Sample Problem You want to sample a dataset randomly. 8.5 Generating a Random Sample | 183 Solution The sample function will randomly select n items from a vector: > sample(vec, n) Discussion Suppose your World Series data contains a vector of years when the Series was played. You can select 10 years at random using sample: > sample(world.series$year, 10) [1] 1906 1963 1966 1928 1905 1924 1961 1959 1927 1934 The items are randomly selected, so running sample again (usually) produces a different result: > sample(world.series$year, 10) [1] 1968 1947 1966 1916 1970 1961 1936 1913 1914 1958 The sample function normally samples without replacement, meaning it will not select the same item twice. Some statistical procedures (especially the bootstrap) require sampling with replacement, which means that one item can appear multiple times in the sample. Specify replace=TRUE to sample with replacement. It’s easy to implement a simple bootstrap using sampling with replacement. This code fragment repeatedly samples a dataset x and calculates the sample median: medians <- numeric(1000) for (i in 1:1000) { medians[i] <- median(sample(x, replace=TRUE)) } From the bootstrap estimates, we can estimate the confidence interval for the median: ci <- quantile(medians, c(0.025, 0.975)) cat("95% confidence interval is (", ci, ")\n") Typical output would be: 95% confidence interval is ( 1.642021 1.702843 ) See Also See Recipe 8.7 for randomly permuting a vector and Recipe 13.8 for more about bootstrapping. 8.6 Generating Random Sequences Problem You want to generate a random sequence, such as a series of simulated coin tosses or a simulated sequence of Bernoulli trials. 184 | Chapter 8: Probability Solution Use the sample function. Sample from the set of possible values, and set replace=TRUE: > sample(set, n, replace=TRUE) Discussion The sample function randomly selects items from a set. It normally samples without replacement, which means that it will not select the same item twice. With replace=TRUE, however, sample can select items over and over; this allows you to gen- erate long, random sequences of items. The following example generates a random sequence of 10 simulated flips of a coin: > sample(c("H","T"), 10, replace=TRUE) [1] "H" "H" "H" "T" "T" "H" "T" "H" "H" "T" The next example generates a sequence of 20 Bernoulli trials—random successes or failures. We use TRUE to signify a success: > sample(c(FALSE,TRUE), 20, replace=TRUE) [1] TRUE FALSE FALSE FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE TRUE [13] TRUE TRUE FALSE TRUE TRUE FALSE FALSE TRUE By default, sample will choose equally among the set elements and so the probability of selecting either TRUE or FALSE is 0.5. With a Bernoulli trial, the probability p of success is not necessarily 0.5. You can bias the sample by using the prob argument of sample; this argument is a vector of probabilities, one for each set element. Suppose we want to generate 20 Bernoulli trials with a probability of success p = 0.8. We set the proba- bility of FALSE to be 0.2 and the probability of TRUE to 0.8: > sample(c(FALSE,TRUE), 20, replace=TRUE, prob=c(0.2,0.8)) [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE TRUE [13] TRUE TRUE TRUE TRUE TRUE TRUE FALSE TRUE The resulting sequence is clearly biased toward TRUE. I chose this example because it’s a simple demonstration of a general technique. For the special case of a binary-valued sequence you can use rbinom, the random generator for binomial variates: > rbinom(10, 1, 0.8) [1] 1 0 1 1 1 1 1 1 1 1 8.7 Randomly Permuting a Vector Problem You want to generate a random permutation of a vector. Solution If v is your vector, then sample(v) returns a random permutation. 8.7 Randomly Permuting a Vector | 185 Discussion I associate the sample function with sampling from large datasets. However, the default parameters enable you to create a random rearrangement of the dataset. The function call sample(v) is equivalent to: sample(v, size=length(v), replace=FALSE) which means “select all the elements of v in random order while using each element exactly once. That is a random permutation.” Here is a random permutation of 1, ..., 10: > sample(1:10) [1] 5 8 7 4 3 9 2 6 1 10 See Also See Recipe 8.5 for more about sample. 8.8 Calculating Probabilities for Discrete Distributions Problem You want to calculate either the simple or the cumulative probability associated with a discrete random variable. Solution For a simple probability, P(X = x), use the density function. All built-in probability distributions have a density function whose name is “d” prefixed to the distribution name—for example, dbinom for the binomial distribution. For a cumulative probability, P(X ≤ x), use the distribution function. All built-in prob- ability distributions have a distribution function whose name is “p” prefixed to the distribution name; thus, pbinom is the distribution function for the binomial distribution. Discussion Suppose we have a binomial random variable X over 10 trials, where each trial has a success probability of 1/2. Then we can calculate the probability of observing x = 7 by calling dbinom: > dbinom(7, size=10, prob=0.5) [1] 0.1171875 That calculates a probability of about 0.117. R calls dbinom the density function. Some textbooks call it the probability mass function or the probability function. Calling it a density function keeps the terminology consistent between discrete and continuous distributions (Recipe 8.9). 186 | Chapter 8: Probability The cumulative probability, P(X ≤ x), is given by the distribution function, which is sometimes called the cumulative probability function. The distribution function for the binomial distribution is pbinom. Here is the cumulative probability for x = 7 (i.e., P(X ≤ 7)): > pbinom(7, size=10, prob=0.5) [1] 0.9453125 It appears the probability of observing X ≤ 7 is about 0.945. Here are the density functions and distribution functions for some common discrete distributions: Distribution Density function: P(X = x) Distribution function: P(X ≤ x) Binomial dbinom(x, size, prob) pbinom(x, size, prob) Geometric dgeom(x, prob) pgeom(x, prob) Poisson dpois(x, lambda) ppois(x, lambda) The complement of the cumulative probability is the survival function, P(X > x). All of the distribution functions let you find this right-tail probability simply by specifying lower.tail=FALSE: > pbinom(7, size=10, prob=0.5, lower.tail=FALSE) [1] 0.0546875 Thus we see that the probability of observing X > 7 is about 0.055. The interval probability, P(x1 < X ≤ x2), is the probability of observing X between the limits x1 and x2. It is simply calculated as the difference between two cumulative prob- abilities: P(X ≤ x2) − P(X ≤ x1). Here is P(3 < X ≤ 7) for our binomial variable: > pbinom(7,size=10,prob=0.5) - pbinom(3,size=10,prob=0.5) [1] 0.7734375 R lets you specify multiple values of x for these functions and will return a vector of the corresponding probabilities. Here we calculate two cumulative probabilities, P(X ≤ 3) and P(X ≤ 7), in one call to pbinom: > pbinom(c(3,7), size=10, prob=0.5) [1] 0.1718750 0.9453125 This leads to a one-liner for calculating interval probabilities. The diff function cal- culates the difference between successive elements of a vector. We apply it to the output of pbinom to obtain the difference in cumulative probabilities—in other words, the in- terval probability: > diff(pbinom(c(3,7), size=10, prob=0.5)) [1] 0.7734375 8.8 Calculating Probabilities for Discrete Distributions | 187 See Also See this chapter’s “Introduction” for more about the built-in probability distributions. 8.9 Calculating Probabilities for Continuous Distributions Problem You want to calculate the distribution function (DF) or cumulative distribution func- tion (CDF) for a continuous random variable. Solution Use the distribution function, which calculates P(X ≤ x). All built-in probability dis- tributions have a distribution function whose name is “p” prefixed to the distribution’s abbreviated name—for instance, pnorm for the Normal distribution. Discussion The R functions for probability distributions follow a consistent pattern, so the solution to this recipe is essentially identical to the solution for discrete random variables (Recipe 8.8). The significant difference is that continuous variables have no “proba- bility” at a single point, P(X = x). Instead, they have a density at a point. Given that consistency, the discussion of distribution functions in Recipe 8.8 is applicable here, too. The following table gives the distribution function for several continuous distributions: Distribution Distribution function: P(X ≤ x) Normal pnorm(x, mean, sd) Student’s t pt(x, df) Exponential pexp(x, rate) Gamma pgamma(x, shape, rate) Chi-squared (χ2) pchisq(x, df) We can use pnorm to calculate the probability that a man is shorter than 66 inches, assuming that men’s heights are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. Mathematically speaking, we want P(X ≤ 66) given that X ~ N(70, 3): > pnorm(66, mean=70, sd=3) [1] 0.09121122 188 | Chapter 8: Probability Likewise, we can use pexp to calculate the probability that an exponential variable with a mean of 40 could be less than 20: > pexp(20, rate=1/40) [1] 0.3934693 Just as for discrete probabilities, the functions for continuous probabilities use lower.tail=FALSE to specify the survival function, P(X > x). This call to pexp gives the probability that the same exponential variable could be greater than 50: > pexp(50, rate=1/40, lower.tail=FALSE) [1] 0.2865048 Also like discrete probabilities, the interval probability for a continuous variable, P(x1 < X < x2), is computed as the difference between two cumulative probabilities, P(X < x2) − P(X < x1). For the same exponential variable, here is P(20 < X < 50), the probability that it could fall between 20 and 50: > pexp(50,rate=1/40) - pexp(20,rate=1/40) [1] 0.3200259 See Also See this chapter’s “Introduction” for more about the built-in probability distributions. 8.10 Converting Probabilities to Quantiles Problem Given a probability p and a distribution, you want to determine the corresponding quantile for p: the value x such that P(X ≤ x) = p. Solution Every built-in distribution includes a quantile function that converts probabilities to quantiles. The function’s name is “q” prefixed to the distribution name; thus, for in- stance, qnorm is the quantile function for the Normal distribution. The first argument of the quantile function is the probability. The remaining arguments are the distribution’s parameters, such as mean, shape, or rate: > qnorm(0.05, mean=100, sd=15) [1] 75.3272 Discussion A common example of computing quantiles is when we compute the limits of a con- fidence interval. If we want to know the 95% confidence interval (α = 0.05) of a standard normal variable, then we need the quantiles with probabilities of α/2 = 0.025 and (1 − α)/2 = 0.975: 8.10 Converting Probabilities to Quantiles | 189 > qnorm(0.025) [1] -1.959964 > qnorm(0.975) [1] 1.959964 In the true spirit of R, the first argument of the quantile functions can be a vector of probabilities, in which case we get a vector of quantiles. We can simplify this example into a one-liner: > qnorm(c(0.025,0.975)) [1] -1.959964 1.959964 All the built-in probability distributions provide a quantile function. Here are the quantile functions for some common discrete distributions: Distribution Quantile function Binomial qbinom(p, size, prob) Geometric qgeom(p, prob) Poisson qpois(p, lambda) And here are the quantile functions for common continuous distributions: Distribution Quantile function Normal qnorm(p, mean, sd) Student’s t qt(p, df) Exponential qexp(p, rate) Gamma qgamma(p, shape, rate=rate) or qgamma(p, shape, scale=scale) Chi-squared (χ2) qchisq(p, df) See Also Determining the quantiles of a data set is different from determining the quantiles of a distribution—see Recipe 9.5. 8.11 Plotting a Density Function Problem You want to plot the density function of a probability distribution. Solution Define a vector x over the domain. Apply the distribution’s density function to x and then plot the result. This code snippet plots the standard normal distribution: 190 | Chapter 8: Probability > x <- seq(from=-3, to=+3, length.out=100) > plot(x, dnorm(x)) Discussion All the built-in probability distributions include a density function. For a particular density, the function name is “d” prepended to the density name. The density function for the Normal distribution is dnorm, the density for the gamma distribution is dgamma, and so forth. If the first argument of the density function is a vector, then the function calculates the density at each point and returns the vector of densities. The following code creates a 2 × 2 plot of four densities, as shown in Figure 8-1: x <- seq(from=0, to=6, length.out=100) # Define the density domains ylim <- c(0, 0.6) par(mfrow=c(2,2)) # Create a 2x2 plotting area plot(x, dunif(x,min=2,max=4), main="Uniform", # Plot a uniform density type='l', ylim=ylim) plot(x, dnorm(x,mean=3,sd=1), main="Normal", # Plot a Normal density type='l', ylim=ylim) plot(x, dexp(x,rate=1/2), main="Exponential", # Plot an exponential density type='l', ylim=ylim) plot(x, dgamma(x,shape=2,rate=1), main="Gamma", # Plot a gamma density type='l', ylim=ylim) A raw density plot is rarely useful or interesting by itself, and we usually shade a region of interest. Figure 8-2 shows a standard normal distribution with shading for the region of 1 ≤ z ≤ 2. We create the plot by first plotting the density and then shading a region with the polygon function. The polygon function draws a series of lines around the highlighted area and fills it. First, we draw the density curve: x <- seq(from=-3, to=+3, length.out=100) y <- dnorm(x) plot(x, y, main="Standard Normal Distribution", type='l', ylab="Density", xlab="Quantile") abline(h=0) Next, we define the region of interest by a series of line segments. The line segments, in turn, are defined by a series of (x, y) coordinates. The polygon function will connect the first and last (x, y) points to close the polygon: # The body of the polygon follows the density curve where 1 <= z <= 2 region.x <- x[1 <= x & x <= 2] region.y <- y[1 <= x & x <= 2] # We add initial and final segments, which drop down to the Y axis 8.11 Plotting a Density Function | 191 region.x <- c(region.x[1], region.x, tail(region.x,1)) region.y <- c( 0, region.y, 0) Finally, we call polygon to plot the boundary of the region and fill it: polygon(region.x, region.y, density=10) By default, polygon does not fill the region. Setting density=10 turns on filling with thin lines at a 45° angle. To fill with a color instead, we would set density to −1 and set col to the desired color: polygon(region.x, region.y, density=-1, col="red") Figure 8-1. Plotting density functions 192 | Chapter 8: Probability Figure 8-2. Density plot with shading 8.11 Plotting a Density Function | 193 CHAPTER 9 General Statistics Introduction Any significant application of R includes statistics or models or graphics. This chapter addresses the statistics. Some recipes simply describe how to calculate a statistic, such as relative frequency. Most recipes involve statistical tests or confidence intervals. The statistical tests let you choose between two competing hypotheses; that paradigm is described next. Confidence intervals reflect the likely range of a population parameter and are calculated based on your data sample. Null Hypotheses, Alternative Hypotheses, and p-Values Many of the statistical tests in this chapter use a time-tested paradigm of statistical inference. In the paradigm, we have one or two data samples. We also have two com- peting hypotheses, either of which could reasonably be true. One hypothesis, called the null hypothesis, is that nothing happened: the mean was unchanged; the treatment had no effect; you got the expected answer; the model did not improve; and so forth. The other hypothesis, called the alternative hypothesis, is that something happened: the mean rose; the treatment improved the patients’ health; you got an unexpected answer; the model fit better; and so forth. We want to determine which hypothesis is more likely in light of the data: 1. To begin, we assume that the null hypothesis is true. 2. We calculate a test statistic. It could be something simple, such as the mean of the sample, or it could be quite complex. The critical requirement is that we must know the statistic’s distribution. We might know the distribution of the sample mean, for example, by invoking the Central Limit Theorem. 3. From the statistic and its distribution we can calculate a p-value, the probability of a test statistic value as extreme or more extreme than the one we observed, while assuming that the null hypothesis is true. 195 4. If the p-value is too small, we have strong evidence against the null hypothesis. This is called rejecting the null hypothesis. 5. If the p-value is not small then we have no such evidence. This is called failing to reject the null hypothesis. There is one necessary decision here: When is a p-value “too small”? In this book, I follow the common convention that we reject the null hypothesis when p < 0.05 and fail to reject it when p > 0.05. In statistical terminology, I chose a significance level of α = 0.05 to define the border between strong evidence and insufficient evidence against the null hypothesis. But the real answer is, “it depends”. Your chosen significance level depends on your problem domain. The conventional limit of p < 0.05 works for many problems. In my work, the data is especially noisy and so I am often satisfied with p < 0.10. For someone working in high-risk areas, p < 0.01 or p < 0.001 might be necessary. In the recipes, I mention which tests include a p-value so that you can compare the p-value against your chosen significance level of α. I worded the recipes to help you interpret the comparison. Here is the wording from Recipe 9.4, a test for the inde- pendence of two factors: Conventionally, a p-value of less than 0.05 indicates that the variables are likely not independent whereas a p-value exceeding 0.05 fails to provide any such evidence. This is a compact way of saying: • The null hypothesis is that the variables are independent. • The alternative hypothesis is that the variables are not independent. • For α = 0.05, if p < 0.05 then we reject the null hypothesis, giving strong evidence that the variables are not independent; if p > 0.05, we fail to reject the null hypothesis. • You are free to choose your own α, of course, in which case your decision to reject or fail to reject might be different. Remember, the recipe states the informal interpretation of the test results, not the rig- orous mathematical interpretation. I use colloquial language in the hope that it will guide you toward a practical understanding and application of the test. If the precise semantics of hypothesis testing is critical for your work, I urge you to consult the ref- erence cited under “See Also” on page 197 or one of the other fine textbooks on math- ematical statistics. 196 | Chapter 9: General Statistics Confidence Intervals Hypothesis testing is a well-understood mathematical procedure, but it can be frus- trating. First, the semantics is tricky. The test does not reach a definite, useful conclu- sion. You might get strong evidence against the null hypothesis, but that’s all you’ll get. Second, it does not give you a number, only evidence. If you want numbers then use confidence intervals, which bound the estimate of a population parameter at a given level of confidence. Recipes in this chapter can calcu- late confidence intervals for means, medians, and proportions of a population. For example, Recipe 9.9 calculates a 95% confidence interval for the population mean based on sample data. The interval is 97.16 < μ < 103.98, which means there is a 95% probability that the population’s mean, μ, is between 97.16 and 103.98. See Also Statistical terminology and conventions can vary. This book generally follows the con- ventions of Mathematical Statistics with Applications, 6th ed., by Wackerly et al. (Dux- bury Press). I recommend this book also for learning more about the statistical tests described in this chapter. 9.1 Summarizing Your Data Problem You want a basic statistical summary of your data. Solution The summary function gives some useful statistics for vectors, matrices, factors, and data frames: > summary(vec) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.1803 1.1090 1.9520 2.1750 2.6920 7.3290 Discussion The Solution exhibits the summary of a vector. The 1st Qu. and 3rd Qu. are the first and third quartile, respectively. Having both the median and mean is useful because you can quickly detect skew. The Solution, for example, shows a mean that is larger than the median; this indicates a possible skew to the right. 9.1 Summarizing Your Data | 197 D o wnload from Wow! eBook The summary of a matrix works column by column. Here we see the summary of a matrix, mat, with three columns named Samp1, Samp2, and Samp3: > summary(mat) Samp1 Samp2 Samp3 Min. :0.1803 Min. : 0.1378 Min. :0.1605 1st Qu.:1.1094 1st Qu.: 1.2686 1st Qu.:1.0056 Median :1.9517 Median : 1.8733 Median :1.8466 Mean :2.1748 Mean : 2.2436 Mean :2.0562 3rd Qu.:2.6916 3rd Qu.: 2.9373 3rd Qu.:2.7666 Max. :7.3289 Max. :11.4175 Max. :6.3870 The summary of a factor gives counts: > summary(fac) Yes No Maybe 44 20 36 The summary of a data frame incorporates all these features. It works column by col- umn, giving an appropriate summary according to the column type. Numeric values receive a statistical summary and factors are counted (character strings are not summarized): > summary(suburbs) city county state pop Length:16 Cook :7 IL:12 Min. : 63348 Class :character Kane :2 IN: 2 1st Qu.: 73833 Mode :character Lake(IN):2 WI: 2 Median : 86700 DuPage :1 Mean : 265042 Kendall :1 3rd Qu.: 103615 Kenosha :1 Max. :2853114 (Other) :2 The “summary” of a list is pretty funky: just the data type of each list member. Here is a summary of a list of vectors: > summary(vec.list) Length Class Mode Samp1 100 -none- numeric Samp2 100 -none- numeric Samp3 100 -none- numeric To summarize a list of vectors, apply summary to each list element: > lapply(vec.list, summary) $Samp1 Min. 1st Qu. Median Mean 3rd Qu. Max. 0.1803 1.1090 1.9520 2.1750 2.6920 7.3290 $Samp2 Min. 1st Qu. Median Mean 3rd Qu. Max. 0.1378 1.2690 1.8730 2.2440 2.9370 11.4200 $Samp3 Min. 1st Qu. Median Mean 3rd Qu. Max. 0.1605 1.0060 1.8470 2.0560 2.7670 6.3870 198 | Chapter 9: General Statistics Unfortunately, the summary function does not compute any measure of variability, such as standard deviation or median absolute deviation. This is a serious shortcoming, so I usually call sd or mad right after calling summary. See Also See Recipes 2.6 and 6.2. 9.2 Calculating Relative Frequencies Problem You want to count the relative frequency of certain observations in your sample. Solution Identify the interesting observations by using a logical expression; then use the mean function to calculate the fraction of observations it identifies. For example, given a vector x, you can find the relative frequency of positive values in this way: > mean(x > 0) Discussion A logical expression, such as x > 0, produces a vector of logical values (TRUE and FALSE), one for each element of x. The mean function converts those values to 1s and 0s, respectively, and computes the average. This gives the fraction of values that are TRUE—in other words, the relative frequency of the interesting values. In the Solution, for example, that’s the relative frequency of positive values. The concept here is pretty simple. The tricky part is dreaming up a suitable logical expression. Here are some examples: mean(lab == “NJ”) Fraction of lab values that are New Jersey mean(after > before) Fraction of observations for which the effect increases mean(abs(x-mean(x)) > 2*sd(x)) Fraction of observations that exceed two standard deviations from the mean mean(diff(ts) > 0) Fraction of observations in a time series that are larger than the previous observation 9.2 Calculating Relative Frequencies | 199 9.3 Tabulating Factors and Creating Contingency Tables Problem You want to tabulate one factor or to build a contingency table from multiple factors. Solution The table function produces counts of one factor: > table(f) It can also produce contingency tables (cross-tabulations) from two or more factors: > table(f1, f2) Discussion The table function counts the levels of one factor, such as these counts of initial and outcome (which are factors): > table(initial) initial Yes No Maybe 37 36 27 > table(outcome) outcome Fail Pass 47 53 The greater power of table is in producing contingency tables, also known as cross- tabulations. Each cell in a contingency table counts how many times that row–column combination occurred: > table(initial, outcome) outcome initial Fail Pass Yes 13 24 No 24 12 Maybe 10 17 This table shows that the combination of initial = Yes and outcome = Fail occurred 13 times, the combination of initial = Yes and outcome = Pass occurred 24 times, and so forth. See Also The xtabs function can also produce a contingency table. It has a formula interface, which some people prefer. 200 | Chapter 9: General Statistics 9.4 Testing Categorical Variables for Independence Problem You have two categorical variables that are represented by factors. You want to test them for independence using the chi-squared test. Solution Use the table function to produce a contingency table from the two factors. Then use the summary function to perform a chi-squared test of the contingency table: > summary(table(fac1,fac2)) The output includes a p-value. Conventionally, a p-value of less than 0.05 indicates that the variables are likely not independent whereas a p-value exceeding 0.05 fails to provide any such evidence. Discussion This example performs a chi-squared test on the contingency table of Recipe 9.3 and yields a p-value of 0.01255: > summary(table(initial,outcome)) Number of cases in table: 100 Number of factors: 2 Test for independence of all factors: Chisq = 8.757, df = 2, p-value = 0.01255 The small p-value indicates that the two factors, initial and outcome, are probably not independent. Practically speaking, we conclude there is some connection between the variables. See Also The chisq.test function can also perform this test. 9.5 Calculating Quantiles (and Quartiles) of a Dataset Problem Given a fraction f, you want to know the corresponding quantile of your data. That is, you seek the observation x such that the fraction of observations below x is f. Solution Use the quantile function. The second argument is the fraction, f: > quantile(vec, f) 9.5 Calculating Quantiles (and Quartiles) of a Dataset | 201 For quartiles, simply omit the second argument altogether: > quantile(vec) Discussion Suppose vec contains 1,000 observations between 0 and 1. The quantile function can tell you which observation delimits the lower 5% of the data: > quantile(vec, .05) 5% 0.04575003 The quantile documentation refers to the second argument as a “probability”, which is natural when we think of probability as meaning relative frequency. In true R style, the second argument can be a vector of probabilities; in this case, quantile returns a vector of corresponding quantiles, one for each probability: > quantile(vec, c(.05, .95)) 5% 95% 0.04575003 0.95122306 That is a handy way to identify the middle 90% (in this case) of the observations. If you omit the probabilities altogether then R assumes you want the probabilities 0, 0.25, 0.50, 0.75, and 1.0—in other words, the quartiles: > quantile(vec) 0% 25% 50% 75% 100% 0.001285589 0.260075658 0.479866042 0.734801500 0.997817661 Amazingly, the quantile function implements nine (yes, nine) different algorithms for computing quantiles. Study the help page before assuming that the default algorithm is the best one for you. 9.6 Inverting a Quantile Problem Given an observation x from your data, you want to know its corresponding quantile. That is, you want to know what fraction of the data is less than x. Solution Assuming your data is in a vector vec, compare the data against the observation and then use mean to compute the relative frequency of values less than x: > mean(vec < x) 202 | Chapter 9: General Statistics Discussion The expression vec < x compares every element of vec against x and returns a vector of logical values, where the nth logical value is TRUE if vec[n] < x. The mean function converts those logical values to 0 and 1: 0 for FALSE and 1 for TRUE. The average of all those 1s and 0s is the fraction of vec that is less than x, or the inverse quantile of x. See Also This is an application of the general approach described in Recipe 9.2. 9.7 Converting Data to Z-Scores Problem You have a dataset, and you want to calculate the corresponding z-scores for all data elements. (This is sometimes called normalizing the data.) Solution Use the scale function: > scale(x) This works for vectors, matrices, and data frames. In the case of a vector, scale returns the vector of normalized values. In the case of matrices and data frames, scale nor- malizes each column independently and returns columns of normalized values in a matrix. Discussion You might also want to normalize a single value y relative to a dataset x. That can be done by using vectorized operations as follows: > (y - mean(x)) / sd(x) 9.8 Testing the Mean of a Sample (t Test) Problem You have a sample from a population. Given this sample, you want to know if the mean of the population could reasonably be a particular value m. Solution Apply the t.test function to the sample x with the argument mu=m: > t.test(x, mu=m) 9.8 Testing the Mean of a Sample (t Test) | 203 The output includes a p-value. Conventionally, if p < 0.05 then the population mean is unlikely to be m whereas p > 0.05 provides no such evidence. If your sample size n is small, then the underlying population must be normally dis- tributed in order to derive meaningful results from the t test. A good rule of thumb is that “small” means n < 30. Discussion The t test is a workhorse of statistics, and this is one of its basic uses: making inferences about a population mean from a sample. The following example simulates sampling from a normal population with mean μ = 100. It uses the t test to ask if the population mean could be 95, and t.test reports a p-value of 0.001897: > x <- rnorm(50, mean=100, sd=15) > t.test(x, mu=95) One Sample t-test data: x t = 3.2832, df = 49, p-value = 0.001897 alternative hypothesis: true mean is not equal to 95 95 percent confidence interval: 97.16167 103.98297 sample estimates: mean of x 100.5723 The p-value is small and so it’s unlikely (based on the sample data) that 95 could be the mean of the population. Informally, we could interpret the low p-value as follows. If the population mean were really 95, then the probability of observing our test statistic (t = 3.2832 or something more extreme) would be only 0.001897. That is very improbable, yet that is the value we observed. Hence we conclude that the null hypothesis is wrong; therefore, the sam- ple data does not support the claim that the population mean is 95. In sharp contrast, testing for a mean of 100 gives a p-value of 0.7374: > t.test(x, mu=100) One Sample t-test data: x t = 0.3372, df = 49, p-value = 0.7374 alternative hypothesis: true mean is not equal to 100 95 percent confidence interval: 97.16167 103.98297 sample estimates: mean of x 100.5723 204 | Chapter 9: General Statistics The large p-value indicates that the sample is consistent with assuming a population mean μ of 100. In statistical terms, the data does not provide evidence against the true mean being 100. A common case is testing for a mean of zero. If you omit the mu argument, it defaults to zero. See Also The t.test is a many-splendored thing. See Recipes 9.9 and 9.15 for other uses. 9.9 Forming a Confidence Interval for a Mean Problem You have a sample from a population. Given that sample, you want to determine a confidence interval for the population’s mean. Solution Apply the t.test function to your sample x: > t.test(x) The output includes a confidence interval at the 95% confidence level. To see intervals at other levels, use the conf.level argument. As in Recipe 9.8, if your sample size n is small then the underlying population must be normally distributed for there to be a meaningful confidence interval. Again, a good rule of thumb is that “small” means n < 30. Discussion Applying the t.test function to a vector yields a lot of output. Buried in the output is a confidence interval: > x <- rnorm(50, mean=100, sd=15) > t.test(x) One Sample t-test data: x t = 59.2578, df = 49, p-value < 2.2e-16 alternative hypothesis: true mean is not equal to 0 95 percent confidence interval: 97.16167 103.98297 sample estimates: mean of x 100.5723 9.9 Forming a Confidence Interval for a Mean | 205 In this example, the confidence interval is approximately 97.16 < μ < 103.98, which is sometimes written simply as (97.16, 103.98). We can raise the confidence level to 99% by setting conf.level=0.99: > t.test(x, conf.level=0.99) One Sample t-test data: x t = 59.2578, df = 49, p-value < 2.2e-16 alternative hypothesis: true mean is not equal to 0 99 percent confidence interval: 96.0239 105.1207 sample estimates: mean of x 100.5723 That change widens the confidence interval to 96.02 < μ < 105.12. 9.10 Forming a Confidence Interval for a Median Problem You have a data sample, and you want to know the confidence interval for the median. Solution Use the wilcox.test function, setting conf.int=TRUE: > wilcox.test(x, conf.int=TRUE) The output will contain a confidence interval for the median. Discussion The procedure for calculating the confidence interval of a mean is well-defined and widely known. The same is not true for the median, unfortunately. There are several procedures for calculating the median’s confidence interval. None of them is “the” procedure, as far as I know, but the Wilcoxon signed rank test is pretty standard. The wilcox.test function implements that procedure. Buried in the output is the 95% confidence interval, which is approximately (0.424, 0.892) in this case: > wilcox.test(x, conf.int=TRUE) Wilcoxon signed rank test data: x V = 465, p-value = 1.863e-09 alternative hypothesis: true location is not equal to 0 95 percent confidence interval: 0.4235421 0.8922106 206 | Chapter 9: General Statistics sample estimates: (pseudo)median 0.6249297 You can change the confidence level by setting conf.level, such as conf.level=0.99 or other such values. The output also includes something called the pseudomedian, which is defined on the help page. Don’t assume it equals the median; they are different: > median(x) [1] 0.547129 See Also The bootstrap procedure is also useful for estimating the median’s confidence interval; see Recipes 8.5 and 13.8. 9.11 Testing a Sample Proportion Problem You have a sample of values from a population consisting of successes and failures. You believe the true proportion of successes is p, and you want to test that hypothesis using the sample data. Solution Use the prop.test function. Suppose the sample size is n and the sample contains x successes: > prop.test(x, n, p) The output includes a p-value. Conventionally, a p-value of less than 0.05 indicates that the true proportion is unlikely to be p whereas a p-value exceeding 0.05 fails to provide such evidence. Discussion Suppose you encounter some loudmouthed fan of the Chicago Cubs early in the base- ball season. The Cubs have played 20 games and won 11 of them, or 55% of their games. Based on that evidence, the fan is “very confident” that the Cubs will win more than half of their games this year. Should he be that confident? The prop.test function can evaluate the fan’s logic. Here, the number of observations is n = 20, the number of successes is x = 11, and p is the true probability of winning a game. We want to know whether it is reasonable to conclude, based on the data, that p > 0.5. Normally, prop.test would check for p ≠ 0.05 but we can check for p > 0.5 instead by setting alternative="greater": 9.11 Testing a Sample Proportion | 207 > prop.test(11, 20, 0.5, alternative="greater") 1-sample proportions test with continuity correction data: 11 out of 20, null probability 0.5 X-squared = 0.05, df = 1, p-value = 0.4115 alternative hypothesis: true p is greater than 0.5 95 percent confidence interval: 0.3496150 1.0000000 sample estimates: p 0.55 The prop.test output shows a large p-value, 0.4115, so we cannot reject the null hy- pothesis; that is, we cannot reasonably conclude that p is greater than 1/2. The Cubs fan is being overly confident based on too little data. No surprise there. 9.12 Forming a Confidence Interval for a Proportion Problem You have a sample of values from a population consisting of successes and failures. Based on the sample data, you want to form a confidence interval for the population’s proportion of successes. Solution Use the prop.test function. Suppose the sample size is n and the sample contains x successes: > prop.test(n, x) The function output includes the confidence interval for p. Discussion I subscribe to a stock market newsletter that is well written for the most part but in- cludes a section purporting to identify stocks that are likely to rise. It does this by looking for a certain pattern in the stock price. It recently reported, for example, that a certain stock was following the pattern. It also reported that the stock rose six times after the last nine times that pattern occurred. The writers concluded that the proba- bility of the stock rising again was therefore 6/9 or 66.7%. Using prop.test, we can obtain the confidence interval for the true proportion of times the stock rises after the pattern. Here, the number of observations is n = 9 and the number of successes is x = 6. The output shows a confidence interval of (0.309, 0.910) at the 95% confidence level: 208 | Chapter 9: General Statistics > prop.test(6, 9) 1-sample proportions test with continuity correction data: 6 out of 9, null probability 0.5 X-squared = 0.4444, df = 1, p-value = 0.505 alternative hypothesis: true p is not equal to 0.5 95 percent confidence interval: 0.3091761 0.9095817 sample estimates: p 0.6666667 The writers are pretty foolish to say the probability of rising is 66.7%. They could be leading their readers into a very bad bet. By default, prop.test calculates a confidence interval at the 95% confidence level. Use the conf.level argument for other confidence levels: > prop.test(n, x, p, conf.level=0.99) # 99% confidence level See Also See Recipe 9.11. 9.13 Testing for Normality Problem You want a statistical test to determine whether your data sample is from a normally distributed population. Solution Use the shapiro.test function: > shapiro.test(x) The output includes a p-value. Conventionally, p < 0.05 indicates that the population is likely not normally distributed whereas p > 0.05 provides no such evidence. Discussion This example reports a p-value of 0.4151 for x: > shapiro.test(x) Shapiro-Wilk normality test data: x W = 0.9651, p-value = 0.4151 9.13 Testing for Normality | 209 The large p-value suggests the underlying population could be normally distributed. The next example reports a small p-value for y, so it is unlikely that this sample came from a normal population: > shapiro.test(y) Shapiro-Wilk normality test data: y W = 0.9503, p-value = 0.03520 I have highlighted the Shapiro–Wilk test because it is a standard R function. You can also install the package nortest, which is dedicated entirely to tests for normality. This package includes: • Anderson–Darling test (ad.test) • Cramer–von Mises test (cvm.test) • Lilliefors test (lillie.test) • Pearson chi-squared test for the composite hypothesis of normality (pearson.test) • Shapiro–Francia test (sf.test) The problem with all these tests is their null hypothesis: they all assume that the pop- ulation is normally distributed until proven otherwise. As a result, the population must be decidedly nonnormal before the test reports a small p-value and you can reject that null hypothesis. That makes the tests quite conservative, tending to err on the side of normality. Instead of depending solely upon a statistical test, I suggest also using histograms (Recipe 10.18) and quantile-quantile plots (Recipe 10.21) to evaluate the normality of any data. Are the tails too fat? Is the peak to peaked? Your judgment is likely better than a single statistical test. See Also See Recipe 3.9 for how to install the nortest package. 9.14 Testing for Runs Problem Your data is a sequence of binary values: yes–no, 0–1, true–false, or other two-valued data. You want to know: Is the sequence random? 210 | Chapter 9: General Statistics Solution The tseries package contains the runs.test function, which checks a sequence for randomness. The sequence should be a factor with two levels: > library(tseries) > runs.test(as.factor(s)) The runs.test function reports a p-value. Conventionally, a p-value of less than 0.05 indicates that the sequence is likely not random whereas a p-value exceeding 0.05 pro- vides no such evidence. Discussion A run is a subsequence composed of identical values, such as all 1s or all 0s. A random sequence should be properly jumbled up, without too many runs. Similarly, it shouldn’t contain too few runs, either. A sequence of perfectly alternating values (0, 1, 0, 1, 0, 1, ...) contains no runs, but would you say that it’s random? The runs.test function checks the number of runs in your sequence. If there are too many or too few, it reports a small p-value. This first example generates a random sequence of 0s and 1s and then tests the sequence for runs. Not surprisingly, runs.test reports a large p-value, indicating the sequence is likely random: > library(tseries) > s <- sample(c(0,1), 100, replace=T) > runs.test(as.factor(s)) Runs Test data: as.factor(s) Standard Normal = 0.2175, p-value = 0.8279 alternative hypothesis: two.sided This next sequence, however, consists of three runs and so the reported p-value is quite low: > s <- c(0,0,0,0,1,1,1,1,0,0,0,0) > runs.test(as.factor(s)) Runs Test data: as.factor(s) Standard Normal = -2.2997, p-value = 0.02147 alternative hypothesis: two.sided See Also See Recipes 5.4 and 8.6. 9.14 Testing for Runs | 211 D o wnload from Wow! eBook 9.15 Comparing the Means of Two Samples Problem You have one sample each from two populations. You want to know if the two popu- lations could have the same mean. Solution Perform a t test by calling the t.test function: > t.test(x, y) By default, t.test assumes that your data are not paired. If the observations are paired (i.e., if each xi is paired with one yi), then specify paired=TRUE: > t.test(x, y, paired=TRUE) In either case, t.test will compute a p-value. Conventionally, if p < 0.05 then the means are likely different whereas p > 0.05 provides no such evidence: • If either sample size is small, then the populations must be normally distributed. Here, “small” means fewer than 20 data points. • If the two populations have the same variance, specify var.equal=TRUE to obtain a less conservative test. Discussion I often use the t test to get a quick sense of the difference between two population means. It requires that the samples be large enough (both samples have 20 or more observations) or that the underlying populations be normally distributed. I don’t take the “normally distributed” part too literally. Being bell-shaped should be good enough. A key distinction here is whether or not your data contains paired observations, since the results may differ in the two cases. Suppose we want to know if coffee in the morning improves scores on SAT tests. We could run the experiment two ways: 1. Randomly select one group of people. Give them the SAT test twice, once with morning coffee and once without morning coffee. For each person, we will have two SAT scores. These are paired observations. 2. Randomly select two groups of people. One group has a cup of morning coffee and takes the SAT test. The other group just takes the test. We have a score for each person, but the scores are not paired in any way. Statistically, these experiments are quite different. In experiment 1, there are two observations for each person (caffeinated and decaf) and they are not statistically in- dependent. In experiment 2, the data are independent. 212 | Chapter 9: General Statistics If you have paired observations (experiment 1) and erroneously analyze them as un- paired observations (experiment 2), then you could get this result with a p-value of 0.9867: > t.test(x,y) Welch Two Sample t-test data: x and y t = -0.0166, df = 198, p-value = 0.9867 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -30.56737 30.05605 sample estimates: mean of x mean of y 501.2008 501.4565 The large p-value forces you to conclude there is no difference between the groups. Contrast that result with the one that follows from analyzing the same data but correctly identifying it as paired: > t.test(x, y, paired=TRUE) Paired t-test data: x and y t = -2.3636, df = 99, p-value = 0.02005 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -0.4702824 -0.0410375 sample estimates: mean of the differences -0.2556599 The p-value plummets to 0.02005, and we reach the exactly opposite conclusion. See Also If the populations are not normally distributed (bell-shaped) and either sample is small, consider using the Wilcoxon–Mann–Whitney test described in Recipe 9.16. 9.16 Comparing the Locations of Two Samples Nonparametrically Problem You have samples from two populations. You don’t know the distribution of the pop- ulations, but you know they have similar shapes. You want to know: Is one population shifted to the left or right compared with the other? 9.16 Comparing the Locations of Two Samples Nonparametrically | 213 Solution You can use a nonparametric test, the Wilcoxon–Mann–Whitney test, which is im- plemented by the wilcox.test function. For paired observations (every xi is paired with yi), set paired=TRUE: > wilcox.test(x, y, paired=TRUE) For unpaired observations, let paired default to FALSE: > wilcox.test(x, y) The test output includes a p-value. Conventionally, a p-value of less than 0.05 indicates that the second population is likely shifted left or right with respect to the first popu- lation whereas a p-value exceeding 0.05 provides no such evidence. Discussion When we stop making assumptions regarding the distributions of populations, we enter the world of nonparametric statistics. The Wilcoxon–Mann–Whitney test is nonpara- metric and so can be applied to more datasets than the t test, which requires that the data be normally distributed (for small samples). This test’s only assumption is that the two populations have the same shape. In this recipe, we are asking: Is the second population shifted left or right with respect to the first? This is similar to asking whether the average of the second population is smaller or larger than the first. However, the Wilcoxon–Mann–Whitney test answers a different question: it tells us whether the central locations of the two populations are significantly different or, equivalently, whether their relative frequencies are different. Suppose we randomly select a group of employees and ask each one to complete the same task under two different circumstances: under favorable conditions and under unfavorable conditions, such as a noisy environment. We measure their completion times under both conditions, so we have two measurements for each employee. We want to know if the two times are significantly different, but we can’t assume they are normally distributed. The data are paired, so we must set paired=TRUE: > wilcox.test(fav, unfav, paired=TRUE) Wilcoxon signed rank test with continuity correction data: fav and unfav V = 0, p-value < 2.2e-16 alternative hypothesis: true location shift is not equal to 0 The p-value is essentially zero. Statistically speaking, we reject the assumption that the completion times were equal. Practically speaking, it’s reasonable to conclude that the times were different. 214 | Chapter 9: General Statistics In this example, setting paired=TRUE is critical. Treating the data as unpaired would be wrong because the observations are not independent; and this, in turn, would produce bogus results. Running the example with paired=FALSE produces a p-value of 0.2298, which leads to the wrong conclusion. See Also See Recipe 9.15 for the parametric test. 9.17 Testing a Correlation for Significance Problem You calculated the correlation between two variables, but you don’t know if the cor- relation is statistically significant. Solution The cor.test function can calculate both the p-value and the confidence interval of the correlation. If the variables came from normally distributed populations then use the default measure of correlation, which is the Pearson method: > cor.test(x, y) For nonnormal populations, use the Spearman method instead: > cor.test(x, y, method="Spearman") The function returns several values, including the p-value from the test of significance. Conventionally, p < 0.05 indicates that the correlation is likely significant whereas p > 0.05 indicates it is not. Discussion In my experience, people often fail to check a correlation for significance. In fact, many people are unaware that a correlation can be insignificant. They jam their data into a computer, calculate the correlation, and blindly believe the result. However, they should ask themselves: Was there enough data? Is the magnitude of the correlation large enough? Fortunately, the cor.test function answers those questions. Suppose we have two vectors, x and y, with values from normal populations. We might be very pleased that their correlation is greater than 0.83: > cor(x, y) [1] 0.8352458 But that is naïve. If we run cor.test, it reports a relatively large p-value of 0.1648: 9.17 Testing a Correlation for Significance | 215 > cor.test(x, y) Pearson's product-moment correlation data: x and y t = 2.1481, df = 2, p-value = 0.1648 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: -0.6379590 0.9964437 sample estimates: cor 0.8352458 The p-value is above the conventional threshold of 0.05, so we conclude that the cor- relation is unlikely to be significant. You can also check the correlation by using the confidence interval. In this example, the confidence interval is (−0.638, 0.996). The interval contains zero and so it is possible that the correlation is zero, in which case there would be no correlation. Again, you could not be confident that the reported correlation is significant. The cor.test output also includes the point estimate reported by cor (at the bottom, labeled “sample estimates”), saving you the additional step of running cor. By default, cor.test calculates the Pearson correlation, which assumes that the under- lying populations are normally distributed. The Spearman method makes no such as- sumption because it is nonparametric. Use method="Spearman" when working with nonnormal data. See Also See Recipe 2.6 for calculating simple correlations. 9.18 Testing Groups for Equal Proportions Problem You have samples from two or more groups. The group’s elements are binary-valued: either success or failure. You want to know if the groups have equal proportions of successes. Solution Use the prop.test function with two vector arguments: > ns <- c(ns1, ns2, ..., nsN) > nt <- c(nt1, nt2, ..., ntN) > prop.test(ns, nt) 216 | Chapter 9: General Statistics These are parallel vectors. The first vector, ns, gives the number of successes in each group. The second vector, nt, gives the size of the corresponding group (often called the number of trials). The output includes a p-value. Conventionally, a p-value of less than 0.05 indicates that it is likely the groups’ proportions are different whereas a p-value exceeding 0.05 provides no such evidence. Discussion In Recipe 9.11 we tested a proportion based on one sample. Here, we have samples from several groups and want to compare the proportions in the underlying groups. I recently taught statistics to 38 students and awarded a grade of A to 14 of them. A colleague taught the same class to 40 students and awarded an A to only 10. I wanted to know: Am I fostering grade inflation by awarding significantly more A grades than she did? I used prop.test. “Success” means awarding an A, so the vector of successes contains two elements: the number awarded by me and the number awarded by my colleague: > successes <- c(14,10) The number of trials is the number of students in the corresponding class: > trials <- c(38,40) The prop.test output yields a p-value of 0.3749: > prop.test(successes, trials) 2-sample test for equality of proportions with continuity correction data: successes out of trials X-squared = 0.7872, df = 1, p-value = 0.3749 alternative hypothesis: two.sided 95 percent confidence interval: -0.1110245 0.3478666 sample estimates: prop 1 prop 2 0.3684211 0.2500000 The relatively large p-value means that we cannot reject the null hypothesis: the evi- dence does not suggest any difference between my grading and hers. See Also See Recipe 9.11. 9.18 Testing Groups for Equal Proportions | 217 9.19 Performing Pairwise Comparisons Between Group Means Problem You have several samples, and you want to perform a pairwise comparison between the sample means. That is, you want to compare the mean of every sample against the mean of every other sample. Solution Place all data into one vector and create a parallel factor to identify the groups. Use pairwise.t.test to perform the pairwise comparison of means: > pairwise.t.test(x,f) # x contains the data, f is the grouping factor The output contains a table of p-values, one for each pair of groups. Conventionally, if p < 0.05 then the two groups likely have different means whereas p > 0.05 provides no such evidence. Discussion This is more complicated than Recipe 9.15, where we compared the means of two samples. Here we have several samples and want to compare the mean of every sample against the mean of every other sample. Statistically speaking, pairwise comparisons are tricky. It is not the same as simply performing a t test on every possible pair. The p-values must be adjusted, for otherwise you will get an overly optimistic result. The help pages for pairwise.t.test and p.adjust describe the adjustment algorithms available in R. Anyone doing serious pair- wise comparisons is urged to review the help pages and consult a good textbook on the subject. Suppose we are using the data of Recipe 5.5, where we combined data for freshmen, sophomores, and juniors into a data frame called comb. The data frame has two columns: the data in a column called values, and the grouping factor in a column called ind. We can use pairwise.t.test to perform pairwise comparisons between the groups: > pairwise.t.test(comb$values, comb$ind) Pairwise comparisons using t tests with pooled SD data: comb$values and comb$ind fresh jrs jrs 0.0043 - soph 0.0193 0.1621 P value adjustment method: holm 218 | Chapter 9: General Statistics Notice the table of p-values. The comparisons of juniors versus freshmen and of soph- omores versus freshmen produced small p-values: 0.0043 and 0.0193, respectively. We can conclude there are significant differences between those groups. However, the comparison of sophomores versus juniors produced a (relatively) large p-value of 0.1621, so they are not significantly different. See Also See Recipes 5.5 and 9.15. 9.20 Testing Two Samples for the Same Distribution Problem You have two samples, and you are wondering: Did they come from the same distribution? Solution The Kolmogorov–Smirnov test compares two samples and tests them for being drawn from the same distribution. The ks.test function implements that test: > ks.test(x, y) The output includes a p-value. Conventionally, a p-value of less than 0.05 indicates that the two samples (x and y) were drawn from different distributions whereas a p-value exceeding 0.05 provides no such evidence. Discussion The Kolmogorov–Smirnov test is wonderful for two reasons. First, it is a nonparametric test and so you needn’t make any assumptions regarding the underlying distributions: it works for all distributions. Second, it checks the location, dispersion, and shape of the populations, based on the samples. If these characteristics disagree then the test will detect that, allowing us to conclude that the underlying distributions are different. Suppose we suspect that the vectors x and y come from differing distributions. Here, ks.test reports a p-value of 0.01297: > ks.test(x, y) Two-sample Kolmogorov-Smirnov test data: x and y D = 0.3333, p-value = 0.01297 alternative hypothesis: two-sided 9.20 Testing Two Samples for the Same Distribution | 219 From the small p-value we can conclude that the samples are from different distribu- tions. However, when we test x against another sample, z, the p-value is much larger (0.8245); this suggests that x and z could have the same underlying distribution: > ks.test(x, z) Two-sample Kolmogorov-Smirnov test data: x and z D = 0.1333, p-value = 0.8245 alternative hypothesis: two-sided 220 | Chapter 9: General Statistics CHAPTER 10 Graphics Introduction Graphics is a great strength of R. The graphics package is part of the standard distri- bution and contains many useful functions for creating a variety of graphic displays. This chapter focuses on those functions, although it occasionally suggests other pack- ages. In this chapter’s See Also sections I mention functions in other packages that do the same job in a different way. I suggest that you explore those alternatives if you are dissatisfied with the basic function. Graphics is a vast subject, and I can only scratch the surface here. If you want to delve deeper, I recommend R Graphics by Paul Murrell (Chapman & Hall, 2006). That book discusses the paradigms behind R graphics, explains how to use the graphics functions, and contains numerous examples—including the code to recreate them. Some of the examples are pretty amazing. The Illustrations The graphs in this chapter are mostly plain and unadorned. I did that intentionally. When you call the plot function, as in: > plot(x) you get a plain, graphical representation of x. You could adorn the graph with colors, a title, labels, a legend, text, and so forth, but then the call to plot becomes more and more crowded, obscuring the basic intention: > plot(x, main="Forecast Results", xlab="Month", ylab="Production", + col=c("red", "black", "green")) I want to keep the recipes clean, so I emphasize the basic plot and then show later (as in Recipe 10.2) how to add adornments. 221 Notes on Graphics Functions It is important to understand the distinction between high-level and low-level graphics functions. A high-level graphics function starts a new graph. It initializes the graphics window (creating it if necessary); sets the scale; maybe draws some adornments, such as a title and labels; and renders the graphic. Examples include: plot Generic plotting function boxplot Create a box plot hist Create a histogram qqnorm Create a quantile-quantile (Q-Q) plot curve Graph a function A low-level graphics function cannot start a new graph. Rather, it adds something to an existing graph: points, lines, text, adornments, and so forth. Examples include: points Add points lines Add lines abline Add a straight line segments Add line segments polygon Add a closed polygon text Add text You must call a high-level graphics routine before calling a low-level graphics routine. The low-level routine needs to have the graph initialized; otherwise, you get an error like this: > abline(a=0, b=1) Error in int_abline(a = a, b = b, h = h, v = v, untf = untf, ...) : plot.new has not been called yet 222 | Chapter 10: Graphics The Generic plot Function In this chapter, we come face to face with polymorphism in R. A polymorphic func- tion or generic function is one whose behavior changes depending on the type of argument. The plot function is polymorphic, so plot(x) produces different results de- pending on whether x is a vector, a factor, a data frame, a linear regression model, a table, or whatever. As you read the recipes, you’ll see plot used over and over. Each time it’s used, carefully note the type of arguments. That will help you understand and remember why the function behaved as it did. Graphics in Other Packages R is highly programmable, and many people have extended its graphics machinery with additional features. Quite often, packages include specialized functions for plotting their results and objects. The zoo package, for example, implements a time series object. If you create a zoo object z and call plot(z), then the zoo package does the plotting; it creates a graphic that is customized for displaying a time series. There are even entire packages devoted to extending R with new graphics paradigms. The lattice package is an alternative to traditional graphics. It uses a more powerful graphics paradigm that enables you to create informative graphics more easily. The results are generally better looking, too. It was implemented by Deepayan Sarkar, who also wrote Lattice: Multivariate Data Visualization with R (Springer, 2008), which ex- plains the package and how to use it. The lattice package is also described in R in a Nutshell (O’Reilly). The ggplot2 package provides yet another graphics paradigm, which is called the Grammar of Graphics. It uses a highly modular approach to graphics, which lets you construct and customize your plots more easily. These graphics, too, are generally more attractive than the traditional ones. The ggplot2 package was created by Hadley Wick- ham, who also wrote ggplot2: Elegant Graphics for Data Analysis (Springer, 2009). That book explains the paradigm behind ggplot2 and how to use the package functions. 10.1 Creating a Scatter Plot Problem You have paired observations: (x1, y1), (x2, y2), ..., (xn, yn). You want to create a scatter plot of the pairs. Solution If your data are held in two parallel vectors, x and y, then use them as arguments of plot: > plot(x, y) 10.1 Creating a Scatter Plot | 223 If your data is held in a (two-column) data frame, plot the data frame: > plot(dfrm) Discussion A scatter plot is usually my first attack on a new dataset. It’s a quick way to see the relationship, if any, between x and y. Creating the scatter plot is easy: > plot(x, y) The plot function does not return anything. Rather, its purpose is to draw a plot of the (x, y) pairs in the graphics window. Life is even easier if your data is captured in a two-column data frame. If you plot a two-column data frame, the function assumes you want a scatter plot created from the two columns. The scatter plot shown in Figure 10-1 was created by one call to plot: > plot(cars) Figure 10-1. Scatter plot The cars dataset contains two columns, speed and dist. The first column is speed, so that becomes the x-axis and dist becomes the y-axis. 224 | Chapter 10: Graphics If your data frame contains more than two columns then you will get multiple scatter plots, which might or might not be useful (Recipe 10.7). To get a scatter plot, your data must be numeric. Recall that plot is a polymorphic function and so, if the arguments are nonnumeric, it will create some other plot. See Recipe 10.16, for example, which creates box plots from factors. See Also See Recipe 10.2 for adding a title and labels; see Recipes 10.3 and 10.5 for adding a grid and a legend (respectively). See Recipe 10.7 for plotting multiple variables. 10.2 Adding a Title and Labels Problem You want to add a title to your plot or add labels for the axes. Solution When calling plot: • Use the main argument for a title. • Use the xlab argument for an x-axis label. • Use the ylab argument for a y-axis label. > plot(x, main="The Title", xlab="X-axis Label", ylab="Y-axis Label") Alternatively: plot your data but set ann=FALSE to inhibit annotations; then call the title function to add a title and labels: > plot(x, ann=FALSE) > title(main="The Title", xlab="X Axis Label", ylab="Y Axis Label") Discussion The graphic in Recipe 10.1 is quite plain. It needs a title and better labels to make it more interesting and easier to interpret. We can add them via arguments to plot, yield- ing the graphic shown in Figure 10-2: > plot(cars, + main="cars: Speed vs. Stopping Distance (1920)", + xlab="Speed (MPH)", + ylab="Stopping Distance (ft)") This is a clear improvement over the defaults. Now the title tells us something about the plot, and the labels give the measurement units. 10.2 Adding a Title and Labels | 225 10.3 Adding a Grid Problem You want to add a grid to your graphic. Solution • Call plot with type="n" to initialize the graphics frame without displaying the data. • Call the grid function to draw the grid. • Call low-level graphics functions, such as points and lines, to draw the graphics overlaid on the grid. > plot(x, y, type="n") > grid() > points(x, y) Figure 10-2. Adding a title and labels 226 | Chapter 10: Graphics Discussion The plot function does not automatically draw a grid. For some plots, however, a grid is helpful to the viewer. The Solution shows the proper way to add a grid. The grid must be drawn first, before the graphics. If you draw the graphics first, the grid will overlay them and partially hide them. We can add a grid to the graphic of Recipe 10.2. First, we draw the graphic but with type="n". That will create a new plot, including title and axes, but will not (yet) plot the data: > plot(cars, + main="cars: Speed vs. Stopping Distance (1920)", + xlab="Speed (MPH)", + ylab="Stopping Distance (ft)", + type="n") Next we draw the grid: > grid() Finally, we draw the scatter plot by calling points, a low-level graphics function, thus creating the graph shown in Figure 10-3: > points(cars) In a hurry? Looking for a shortcut? You can call plot as you normally would and then call grid. The only problem is that the grid will be overlaid on your graphics, partially overwriting them. Sometimes this creates a visible problem and sometimes not. Try it. 10.4 Creating a Scatter Plot of Multiple Groups Problem You have paired observations in two vectors, x and y, and a parallel factor f that indi- cates their groups. You want to create a scatter plot of x and y that distinguishes among the groups. Solution Use the pch argument of plot. It will plot each point with a different plotting character, according to its group: > plot(x, y, pch=as.integer(f)) 10.4 Creating a Scatter Plot of Multiple Groups | 227 Discussion Plotting multiple groups in one scatter plot creates an uninformative mess unless we distinguish one group from another. The pch argument of plot lets us plot points using a different plotting character for each (x, y) pair. The iris dataset contains paired measures of Petal.Length and Petal.Width. Each measurement also has a Species property indicating the species of the flower that was measured. If we plot all the data at once, we just get the scatter plot shown in the lefthand panel of Figure 10-4: > with(iris, plot(Petal.Length, Petal.Width)) The graphic would be far more informative if we distinguished the points by species. The pch argument is a vector of integers, one integer for each (x, y) point, with values between 0 and 18. The point will be drawn according to its corresponding value in pch. This call to plot draws each point according to its group: > with(iris, plot(Petal.Length, Petal.Width, pch=as.integer(Species))) The result is displayed in the righthand panel of Figure 10-4, which clearly shows the points in groups according to their species. Figure 10-3. Adding a grid 228 | Chapter 10: Graphics See Also See Recipe 10.5 to add a legend. 10.5 Adding a Legend Problem You want your plot to include a legend, the little box that decodes the graphic for the viewer. Solution After calling plot, call the legend function. The first three arguments are always the same, but the arguments after that vary as a function of what you are labeling. Here is how you create legends for points, lines, and colors: Legend for points legend(x, y, labels, pch=c(pointtype1, pointtype2, ...)) Legend for lines according to line type legend(x, y, labels, lty=c(linetype1, linetype2, ...)) Legend for lines according to line width legend(x, y, labels, lwd=c(width1, width2, ...)) Figure 10-4. Scatter plots of multiple groups 10.5 Adding a Legend | 229 D o wnload from Wow! eBook Legend for colors legend(x, y, labels, col=c(color1, color2, ...)) Here, x and y are the coordinates of the legend box, and labels is a vector of the char- acter strings to appear in the legend. The pch, lty, lwd, and col arguments are vectors that parallel the labels. Each label will be shown next to its corresponding point type, line type, line width, or color according to which one you specified. Discussion The lefthand panel of Figure 10-5 shows a legend added to the scatter plot of Rec- ipe 10.4. The legend tells the viewer which points are associated with which species. I could have added the legend in this way (after calling plot): > legend(1.5, 2.4, c("setosa","versicolor","virginica"), pch=1:3) Figure 10-5. Examples of legends The first two arguments give the x and y coordinates of the legend’s box. The third argument is a vector of labels; in this case, names of species. The last argument indicates that we are labeling points by their type—specifically, types 1 through 3. In fact, I created the plot and legend this way: > f <- factor(iris$Species) > with(iris, plot(Petal.Length, Petal.Width, pch=as.integer(f))) > legend(1.5, 2.4, as.character(levels(f)), pch=1:length(levels(f))) 230 | Chapter 10: Graphics The beauty here is that species names are not hard-coded into the R code. Rather, everything is driven by the species factor, f. I like that for two reasons. First, it means that I won’t forget or misspell a species name. Second, if someone adds another species then the code automatically adapts to the addition. That second virtue is especially valuable inside an R script. The righthand panel of Figure 10-5 shows a legend for lines based on the line type (solid, dashed, or dotted). It was created like this: > legend(0.5, 95, c("Estimate","Lower conf lim","Upper conf lim"), + lty=c("solid","dashed","dotted")) The two examples in this recipe show legends for point types and line types. The legend function can also create legends for colors, fills, and line widths. The function also has a bewildering number of arguments and options for controlling the exact ap- pearance of the legend. See the help page for details. Legends are a wonderful feature in general, but the legend function comes with two annoyances. First, you must pick the legend’s x and y coordinates yourself. R cannot pick them for you. Plot the data without a legend; then pick a nice, empty area and put the legend there. Otherwise, the legend might obscure the graphics. Second, the correspondence between the labels and the point types (or line types or colors) is not automatic. You are responsible for getting that right. The task is prone to errors and is not easy to automate. 10.6 Plotting the Regression Line of a Scatter Plot Problem You are plotting pairs of data points, and you want to add a line that illustrates their linear regression. Solution Create a model object, plot the (x, y) pairs, and then plot the model object using the abline function: > m <- lm(y ~ x) > plot(y ~ x) > abline(m) Discussion When you plot a model object using abline(m), it draws the fitted regression line. This is very different from plot(m), which produces diagnostic plots (Recipe 11.15). 10.6 Plotting the Regression Line of a Scatter Plot | 231 Suppose we are modeling the strongx dataset found in the faraway package: > library(faraway) > data(strongx) > m <- lm(crossx ~ energy, data=strongx) We can display the data in a scatter plot and then add the regression line using abline, producing Figure 10-6: > plot(crossx ~ energy, data=strongx) > abline(m) Notice that we use a formula to define the plot (crossx ~ energy). I like using that syntax because we quickly see the parallel with the regression formula. The alternative is calling plot(strongx$energy, strongx$crossx), which trips me up because the order of arguments is reversed from the lm formula. See Also See Chapter 11 for more about linear regression and the lm function. Figure 10-6. Scatter plot with regression line 232 | Chapter 10: Graphics 10.7 Plotting All Variables Against All Other Variables Problem Your dataset contains multiple numeric variables. You want to see scatter plots for all pairs of variables. Solution Place your data in a data frame and then plot the data frame. R will create one scatter plot for every pair of columns: > plot(dfrm) Discussion When you have a large number of variables, finding interrelationships between them is difficult. One useful technique is looking at scatter plots of all pairs of variables. This would be quite tedious if done pair-by-pair, but R provides an easy way to produce all those scatter plots at once. The iris dataset contains four numeric variables and one categorical variable: > head(iris) Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.1 3.5 1.4 0.2 setosa 2 4.9 3.0 1.4 0.2 setosa 3 4.7 3.2 1.3 0.2 setosa 4 4.6 3.1 1.5 0.2 setosa 5 5.0 3.6 1.4 0.2 setosa 6 5.4 3.9 1.7 0.4 setosa What is the relationship, if any, between the numeric columns? Plotting those four columns produces the multiple scatter plots shown in Figure 10-7: > plot(iris[,1:4]) 10.8 Creating One Scatter Plot for Each Factor Level Problem Your dataset contains (at least) two numeric variables and a factor. You want to create several scatter plots for the numeric variables, with one scatter plot for each level of the factor. Solution This kind of plot is called a conditioning plot, which is produced by the coplot function: > coplot(y ~ x | f) 10.8 Creating One Scatter Plot for Each Factor Level | 233 This will produce scatter plots of x versus y, with one plot for every level of f. Discussion Conditioning plots are another way to explore and illustrate the effect of a factor on your data. The Cars93 dataset contains 27 variables describing 93 car models as of 1993. Two numeric variables are MPG.city, the miles per gallon in the city, and Horsepower, the engine horsepower. One categorical variable is Origin, which can be USA or non-USA according to where the model was built. Figure 10-7. Multiple scatter plots 234 | Chapter 10: Graphics Exploring the relationship between MPG and horsepower, we might ask: Is there a different relationship for USA models and non-USA models? The conditioning plot in Figure 10-8 shows two scatter plots of MPG versus horsepower that were created si- multaneously by one call to coplot. The lefthand plot is for USA vehicles and the right- hand plot is for non-USA vehicles. It was produced like this: > data(Cars93, package="MASS") > coplot(Horsepower ~ MPG.city | Origin, data=Cars93) Figure 10-8. A conditioning plot The data argument of coplot lets us take variables from the Cars93 data frame. 10.8 Creating One Scatter Plot for Each Factor Level | 235 The plot reveals a few insights. If we really crave that 300-horsepower monster then we’ll have to buy a car built in the USA; but if we want high MPG, we have more choices among non-USA models. These insights could be teased out of a statistical analysis, but the visual presentation reveals them much more quickly. 10.9 Creating a Bar Chart Problem You want to create a bar chart. Solution Use the barplot function. The first argument is a vector of bar heights: > barplot(c(height1, height2, ..., heightn)) Discussion The barplot function produces a simple bar chart. It assumes that the heights of your bars are conveniently stored in a vector. That is not always the case, however. Often you have a vector of numeric data and a parallel factor that groups the data, and you want to produce a bar chart of the group means or the group totals. For example, the airquality dataset contains a numeric Temp column and a Month column. We can create a bar chart of the mean temperature by month in two steps. First, we compute the means: > heights <- tapply(airquality$Temp, airquality$Month, mean) That gives the heights of the bars, from which we create the bar chart: > barplot(heights) The result is shown in the lefthand panel of Figure 10-9. The result is pretty bland, as you can see, so it’s common to add some simple adornments: a title, labels for the bars, and a label for the y-axis: > barplot(heights, + main="Mean Temp. by Month", + names.arg=c("May", "Jun", "Jul", "Aug", "Sep"), + ylab="Temp (deg. F)") The righthand panel of Figure 10-9 shows the improved bar chart. See Also See Recipe 10.10 for adding confidence intervals and Recipe 10.11 for adding color; see the barchart function in the lattice package. 236 | Chapter 10: Graphics 10.10 Adding Confidence Intervals to a Bar Chart Problem You want to augment a bar chart with confidence intervals. Solution Suppose that x is a vector of statistics, such as a vector of means, and that lower and upper are vectors of the corresponding limits for the confidence intervals. The barplot2 function of the gplots library can display a bar chart of x and its confidence intervals: > library(gplots) > barplot2(x, plot.ci=TRUE, ci.l=lower, ci.u=upper) Discussion Most bar charts display point estimates, which are shown by the heights of the bars, but rarely do they include confidence intervals. The statistician in me dislikes this in- tensely. The point estimate is only half of the story; the confidence interval gives the full story. Fortunately, the barplot2 function can plot a bar chart with confidence intervals. The hard part is calculating the intervals. In Recipe 10.9, we calculated group means before plotting them. We can calculate the confidence intervals for those means in a similar way. The t.test function returns a list of values. That list includes an element called Figure 10-9. Bar charts, plain and adorned 10.10 Adding Confidence Intervals to a Bar Chart | 237 conf.int, which is itself a vector of two elements: the lower and upper bounds of the confidence interval for the mean: > attach(airquality) > heights <- tapply(Temp, Month, mean) > lower <- tapply(Temp, Month, function(v) t.test(v)$conf.int[1]) > upper <- tapply(Temp, Month, function(v) t.test(v)$conf.int[2]) We could create a basic bar chart with confidence intervals in this way: > barplot2(heights, plot.ci=TRUE, ci.l=lower, ci.u=upper) But as noted in Recipe 10.9, that would create a very bland graphic. The bar chart is more attractive if we limit the y range (ylim), trim the bars (xpd), add a title (main), label the bars (names.arg), and label the y-axis. The bar chart in Figure 10-10 was produced by this call to barplot2: > barplot2(heights, plot.ci=TRUE, ci.l=lower, ci.u=upper, + ylim=c(50,90), xpd=FALSE, + main="Mean Temp. By Month", + names.arg=c("May","Jun","Jul","Aug","Sep"), + ylab="Temp (deg. F)") Figure 10-10. Bar chart with confidence intervals 238 | Chapter 10: Graphics See Also See Recipe 3.9 to install the gplots package, Recipe 6.5 for more about tapply, and Recipe 9.9 for more about t.test. 10.11 Coloring a Bar Chart Problem You want to color or shade the bars of a bar chart. Solution Use the col argument of barplot: > barplot(heights, col=colors) Here, heights is the vector of bar heights and colors is a vector of corresponding colors. To generate the vector of colors, you would typically use the gray function to generate a vector of grays or the rainbow function to generate a vector of colors. Discussion Building a vector of colors can be tricky. Using explicit colors is a simple solution. This little example plots a three-bar chart and colors the bars red, white, and blue (respectively): > barplot(c(3,5,4), col=c("red","white","blue")) Life is rarely that simple, however. More likely, you want colors that convey some information about your dataset. A typical effect is to shade the bars according to their rank: shorter bars are light colored; taller bars are darker. In this example, I’ll use gray- scale colors generated by the gray function. The one argument of gray is a vector of numeric values between 0 and 1. The function returns one shade of gray for each vector element, ranging from pure black for 0.0 to pure white for 1.0. To shade the bar chart, we first convert the bars’ ranks to relative heights, expressed as a value between zero and one: > rel.hts <- rank(heights) / length(heights) Then we convert the relative heights into a vector of grayscale colors while inverting the relative heights so that the taller bars are dark, not light: > grays <- gray(1 - rel.hts) We could easily create a shaded bar chart in this way: > barplot(heights, col=grays) 10.11 Coloring a Bar Chart | 239 However, we’ll add adornments to make the chart easier to interpret. Here is the com- plete solution, with the result shown in Figure 10-11: > rel.hts <- (heights - min(heights)) / (max(heights) - min(heights)) > grays <- gray(1 - rel.hts) > barplot(heights, + col=grays, + ylim=c(50,90), xpd=FALSE, + main="Mean Temp. By Month", + names.arg=c("May", "Jun", "Jul", "Aug", "Sep"), + ylab="Temp (deg. F)") Figure 10-11. Bar chart with shading I used grayscale here because some editions of this book are not printed in color. For a multichromatic effect I suggest using the rainbow function, which can produce a range of colors; use it instead of the gray function used here. See Also See Recipe 10.9 for creating a bar chart. 240 | Chapter 10: Graphics 10.12 Plotting a Line from x and y Points Problem You have paired observations: (x1, y1), (x2, y2), ..., (xn, yn). You want to plot a series of line segments that connect the data points. Solution Use the plot function with a plot type of "l" (the letter ell, not the digit one): > plot(x, y, type="l") If your data is captured in a two-column data frame, you can plot the line segments from data frame contents: > plot(dfrm, type="l") Discussion Observe that this recipe makes the same assumption as Recipe 10.1 for creating a scatter plot—you have paired observations. The only difference between a scatter plot and a line drawing is that one plots points and the other connects them. The plot function handles both cases. The type argument determines scatter plot or line drawing; the default is a scatter plot. We can plot a built-in dataset, pressure, both ways: as a scatter plot and a line plot. The left panel of Figure 10-12 show the scatter plot, created as follows: > plot(pressure) Figure 10-12. Same dataset, two plot types 10.12 Plotting a Line from x and y Points | 241 The right panel is a line plot, which was created simply by changing the plot type to “line” (l): > plot(pressure, type="l") See Also See Recipe 10.1. 10.13 Changing the Type, Width, or Color of a Line Problem You are plotting a line. You want to change the type, width, or color of the line. Solution The plot function (and other graphics functions) include parameters for controlling the appearance of lines. Use the lty parameter to control the line type: • lty="solid" or lty=1 (default) • lty="dashed" or lty=2 • lty="dotted" or lty=3 • lty="dotdash" or lty=4 • lty="longdash" or lty=5 • lty="twodash" or lty=6 • lty="blank" or lty=0 (inhibits drawing) This call to plot, for example, draws lines with a dashed style: > plot(x, y, type="l", lty="dashed") Use the lwd parameter to control the line width or thickness. By default, lines have a width of 1: > plot(x, y, type="l", lwd=2) # Draw a thicker line Use the col parameter to control line color. By default, lines are drawn in black: > plot(x, y, type="l", col="red") # Draw a red line Discussion The Solution shows how to draw one line and specify its style, width, or color. A com- mon scenario involves drawing multiple lines, each with its own style, width, or color. The Solution generalizes to that case, but call the plot function for the first line and call the lines function for subsequent lines. The lines function adds lines to an existing plot and lets you specify their style, width, and/or color: 242 | Chapter 10: Graphics > plot(x, y.democr, type="l", col="blue") > lines(x, y.republ, col="red") > lines(x, y.indeps, col="yellow") Here, the initial call to plot initializes the graphics window and draws the first lines in blue. The subsequent calls to lines draw additional lines, first in red and then in yellow. See Also See Recipe 10.12 for plotting a basic line. 10.14 Plotting Multiple Datasets Problem You want to show multiple datasets in one plot. Solution Initialize the plot using a high-level graphics function such as plot or curve. Then add additional datasets using low-level functions such as lines and points. When initializing the plot, be careful to size the canvas correctly. Using the xlim and ylim parameters, create a canvas large enough to hold all datasets and not merely the first dataset. This code snippet shows how to compute the limits using the range func- tion and then set them using xlim and ylim. It assumes you have two datasets, one represented by x1 and y1 and the other by x2 and y2: > xlim <- range(c(x1,x2)) > ylim <- range(c(y1,y2)) > plot(x1, y1, type="l", xlim=xlim, ylim=ylim) > lines(x2, y2, lty="dashed") Discussion As discussed in the “Introduction” to this chapter, the plot function initializes the graphics window and the lines function adds lines after that. My first try at using plot and lines together looked something like this (I wanted to plot the (x1, y1) dataset and the (x2, y2) dataset together): > plot(x1, y1, type="l") > lines(x2, y2, lty="dashed") The result was something like the lefthand panel of Figure 10-13. The first dataset plotted correctly, but the second dataset just sort of trailed off the page. This behavior was puzzling. The problem was that the plot function sized the canvas based on the data it was given. The second dataset (plotted by lines) had a different, wider range, so it fell outside the canvas and disappeared. 10.14 Plotting Multiple Datasets | 243 Figure 10-13. Plotting multiple datasets, correctly and incorrectly The solution is to size the canvas correctly in the first place. The plot function has two arguments for sizing the canvas—xlim is a two-element vector giving the lower and upper limits of the x-axis, and ylim does the same for the y-axis: > plot(x, y, xlim=c(lower,upper), ylim=c(lower,upper), ...) The range function is handy here. It returns the upper and lower bounds of a vector; hence we can use it to compute xlim and ylim before calling plot, like this: > xlim <- range(c(x1,x2)) > ylim <- range(c(y1,y2)) > plot(x1, y1, type="l", xlim=xlim, ylim=ylim) > lines(x2, y2, lty="dashed") The righthand panel of Figure 10-13 shows the result. Now the canvas is big enough, and no data is lost. This Discussion has been framed in terms of lines, but plotting points presents the same problem and has the same solution. See Also See this chapter’s “Introduction” for comments about high- and low-level graphics functions. 244 | Chapter 10: Graphics 10.15 Adding Vertical or Horizontal Lines Problem You want to add a vertical or horizontal line to your plot, such as an axis through the origin. Solution The abline function will draw a vertical line or horizontal line when given an argument of v or h, respectively: > abline(v=x) # Draw a vertical line at x > abline(h=y) # Draw a horizontal line at y Discussion The abline function draws straight lines. I often use it to draw axes through the origin of my plot: > abline(v=0) # Vertical line at x = 0 > abline(h=0) # Horizontal line at y = 0 The abline function adds lines to an existing plot; it cannot create a new plot. So call a plot creation function first, such as plot(...), and then call abline. The v and h arguments can be vectors, in which case abline draws multiple lines—one for each vector element. A typical use would be drawing regularly spaced lines. Suppose we have a sample of points, samp. First, we plot them with a solid line through the mean: > plot(samp) > m <- mean(samp) > abline(h=m) We want to illustrate the sample standard deviation graphically, so we calculate it and draw dotted lines at ±1 and ±2 standard deviations away from the mean: > stdevs <- m + c(-2,-1,+1,+2)*sd(samp) > abline(h=stdevs, lty="dotted") The result is shown in Figure 10-14. See Also See Recipe 10.13 for more about changing line types. 10.15 Adding Vertical or Horizontal Lines | 245 10.16 Creating a Box Plot Problem You want to create a box plot of your data. Solution Use boxplot(x), where x is a vector of numeric values. Discussion A box plot provides a quick and easy visual summary of a dataset. Figure 10-15 shows a typical box plot: • The thick line in the middle is the median. • The box surrounding the median identifies the first and third quartiles; the bottom of the box is Q1, and the top is Q3. Figure 10-14. Delineating sample standard deviations 246 | Chapter 10: Graphics • The “whiskers” above and below the box show the range of the data, excluding outliers. • The circles identify outliers. By default, an outlier is defined as any value that is farther than 1.5 × IQR away from the box. (IQR is the interquartile range, or Q3 − Q1.) In this example, there are three outliers. Figure 10-15. A typical box plot See Also One box plot alone is pretty boring. See Recipe 10.17 for creating multiple box plots. 10.17 Creating One Box Plot for Each Factor Level Problem Your dataset contains a numeric variable and a factor (categorical variable). You want to create several box plots of the numeric variable broken out by factor levels. 10.17 Creating One Box Plot for Each Factor Level | 247 Solution Use the boxplot function with a formula: > boxplot(x ~ f) Here, x is the numeric variable and f is the factor. You can also use the two-argument form of the plot function, being careful to put the factor first: > plot(f, x) Discussion This recipe is another great way to explore and illustrate the relationship between two variables. In this case, we want to know whether the numeric variable changes accord- ing to the level of the factor. The UScereal dataset contains many variables regarding breakfast cereals. One variable is the amount of sugar per portion and another is the shelf position (counting from the floor). Cereal manufacturers can negotiate for shelf position, placing their product for the best sales potential. I wonder: Where do they put the high-sugar cereals? We can explore that question by creating one box plot per shelf, like this: > data(UScereal, package="MASS") > boxplot(sugars ~ shelf, data=UScereal) The resulting box plot would be quite drab, however, so we add a few adornments: > data(UScereal, package="MASS") > boxplot(sugars ~ shelf, data=UScereal, + main="Sugar Content by Shelf", + xlab="Shelf", ylab="Sugar (grams per portion)") The result is shown in Figure 10-16. The data argument lets us take variables from the UScereal data frame. I supplied xlab (the x-axis label) and ylab (the y-axis label) because the basic box plot has no labels, making it harder to interpret. The box plots suggest that shelf #2 has the most high-sugar cereals. Could it be that this shelf is at eye level for young children who can influence their parent’s choice of cereals? See Also See Recipe 10.16 for creating a basic box plot. 10.18 Creating a Histogram Problem You want to create a histogram of your data. 248 | Chapter 10: Graphics Solution Use hist(x), where x is a vector of numeric values. Discussion The lefthand panel of Figure 10-17 shows a histogram of the MPG.city column taken from the Cars93 dataset, created like this: > data(Cars93, package="MASS") > hist(Cars93$MPG.city) The hist function must decide how many cells (bins) to create for binning the data. In this example, the default algorithm chose seven bins. That creates too few bars for my taste because the shape of the distribution remains hidden. So I would include a second argument for hist—namely, the suggested number of bins: > hist(Cars93$MPG.city, 20) The number is only a suggestion, but hist will expand the number of bins as possible to accommodate that suggestion. Figure 10-16. Box plots by factor level 10.18 Creating a Histogram | 249 The righthand panel of Figure 10-17 shows a histogram for the same data but with more bins and with replacements for the default title and x-axis label. It was created like this: > hist(Cars93$MPG.city, 20, main="City MPG (1993)", xlab="MPG") See Also The histogram function of the lattice package is an alternative to hist. 10.19 Adding a Density Estimate to a Histogram Problem You have a histogram of your data sample, and you want to add a curve to illustrate the apparent density. Solution Use the density function to approximate the sample density; then use lines to draw the approximation: > hist(x, prob=T) # Histogram of x, using a probability scale > lines(density(x)) # Graph the approximate density Figure 10-17. Histograms 250 | Chapter 10: Graphics D o wnload from Wow! eBook Discussion A histogram suggests the density function of your data, but it is rough. A smoother estimate could help you better visualize the underlying distribution. This example takes a sample from a gamma distribution and then plots the histogram and the estimated density. The result is shown in Figure 10-18: > samp <- rgamma(500, 2, 2) > hist(samp, 20, prob=T) > lines(density(samp)) Figure 10-18. Histogram with density estimate You must include the prob=T argument to hist in order for this recipe to work. The result of prob=T is that the y-axis of Figure 10-18 is given in probabilities. Without it, hist gives counts instead; that would be incompatible with plotting the output of density. 10.19 Adding a Density Estimate to a Histogram | 251 See Also The density function approximates the shape of the density nonparametrically. If you know the actual underlying distribution, use instead Recipe 8.11 to plot the density function. 10.20 Creating a Discrete Histogram Problem You want to create a histogram of discrete data. Solution Use the table function to count occurrences. Then use the plot function with type="h" to graph the occurrances as a histogram: > plot(table(x), type="h") Here, x is a vector of discrete values. Discussion You can use the hist function to create a histogram of discrete data, but hist is really oriented toward continuous data. The resulting histograms don’t have that look of “discreteness”. Figure 10-19 shows a histogram created using this function call: > plot(table(x), type="h", lwd=5, ylab="Freq") The lwd=5 parameter widens the bars a bit, giving them a more pleasing appearance. If you prefer a histogram with relative frequencies, not counts, then simply scale the table elements by the number of data items, like this: > plot(table(x)/length(x), type="h", lwd=5, ylab="Freq") See Also See Recipe 10.18 for histograms of continuous data. This recipe was taken straight from the help page for plot. 10.21 Creating a Normal Quantile-Quantile (Q-Q) Plot Problem You want to create a quantile-quantile (Q-Q) plot of your data, typically because you want to know whether the data is normally distributed. 252 | Chapter 10: Graphics Solution Use the qqnorm function to create the basic quantile-quantile plot; then use qqline to augment it with a diagonal line: > qqnorm(x) > qqline(x) Here, x is a numeric vector. Discussion Sometimes it’s important to know if your data is normally distributed. A quantile- quantile (Q-Q) plot is a good first check. The Cars93 dataset contains a Price column. Is it normally distributed? This code snip- pet creates a Q-Q plot of Price, shown in the lefthand panel of Figure 10-20: > data(Cars93, package="MASS") > qqnorm(Cars93$Price, main="Q-Q Plot: Price") > qqline(Cars93$Price) Figure 10-19. Histogram of discrete data 10.21 Creating a Normal Quantile-Quantile (Q-Q) Plot | 253 If the data had a perfect normal distribution, then the points would fall exactly on the diagonal line. Many points are close, especially in the middle section, but the points in the tails are pretty far off. Too many points are above the line, indicating a general skew to the left. The leftward skew might be cured by a logarithmic transformation. We can plot log(Price), which yields the righthand panel of Figure 10-20: > data(Cars93, package="MASS") > qqnorm(log(Cars93$Price), main="Q-Q Plot: log(Price)") > qqline(log(Cars93$Price)) Notice that the points in the new plot are much better behaved, staying close to the line except in the extreme left tail. It appears that log(Price) is approximately Normal. See Also See Recipe 10.22 for creating Q-Q plots for other distributions. See Recipe 11.15 for an application of Normal Q-Q plots to diagnosing linear regression. 10.22 Creating Other Quantile-Quantile Plots Problem You want to view a quantile-quantile plot for your data, but the data is not normally distributed. Figure 10-20. Quantile-quantile (Q-Q) plots 254 | Chapter 10: Graphics Solution For this recipe, you must have some idea of the underlying distribution, of course. The solution is built from the following steps: • Use the ppoints function to generate a sequence of points between 0 and 1. • Transform those points into quantiles, using the quantile function for the assumed distribution. • Sort your sample data. • Plot the sorted data against the computed quantiles. • Use abline to plot the diagonal line. This can all be done in two lines of R code. Here is an example that assumes your data, y, has a Student’s t distribution with 5 degrees of freedom. Recall that the quantile function for Student’s t is qt and that its second argument is the degrees of freedom: > plot(qt(ppoints(y), 5), sort(y)) > abline(a=0, b=1) Discussion The solution looks complicated, but it’s not. The R code will always look like this: > plot(quantileFunction(ppoints(y), ...), sort(y)) > abline(a=0, b=1) where quantileFunction is the quantile function for the assumed distribution and "..." are the parameters needed to quantileFunction beyond y. The abline function draws a line along the diagonal, where all the data points would ideally fall. I’ll illustrate this recipe by taking a random sample from an exponential distribution with a mean of 10 (or, equivalently, a rate of 1/10): > RATE <- 1/10 > y <- rexp(N, rate=RATE) The quantile function for the exponential distribution is qexp, which takes a rate argument. We can create the Q-Q plot in this way: > plot(qexp(ppoints(y), rate=RATE), sort(y)) > abline(a=0, b=1) The resulting plot would look too plain, so we include a title and labels to help the viewer. Here is the actual code that created Figure 10-21: > plot(qexp(ppoints(y), rate=RATE), sort(y), + main="Q-Q Plot", xlab="Theoretical Quantiles", ylab="Sample Quantiles") > abline(a=0, b=1) It is interesting to see how far the Q-Q points can deviate from the diagonal—even when we know the exact distribution of the population. 10.22 Creating Other Quantile-Quantile Plots | 255 10.23 Plotting a Variable in Multiple Colors Problem You want to plot your data in multiple colors, typically to make the plot more infor- mative, readable, or interesting. Solution Use the col argument of the plot function: > plot(x, col=colors) The value of col can be: Figure 10-21. Quantile-quantile plot, exponential distribution 256 | Chapter 10: Graphics • One color, in which case all data points are that color. • A vector of colors, the same length as x, in which case each value of x is colored with its corresponding color. • A short vector, in which case the vector of colors is recycled. Discussion If you plot x like this, the plot is drawn in black: > plot(x) You can draw the entire plot in blue, like this: > plot(x, col="blue") However, it is much more useful (and interesting) to vary the color in a way that illu- minates the data. I’ll illustrate this by plotting a graphic two ways, once in black and white and once with simple shading. This call to plot produces the basic black-and- white graphic shown in the lefthand panel of Figure 10-22: > plot(x, type="h", lwd=3) Figure 10-22. Adding shading to a plot (The lwd=3 argument widens the lines, making them easier to see.) The righthand panel of Figure 10-22 is produced by creating a vector of "gray" and "black" according to the sign of x and then plotting x using those colors: > colors <- ifelse(x >= 0, "black", "gray") > plot(x, type='h', lwd=3, col=colors) 10.23 Plotting a Variable in Multiple Colors | 257 The negative values are now plotted in gray because the corresponding element of colors is "gray". I used grayscale shading because some editions of this book are in black and white, so colors would not be visible to the reader. You could easily use full colors instead—for example, by replacing the colors definition with this assignment statement: > colors <- ifelse(x >= 0, "green", "red") That would plot the positive values in green and the negative values in red. Oddly, this recipe does not work when plotting lines. This call to plot might seem to plot a line whose colors alternate between blue and green: > plot(x, type="l", col=c("blue", "green")) But it doesn’t work. The first color, "blue", is used for all line segments, and the second color, "green", is ignored. See Also See Recipe 5.3 regarding the Recycling Rule. Execute colors() to see a list of available colors, and use the segments function to plot line segments in multiple colors. 10.24 Graphing a Function Problem You want to graph the value of a function. Solution The curve function can graph a function, given the function and the limits of its domain: > curve(sin, -3, +3) # Graph the sine function from -3 to +3 Discussion The lefthand panel of Figure 10-23 shows a graph of the standard normal density func- tion. The graph was created like this: > curve(dnorm, -3.5, +3.5, + main="Std. Normal Density") The curve function calls the dnorm function for a range of arguments from −3.5 to +3.5 and then plots the result. Like any well-behaved, high-level graphics function, it accepts a main argument that specifies a title. 258 | Chapter 10: Graphics curve can graph any function that takes one argument and returns one value. The righthand panel of Figure 10-23 was created by defining a local function and plotting it: > f <- function(x) exp(-abs(x)) * sin(2*pi*x) > curve(f, -5, +5, main="Dampened Sine Wave") See Also See Recipe 2.12 for how to define a function. 10.25 Pausing Between Plots Problem You are creating several plots, and each plot is overwriting the previous one. You want R to pause between plots so you can view each one before it’s overwritten. Solution There is a global graphics option called ask. Set it to TRUE, and R will pause before each new plot: > par(ask=TRUE) When you are tired of R pausing between plots, set it to FALSE: > par(ask=FALSE) Figure 10-23. Graphing a function 10.25 Pausing Between Plots | 259 Discussion When ask is TRUE, R will print this message immediately before starting a new plot: Waiting to confirm page change... When you are ready, click once on the graphics window. R will begin the next plot. See Also If one graph is overwriting another, consider using Recipe 10.26 to plot multiple graphs in one frame. See Recipe 10.29 for more about changing graphical parameters. 10.26 Displaying Several Figures on One Page Problem You want to display several plots side by side on one page. Solution First, divide the graphics window into a matrix of N rows and M columns. This is done by setting the graphics parameter called mfrow. Its value is a two-element vector giving the number of rows and columns: > par(mfrow=(c(N,M)) # Divide the graphics window into N x M matrix Second, fill each cell in the matrix by repeatedly calling plot. After creating a 3 × 2 grid, for example, you would call plot six times to fill the entire grid. Discussion Let’s use a multifigure plot to display four different beta distributions. First, we create four plotting areas in a 2 × 2 grid: > par(mfrow=c(2,2)) Second, we call plot four times, plotting a different beta distribution each time:* > Quantile <- seq(from=0, to=1, length.out=30) > plot(Quantile, dbeta(Quantile, 2, 4), type="l", main="First") > plot(Quantile, dbeta(Quantile, 4, 2), type="l", main="Second") > plot(Quantile, dbeta(Quantile, 1, 1), type="l", main="Third") > plot(Quantile, dbeta(Quantile, 0.5, 0.5), type="l", main="Fourth") The result is shown in Figure 10-24. Observe that each plot has its own title (e.g., main="First"), which I used here to illustrate the plotting order. Each plot could also have its own labels, legend, grid, and other adornments. * Recall that the dbeta function computes the probability density function of the beta distribution. The second and third arguments are the distribution’s shape parameters. 260 | Chapter 10: Graphics Figure 10-24. Plots of several beta distributions When you create the graphical matrix using mfrow, R fills the graphics window row by row. In the example just given, R filled the first row and then the second row. You might prefer to fill it column by column instead. A graphics parameter called mfcol lets you do that. It works like mfrow but forces R to fill the matrix column by column. This call to par divides the graphics window into a matrix of two rows and three columns: > par(mfcol=c(3,3)) # mfcol, not mfrow Successive calls to plot will fill the first column, then the second column, and finally the third column. The mfrow and mfcol parameters let you create a simple matrix of plots. For more elab- orate designs, consider the layout function and the split.screen function. 10.26 Displaying Several Figures on One Page | 261 See Also Recipe 8.11 shows a similar example of plotting multiple figures. Another way to dis- play multiple plots is by opening multiple graphics windows; see Recipe 10.27. See Recipe 10.29 for more about the par function and graphical parameters. The grid package and the lattice package contain additional tools for multifigure layout. 10.27 Opening Additional Graphics Windows Problem You want to open another graphics window so you can simultaneously view two graphs. Solution Use the win.graph function to open a second graphics window: > win.graph() Subsequent graphics output will appear in the new window. Your original graphics window will remain unchanged. Discussion Every time you create a new graph, it overwrites the old contents of the graphics win- dow. Sometimes you want to preserve the existing graph while creating a new graph, usually so you can compare them side by side. The win.graph function creates a new graphics window. The new window becomes the active window, which means that all graphics output is directed there. All other graphics windows become inactive—their contents are frozen. Eventually, you may want to freeze the active window, switch to an inactive window, and overwrite that window. The dev.set function switches between windows. The active window becomes inactive, and the next window (in numeric order) becomes active: > dev.set() windows 2 dev.set returns the number of the now-active window. 262 | Chapter 10: Graphics By default, R creates a square window. You can supply width and height arguments to win.graph, thereby altering the default dimensions. I prefer graph dimensions based on the Golden Ratio (1.618...), for example, so I often create new graphics windows like this:† > win.graph(width=7.0, height=5.5) Finally, note that win.graph is a system-independent convenience function: it works on all platforms but is just a wrapper for the system-dependent function that actually creates the new graphics windows. You’ll have more control if you use the system- dependent function, but your code won’t be portable. See the help page for win.graph. See Also See Recipe 10.26 for creating multiple plots in one window. 10.28 Writing Your Plot to a File Problem You want to save your graphics in a file, such as a PNG, JPEG, or PostScript file. Solution Create the graphics on your screen; then call the savePlot function: > savePlot(filename="filename.ext", type="type") The type argument is platform specific. See the Devices help page for a list of types available on your machine: > help(Devices) Most platforms allows a type of "png" or "jpeg", but each platform has others, too. Discussion On Windows, a shortcut is to select the graphics window and then click on File → Save as... from the menu. That will prompt you for a file type and a file name before writing the file. The procedure is similar on OS X. Remember that the file will be written to your current working directory (unless you use an absolute file path), so be certain you know which directory that is before calling savePlot (Recipe 3.1). † Yes, I know that 7.0/5.5 is not the Golden Ratio. The width and height arguments are the dimensions of the outer frame; owing to adjustments for margins, the (inner) plot area will be smaller. This width and height yield a plot area with the appropriate dimensions on my computer. Your mileage may vary. 10.28 Writing Your Plot to a File | 263 The Solution will differ in a batch script: • Call a function to open a new graphics file, such as png(...) or jpeg(...). • Call plot and its friends to generate the graphics image. • Call dev.off() to close the graphics file. Functions for opening graphics files depend upon your platform and file format. Again, see help(Devices) for a list of available functions on your platform. A common function is png, which creates a PNG file; it is used like this: > png("filename.ext", width=w, height=h) Here w is the desired width and h is the desired height, both in pixels. Using that func- tion, the recipe works like this to create myPlot.png: > png("myPlot.png", width=648, height=432) # Or whatever dimensions work for you > plot(x, y, main="Scatterplot of X, Y") > dev.off() See Also See Recipe 3.1 for more about the current working directory. 10.29 Changing Graphical Parameters Problem You want to change a global parameter of the graphics software, such as line type, background color, or font size. Solution Use the par function, which lets you set values of global graphics parameters. For ex- ample, this call to par will change the default line width from 1 to 2: > par(lwd=2) Discussion Every graphics plot depends on many assumptions. How wide should the lines be? What’s the background color? What’s the foreground color? How large should the font be? How wide must the margins be? The assumed answers are given by global, graphical parameters and their default values. Thank goodness for those defaults; otherwise, the simplest plot would be a nightmare of details. Sometimes, however, the defaults don’t work for you, and you need to change them. The par function does that. There are dozens of such parameters, and serious graphics geeks will want to study the full list that is given on the help page for par. 264 | Chapter 10: Graphics To see a parameter’s current value, call par with the parameter name given as a string: > par("lty") # What is the default line type? [1] "solid" > par("bg") # What is the default background color? [1] "transparent" To set a parameter’s value, pass it to par as a function parameter. Table 10-1 lists some common uses. Table 10-1. Parameters for the par function Parameter and type Purpose Example ask=logical Pause before every new graph if TRUE (Recipe 10.25) par(ask=TRUE) bg="color" Background color par(bg="lightyellow") cex=number Height of text and plotted points, expressed as a multiple of normal size par(cex=1.5) col="color" Default plotting color par(col="blue") fg="color" Foreground color par(fg="gray") lty="linetype" Type of line: solid, dotted, dashed, etc. (Recipe 10.13) par(lty="dotted") lwd=number Line width: 1 = normal, 2 = thicker, 3 = even thicker, etc. par(lwd=2) mfcol=c(nr,nc) or mfrow=c(nr,nc) Create a multifigure plot matrix with nr rows and nc columns (Recipe 10.26) par(mfrow=c(2,2)) new=logical Used to plot one figure on top of another par(new=TRUE) pch=pointtype Default point type (see help page for points function) par(pch=21) xlog=logical Use logarithmic X scale par(xlog=TRUE) ylog=logical Use logarithmic Y scale par(ylog=TRUE) The par function comes with a warning, however. Changing a global parameter has a global effect, naturally. The change will affect all your plots, not just the next one, until you reverse the change. Furthermore, if you call packages that create their own graphics, the global change will also affect them. So think twice before you diddle with those globals: do you really want all lines in all graphics to be (say) magenta, dotted, and three times wider? Probably not, so use local pa- rameters rather than global parameters whenever possible. 10.29 Changing Graphical Parameters | 265 See Also The help page for par lists the global graphics parameters; the chapter of R in a Nutshell on graphics includes the list with useful annotations. R Graphics contains extensive explanations of graphics parameters. 266 | Chapter 10: Graphics CHAPTER 11 Linear Regression and ANOVA Introduction In statistics, modeling is where we get down to business. Models quantify the relation- ships between our variables. Models let us make predictions. A simple linear regression is the most basic model. It’s just two variables and is modeled as a linear relationship with an error term: yi = β0 + β1xi + εi We are given the data for x and y. Our mission is to fit the model, which will give us the best estimates for β0 and β1 (Recipe 11.1). That generalizes naturally to multiple linear regression, where we have multiple vari- ables on the righthand side of the relationship (Recipe 11.2): yi = β0 + β1ui + β2vi + β3wi + εi Statisticians call u, v, and w the predictors and y the response. Obviously, the model is useful only if there is a fairly linear relationship between the predictors and the response, but that requirement is much less restrictive than you might think. Recipe 11.11 dis- cusses transforming your variables into a (more) linear relationship so that you can use the well-developed machinery of linear regression. The beauty of R is that anyone can build these linear models. The models are built by a function, lm, which returns a model object. From the model object, we get the coef- ficients (βi) and regression statistics. It’s easy. The horror of R is that anyone can build these models. Nothing requires you to check that the model is reasonable, much less statistically significant. Before you blindly be- lieve a model, check it. Most of the information you need is in the regression summary (Recipe 11.4): Is the model statistically significant? Check the F statistic at the bottom of the summary. 267 Are the coefficients significant? Check the coefficient’s t statistics and p-values in the summary, or check their confidence intervals (Recipe 11.13). Is the model useful? Check the R2 near the bottom of the summary. Does the model fit the data well? Plot the residuals and check the regression diagnostics (Recipes 11.14 and 11.15). Does the data satisfy the assumptions behind linear regression? Check whether the diagnostics confirm that a linear model is reasonable for your data (Recipe 11.15). ANOVA Analysis of variance (ANOVA) is a powerful statistical technique. First-year graduate students in statistics are taught ANOVA almost immediately because of its importance, both theoretical and practical. I am often amazed, however, at the extent to which people outside the field are unaware of its purpose and value. Regression creates a model, and ANOVA is one method of evaluating such models. The mathematics of ANOVA are intertwined with the mathematics of regression, so statisticians usually present them together; I follow that tradition here. ANOVA is actually a family of techniques that are connected by a common mathe- matical analysis. This chapter mentions several applications: One-way ANOVA This is the simplest application of ANOVA. Suppose you have data samples from several populations and are wondering whether the populations have different means. One-way ANOVA answers that question. If the populations have normal distributions, use the oneway.test function (Recipe 11.20); otherwise, use the nonparametric version, the kruskal.test function (Recipe 11.23). Model comparison When you add or delete a predictor variable from a linear regression, you want to know whether that change did or did not improve the model. The anova function compares two regression models and reports whether they are significantly differ- ent (Recipe 11.24). ANOVA table The anova function can also construct the ANOVA table of a linear regression model, which includes the F statistic needed to gauge the model’s statistical sig- nificance (Recipe 11.3). This important table is discussed in nearly every textbook on regression. The following references contain more about the mathematics of ANOVA. 268 | Chapter 11: Linear Regression and ANOVA See Also There are many good texts on linear regression. One of my favorites is Applied Linear Regression Models (4th ed.) by Kutner, Nachtsheim, and Neter (McGraw-Hill/Irwin). I generally follow their terminology and conventions in this chapter. I also like Practical Regression and ANOVA Using R by Julian Faraway because it illustrates regression using R and is quite readable. It is unpublished, but you can download the free PDF from CRAN. 11.1 Performing Simple Linear Regression Problem You have two vectors, x and y, that hold paired observations: (x1, y1), (x2, y2), ..., (xn, yn). You believe there is a linear relationship between x and y, and you want to create a regression model of the relationship. Solution The lm function performs a linear regression and reports the coefficients: > lm(y ~ x) Call: lm(formula = y ~ x) Coefficients: (Intercept) x 17.72 3.25 Discussion Simple linear regression involves two variables: a predictor variable, often called x; and a response variable, often called y. The regression uses the ordinary least-squares (OLS) algorithm to fit the linear model: yi = β0 + β1xi + εi where β0 and β1 are the regression coefficients and the εi are the error terms. The lm function can perform linear regression. The main argument is a model formu- la, such as y ~ x. The formula has the response variable on the left of the tilde character (~) and the predictor variable on the right. The function estimates the regression coef- ficients, β0 and β1, and reports them as the intercept and the coefficient of x, respectively: Coefficients: (Intercept) x 17.72 3.25 11.1 Performing Simple Linear Regression | 269 In this case, the regression equation is: yi = 17.72 + 3.25xi + εi It is quite common for data to be captured inside a data frame, in which case you want to perform a regression between two data frame columns. Here, x and y are columns of a data frame dfrm: > dfrm x y 1 0.04781401 5.406651 2 1.90857986 19.941568 3 2.79987246 23.922613 4 4.46755305 32.432904 5 3.76490363 44.259268 6 5.92364632 61.151480 7 8.04611587 26.305505 8 7.11097986 43.606087 9 9.73645966 58.262112 10 9.19324543 57.631029 . . (etc.) . The lm function lets you specify a data frame by using the data parameter. If you do, the function will take the variables from the data frame and not from your workspace: > lm(y ~ x, data=dfrm) # Take x and y from dfrm Call: lm(formula = y ~ x, data = dfrm) Coefficients: (Intercept) x 17.72 3.25 11.2 Performing Multiple Linear Regression Problem You have several predictor variables (e.g., u, v, and w) and a response variable y. You believe there is a linear relationship between the predictors and the response, and you want to perform a linear regression on the data. Solution Use the lm function. Specify the multiple predictors on the righthand side of the for- mula, separated by plus signs (+): > lm(y ~ u + v + w) 270 | Chapter 11: Linear Regression and ANOVA Discussion Multiple linear regression is the obvious generalization of simple linear regression. It allows multiple predictor variables instead of one predictor variable and still uses OLS to compute the coefficients of a linear equation. The three-variable regression just given corresponds to this linear model: yi = β0 + β1ui + β2vi + β3wi + εi R uses the lm function for both simple and multiple linear regression. You simply add more variables to the righthand side of the model formula. The output then shows the coefficients of the fitted model: > lm(y ~ u + v + w) Call: lm(formula = y ~ u + v + w) Coefficients: (Intercept) u v w 1.4222 1.0359 0.9217 0.7261 The data parameter of lm is especially valuable when the number of variables increases, since it’s much easier to keep your data in one data frame than in many separate vari- ables. Suppose your data is captured in a data frame, such as the dfrm variable shown here: > dfrm y u v w 1 6.584519 0.79939065 2.7971413 4.366557 2 6.425215 -2.31338537 2.7836201 4.515084 3 7.830578 1.71736899 2.7570401 3.865557 4 2.757777 1.27652888 0.4191765 2.547935 5 5.794566 0.39643488 2.3785468 3.265971 6 7.314611 1.82247760 1.8291302 4.518522 7 2.533638 -1.34186107 2.3472593 2.570884 8 8.696910 0.75946803 3.4028180 4.442560 9 6.304464 0.92000133 2.0654513 2.835248 10 8.095094 1.02341093 2.6729252 3.868573 . . (etc.) . When we supply dfrm to the data parameter of lm, R looks for the regression variables in the columns of the data frame: > lm(y ~ u + v + w, data=dfrm) Call: lm(formula = y ~ u + v + w, data = dfrm) Coefficients: (Intercept) u v w 1.4222 1.0359 0.9217 0.7261 11.2 Performing Multiple Linear Regression | 271 See Also See Recipe 11.1 for simple linear regression. 11.3 Getting Regression Statistics Problem You want the critical statistics and information regarding your regression, such as R2, the F statistic, confidence intervals for the coefficients, residuals, the ANOVA table, and so forth. Solution Save the regression model in a variable, say m: > m <- lm(y ~ u + v + w) Then use functions to extract regression statistics and information from the model: anova(m) ANOVA table coefficients(m) Model coefficients coef(m) Same as coefficients(m) confint(m) Confidence intervals for the regression coefficients deviance(m) Residual sum of squares effects(m) Vector of orthogonal effects fitted(m) Vector of fitted y values residuals(m) Model residuals resid(m) Same as residuals(m) summary(m) Key statistics, such as R2, the F statistic, and the residual standard error (σ) vcov(m) Variance–covariance matrix of the main parameters 272 | Chapter 11: Linear Regression and ANOVA Discussion When I started using R, the documentation said use the lm function to perform linear regression. So I did something like this, getting the output shown in Recipe 11.2: > lm(y ~ u + v + w) Call: lm(formula = y ~ u + v + w) Coefficients: (Intercept) u v w 1.4222 1.0359 0.9217 0.7261 I was so disappointed! The output was nothing compared to other statistics packages such as SAS. Where is R2? Where are the confidence intervals for the coefficients? Where is the F statistic, its p-value, and the ANOVA table? Of course, all that information is available—you just have to ask for it. Other statistics systems dump everything and let you wade through it. R is more minimalist. It prints a bare-bones output and lets you request what more you want. The lm function returns a model object that you can assign to a variable: > m <- lm(y ~ u + v + w) From the model object, you can extract important information using specialized func- tions. The most important function is summary: > summary(m) Call: lm(formula = y ~ u + v + w) Residuals: Min 1Q Median 3Q Max -3.3965 -0.9472 -0.4708 1.3730 3.1283 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.4222 1.4036 1.013 0.32029 u 1.0359 0.2811 3.685 0.00106 ** v 0.9217 0.3787 2.434 0.02211 * w 0.7261 0.3652 1.988 0.05744 . --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 1.625 on 26 degrees of freedom Multiple R-squared: 0.4981, Adjusted R-squared: 0.4402 F-statistic: 8.603 on 3 and 26 DF, p-value: 0.0003915 The summary shows the estimated coefficients. It shows the critical statistics, such as R2 and the F statistic. It shows an estimate of σ, the standard error of the residuals. The summary is so important that there is an entire recipe devoted to understanding it (Recipe 11.4). 11.3 Getting Regression Statistics | 273 D o wnload from Wow! eBook There are specialized extractor functions for other important information: Model coefficients (point estimates) > coef(m) (Intercept) u v w 1.4222050 1.0358725 0.9217432 0.7260653 Confidence intervals for model coefficients > confint(m) 2.5 % 97.5 % (Intercept) -1.46302727 4.307437 u 0.45805053 1.613694 v 0.14332834 1.700158 w -0.02466125 1.476792 Model residuals > resid(m) 1 2 3 4 5 6 -1.41440465 1.55535335 -0.71853222 -2.22308948 -0.60201283 -0.96217874 7 8 9 10 11 12 -1.52877080 0.12587924 -0.03313637 0.34017869 1.28200521 -0.90242817 13 14 15 16 17 18 2.04481731 1.13630451 -1.19766679 -0.60210494 1.79964497 1.25941264 19 20 21 22 23 24 -2.03323530 1.40337142 -1.25605632 -0.84860707 -0.47307439 -0.76335244 25 26 27 28 29 30 2.16275214 1.53483492 1.65085364 -3.39647629 -0.46853750 3.12825629 Residual sum of squares > deviance(m) [1] 68.69616 ANOVA table > anova(m) Analysis of Variance Table Response: y Df Sum Sq Mean Sq F value Pr(>F) u 1 27.916 27.9165 10.5658 0.003178 ** v 1 29.830 29.8299 11.2900 0.002416 ** w 1 10.442 10.4423 3.9522 0.057436 . Residuals 26 68.696 2.6422 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 If you find it annoying to save the model in a variable, you are welcome to use one- liners such as this: > summary(lm(y ~ u + v + w)) 274 | Chapter 11: Linear Regression and ANOVA See Also See Recipe 11.4. See Recipe 11.16 for regression statistics specific to model diagnostics. 11.4 Understanding the Regression Summary Problem You created a linear regression model, m. However, you are confused by the output from summary(m). Discussion The model summary is important because it links you to the most critical regression statistics. Here is the model summary from Recipe 11.3: > summary(m) Call: lm(formula = y ~ u + v + w) Residuals: Min 1Q Median 3Q Max -3.3965 -0.9472 -0.4708 1.3730 3.1283 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.4222 1.4036 1.013 0.32029 u 1.0359 0.2811 3.685 0.00106 ** v 0.9217 0.3787 2.434 0.02211 * w 0.7261 0.3652 1.988 0.05744 . --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 1.625 on 26 degrees of freedom Multiple R-squared: 0.4981, Adjusted R-squared: 0.4402 F-statistic: 8.603 on 3 and 26 DF, p-value: 0.0003915 Let’s dissect this summary by section. We’ll read it from top to bottom—even though the most important statistic, the F statistic, appears at the end: Call Call: lm(formula = y ~ u + v + w) This shows how lm was called when it created the model, which is important for putting this summary into the proper context. 11.4 Understanding the Regression Summary | 275 Residuals statistics Residuals: Min 1Q Median 3Q Max -3.3965 -0.9472 -0.4708 1.3730 3.1283 Ideally, the regression residuals would have a perfect, normal distribution. These statistics help you identify possible deviations from normality. The OLS algorithm is mathematically guaranteed to produce residuals with a mean of zero.* Hence the sign of the median indicates the skew’s direction, and the magnitude of the median indicates the extent. In this case the median is negative, which suggests some skew to the left. If the residuals have a nice, bell-shaped distribution, then the first quartile (1Q) and third quartile (3Q) should have about the same magnitude. In this example, the larger magnitude of 3Q versus 1Q (1.3730 versus 0.9472) indicates a slight skew to the right in our data, although the negative median makes the situation less clear-cut. The Min and Max residuals offer a quick way to detect extreme outliers in the data, since extreme outliers (in the response variable) produce large residuals. Coefficients Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.4222 1.4036 1.013 0.32029 u 1.0359 0.2811 3.685 0.00106 ** v 0.9217 0.3787 2.434 0.02211 * w 0.7261 0.3652 1.988 0.05744 . --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 The column labeled Estimate contains the estimated regression coefficients as cal- culated by ordinary least squares. Theoretically, if a variable’s coefficient is zero then the variable is worthless; it adds nothing to the model. Yet the coefficients shown here are only estimates, and they will never be exactly zero. We therefore ask: Statistically speaking, how likely is it that the true coefficient is zero? That is the purpose of the t statistics and the p-values, which in the summary are labeled (respectively) t value and Pr(>|t|). The p-value is a probability. It gauges the likelihood that the coefficient is not significant, so smaller is better. Big is bad because it indicates a high likelihood of insignificance. In this example, the p-value for the u coefficient is a mere 0.00106, so u is likely significant. The p-value for w, however, is 0.05744; this is just over * Unless you performed the linear regression without an intercept term (Recipe 11.5). 276 | Chapter 11: Linear Regression and ANOVA our conventional limit of 0.05, which suggests that w is likely insignificant.† Var- iables with large p-values are candidates for elimination. A handy feature is that R flags the significant variables for quick identification. Do you notice the extreme righthand column containing double asterisks (**), a single asterisk (*), and a period(.)? That column highlights the significant variables. The line labeled "Signif. codes" at the bottom gives a cryptic guide to the flags’ meanings: *** p-value between 0 and 0.001 ** p-value between 0.001 and 0.01 * p-value between 0.01 and 0.05 . p-value between 0.05 and 0.1 (blank) p-value between 0.1 and 1.0 The column labeled Std. Error is the standard error of the estimated coefficient. The column labeled t value is the t statistic from which the p-value was calculated. Residual standard error Residual standard error: 1.625 on 26 degrees of freedom This reports the standard error of the residuals (σ)—that is, the sample standard deviation of ε. R2 (coefficient of determination) Multiple R-squared: 0.4981, Adjusted R-squared: 0.4402 R2 is a measure of the model’s quality. Bigger is better. Mathematically, it is the fraction of the variance of y that is explained by the regression model. The remain- ing variance is not explained by the model, so it must be due to other factors (i.e., unknown variables or sampling variability). In this case, the model explains 0.4981 (49.81%) of the variance of y, and the remaining 0.5019 (50.19%) is unexplained. That being said, I strongly suggest using the adjusted rather than the basic R2. The adjusted value accounts for the number of variables in your model and so is a more realistic assessment of its effectiveness. In this case, then, I would use 0.4402, not 0.4981. F statistic F-statistic: 8.603 on 3 and 26 DF, p-value: 0.0003915 The F statistic tells you whether the model is significant or insignificant. The model is significant if any of the coefficients are nonzero (i.e., if βi ≠ 0 for some i). It is insignificant if all coefficients are zero (β1 = β2 = … = βn = 0). † The significance level of α = 0.05 is the convention observed in this book. Your application might instead use α = 0.10, α = 0.01, or some other value. See the “Introduction” to Chapter 9. 11.4 Understanding the Regression Summary | 277 Conventionally, a p-value of less than 0.05 indicates that the model is likely sig- nificant (one or more βi are nonzero) whereas values exceeding 0.05 indicate that the model is likely not significant. Here, the probability is only 0.000391 that our model is insignificant. That’s good. Most people look at the R2 statistic first. The statistician wisely starts with the F statistic, for if the model is not significant then nothing else matters. See Also See Recipe 11.3 for more on extracting statistics and information from the model object. 11.5 Performing Linear Regression Without an Intercept Problem You want to perform a linear regression, but you want to force the intercept to be zero. Solution Add "+ 0" to the righthand side of your regression formula. That will force lm to fit the model with a zero intercept: > lm(y ~ x + 0) The corresponding regression equation is: yi = βxi + εi Discussion Linear regression ordinarily includes an intercept term, so that is the default in R. In rare cases, however, you may want to fit the data while assuming that the intercept is zero. In this you make a modeling assumption: when x is zero, y should be zero. When you force a zero intercept, the lm output includes a coefficient for x but no intercept for y, as shown here: > lm(y ~ x + 0) Call: lm(formula = y ~ x + 0) Coefficients: x 2.582 I strongly suggest you check that modeling assumption before proceeding. Perform a regression with an intercept; then see if the intercept could plausibly be zero. Check the intercept’s confidence interval. In this example, the confidence interval is (−12.17, 10.17): 278 | Chapter 11: Linear Regression and ANOVA > confint(lm(y ~ x)) 2.5 % 97.5 % (Intercept) -12.171368 10.170070 x 2.012768 3.248262 Because the confidence interval contains zero, it is statistically plausible that the inter- cept could be zero. Now it is reasonable to rerun the regression while forcing a zero intercept. 11.6 Performing Linear Regression with Interaction Terms Problem You want to include an interaction term in your regression. Solution The R syntax for regression formulas lets you specify interaction terms. The interaction of two variables, u and v, is indicated by separating their names with an asterisk (*): > lm(y ~ u*v) This corresponds to the model yi = β0 + β1ui + β2vi + β3uivi + εi, which includes the first-order interaction term β3uivi. Discussion In regression, an interaction occurs when the product of two predictor variables is also a significant predictor (i.e., in addition to the predictor variables themselves). Suppose we have two predictors, u and v, and want to include their interaction in the regression. This is expressed by the following equation: yi = β0 + β1ui + β2vi + β3uivi + εi Here the product term, β3uivi, is called the interaction term. The R formula for that equation is: y ~ u*v When you write y ~ u*v, R automatically includes u, v, and their product in the model. This is for a good reason. If a model includes an interaction term, such as β3uivi, then regression theory tells us the model should also contain the constituent variables ui and vi. Likewise, if you have three predictors (u, v, and w) and want to include all their inter- actions, separate them by asterisks: y ~ u*v*w This corresponds to the regression equation: yi = β0 + β1ui + β2vi + β3wi + β4uivi + β5uiwi + β6viwi + β7uiviwi + εi 11.6 Performing Linear Regression with Interaction Terms | 279 Now we have all the first-order interactions and a second-order interaction (β7uiviwi). Sometimes, however, you may not want every possible interaction. You can explicitly specify a single product by using the colon operator (:). For example, u:v:w denotes the product term βuiviwi but without all possible interactions. So the R formula: y ~ u + v + w + u:v:w corresponds to the regression equation: yi = β0 + β1ui + β2vi + β3wi + β4uiviwi + εi It might seem odd that colon (:) means pure multiplication while asterisk (*) means both multiplication and inclusion of constituent terms. Again, this is because we nor- mally incorporate the constituents when we include their interaction, so making that the default for asterisk makes sense. There is some additional syntax for easily specifying many interactions: (u + v + ... + w)^2 Include all variables (u, v, ..., w) and all their first-order interactions. (u + v + ... + w)^3 Include all variables, all their first-order interactions, and all their second-order interactions. (u + v + ... + w)^4 And so forth. Both the asterisk (*) and the colon (:) follow a “distributive law”, so the following notations are also allowed: x*(u + v + ... + w) Same as x*u + x*v + ... + x*w (which is the same as x + u + v + ... + w + x:u + x:v + ... + x:w). x:(u + v + ... + w) Same as x:u + x:v + ... + x:w. All this syntax gives you some flexibility in writing your formula. For example, these three formulas are equivalent: y ~ u*v y ~ u + v + u:v y ~ (u + v)^2 They all define the same regression equation, yi = β0 + β1ui + β2vi + β3uivi + εi . See Also The full syntax for formulas is richer than described here. See R in a Nutshell (O’Reilly) or the R Language Definition for more details. 280 | Chapter 11: Linear Regression and ANOVA 11.7 Selecting the Best Regression Variables Problem You are creating a new regression model or improving an existing model. You have the luxury of many regression variables, and you want to select the best subset of those variables. Solution The step function can perform stepwise regression, either forward or backward. Back- ward stepwise regression starts with many variables and removes the underperformers: > full.model <- lm(y ~ x1 + x2 + x3 + x4) > reduced.model <- step(full.model, direction="backward") Forward stepwise regression starts with a few variables and adds new ones to improve the model until it cannot be improved further: > min.model <- lm(y ~ 1) > fwd.model <- step(min.model, direction="forward", scope=( ~ x1 + x2 + x3 + x4 )) Discussion When you have many predictors, it can be quite difficult to choose the best subset. Adding and removing individual variables affects the overall mix, so the search for “the best” can become tedious. The step function automates that search. Backward stepwise regression is the easiest approach. Start with a model that includes all the predictors. We call that the full model. The model summary, shown here, indicates that not all predictors are statisti- cally significant: > full.model <- lm(y ~ x1 + x2 + x3 + x4) > summary(full.model) Call: lm(formula = y ~ x1 + x2 + x3 + x4) Residuals: Min 1Q Median 3Q Max -34.405 -5.153 2.025 6.525 19.186 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 10.274 4.573 2.247 0.0337 * x1 -102.298 47.558 -2.151 0.0413 * x2 4.362 40.237 0.108 0.9145 x3 75.115 34.236 2.194 0.0377 * x4 26.286 42.239 0.622 0.5394 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 11.7 Selecting the Best Regression Variables | 281 Residual standard error: 12.19 on 25 degrees of freedom Multiple R-squared: 0.8848, Adjusted R-squared: 0.8664 F-statistic: 48 on 4 and 25 DF, p-value: 2.235e-11 We want to eliminate the insignificant variables, so we use step to incrementally elim- inate the underperformers. The result is called the reduced model: > reduced.model <- step(full.model, direction="backward") Start: AIC=154.58 y ~ x1 + x2 + x3 + x4 Df Sum of Sq RSS AIC - x2 1 1.75 3718.5 152.60 - x4 1 57.58 3774.3 153.04 3716.7 154.58 - x1 1 687.89 4404.6 157.68 - x3 1 715.67 4432.4 157.87 Step: AIC=152.6 y ~ x1 + x3 + x4 Df Sum of Sq RSS AIC - x4 1 77.82 3796.3 151.22 3718.5 152.60 - x3 1 737.39 4455.9 156.02 - x1 1 787.96 4506.4 156.36 Step: AIC=151.22 y ~ x1 + x3 Df Sum of Sq RSS AIC 3796.3 151.22 - x1 1 884.44 4680.7 155.50 - x3 1 964.24 4760.5 156.01 The output from step shows the sequence of models that it explored. In this case, step removed x2 and x4 and left only x1 and x3 in the final (reduced) model. The sum- mary of the reduced model shows that it contains only significant predictors: > summary(reduced.model) Call: lm(formula = y ~ x1 + x3) Residuals: Min 1Q Median 3Q Max -36.192 -2.677 2.049 7.453 19.704 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 10.267 4.437 2.314 0.0285 * x1 -79.265 31.604 -2.508 0.0185 * x3 82.726 31.590 2.619 0.0143 * --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 282 | Chapter 11: Linear Regression and ANOVA Residual standard error: 11.86 on 27 degrees of freedom Multiple R-squared: 0.8823, Adjusted R-squared: 0.8736 F-statistic: 101.2 on 2 and 27 DF, p-value: 2.842e-13 Backward stepwise regression is easy, but sometimes it’s not feasible to start with “ev- erything” because you have too many candidate variables. In that case use forward stepwise regression, which will start with nothing and incrementally add variables that improve the regression. It stops when no further improvement is possible. A model that “starts with nothing” may look odd at first: > min.model <- lm(y ~ 1) This is a model with a response variable (y) but no predictor variables. (All the fitted values for y are simply the mean of y, which is what you would guess if no predictors were available.) We must tell step which candidate variables are available for inclusion in the model. That is the purpose of the scope argument. The scope is a formula with nothing on the lefthand side of the tilde (~) and candidate variables on the righthand side: > fwd.model <- step(min.model, + direction="forward", + scope=( ~ x1 + x2 + x3 + x4), + trace=0 ) Here we see that x1, x2, x3, and x4 are all candidates for inclusion. (I also included trace=0 to inhibit the voluminous output from step.) The resulting model has two significant predictors and no insignificant predictors: > summary(fwd.model) Call: lm(formula = y ~ x3 + x1) Residuals: Min 1Q Median 3Q Max -36.192 -2.677 2.049 7.453 19.704 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 10.267 4.437 2.314 0.0285 * x3 82.726 31.590 2.619 0.0143 * x1 -79.265 31.604 -2.508 0.0185 * --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 11.86 on 27 degrees of freedom Multiple R-squared: 0.8823, Adjusted R-squared: 0.8736 F-statistic: 101.2 on 2 and 27 DF, p-value: 2.842e-13 11.7 Selecting the Best Regression Variables | 283 The step-forward algorithm reached the same model as the step-backward model by including x1 and x3 but excluding x2 and x4. This is a toy example, so that is not sur- prising. In real applications, I suggest trying both the forward and the backward re- gression and then comparing the results. You might be surprised. Finally, don’t get carried away by stepwise regression. It is not a panacea, it cannot turn junk into gold, and it is definitely not a substitute for choosing predictors carefully and wisely. You might think: “Oh boy! I can generate every possible interaction term for my model, then let step choose the best ones! What a model I’ll get!” You’d be thinking of something like this: > full.model <- lm(y ~ (x1 + x2 + x3 + x4)^4) # All possible interactions > reduced.model <- step(full.model, direction="backward") This does not work well: most of the interaction terms are meaningless. The step func- tion becomes overwhelmed, and you are left with many insignificant terms. See Also See Recipe 11.24. 11.8 Regressing on a Subset of Your Data Problem You want to fit a linear model to a subset of your data, not to the entire dataset. Solution The lm function has a subset parameter that specifies which data elements should be used for fitting. The parameter’s value can be any index expression that could index your data. This shows a fitting that uses only the first 100 observations: > lm(y ~ x, subset=1:100) # Use only x[1:100] Discussion You will often want to regress only a subset of your data. This can happen, for example, when using in-sample data to create the model and out-of-sample data to test it. The lm function has a parameter, subset, that selects the observations used for fitting. The value of subset is a vector. It can be a vector of index values, in which case lm selects only the indicated observations from your data. It can also be a logical vector, the same length as your data, in which case lm selects the observations with a corresponding TRUE. Suppose you have 1,000 observations of (x, y) pairs and want to fit your model using only the first half of those observations. Use a subset parameter of 1:500, indicating lm should use observations 1 through 500: 284 | Chapter 11: Linear Regression and ANOVA > lm(y ~ x, subset=1:500) More generally, you can use the expression 1:floor(length(x)/2) to select the first half of your data, regardless of size: > lm(y ~ x, subset=1:floor(length(x)/2)) Let’s say your data was collected in several labs and you have a factor, lab, that identifies the lab of origin. You can limit your regression to observations collected in New Jersey by using a logical vector that is TRUE only for those observations: > lm(y ~ x, subset=(lab == "NJ")) 11.9 Using an Expression Inside a Regression Formula Problem You want to regress on calculated values, not simple variables, but the syntax of a regression formula seems to forbid that. Solution Embed the expressions for the calculated values inside the I(...) operator. That will force R to calculate the expression and use the calculated value for the regression. Discussion If you want to regress on the sum of u and v, then this is your regression equation: yi = β0 + β1(ui + vi) + εi How do you write that equation as a regression formula? This won’t work: > lm(y ~ u + v) # Not quite right Here R will interpret u and v as two separate predictors, each with its own regression coefficient. Likewise, suppose your regression equation is: yi = β0 + β1ui + β2ui2 + εi This won’t work: > lm(y ~ u + u^2) # That's an interaction, not a quadratic term R will interpret u^2 as an interaction term (Recipe 11.6) and not as the square of u. The solution is to surround the expressions by the I(...) operator, which prevents the expressions from being interpreted as a regression formula. Instead, it forces R to cal- culate the expression’s value and then incorporate that value directly into the regres- sion. Thus the first example becomes: > lm(y ~ I(u + v)) In response to that command, R computes u + v and then regresses y on the sum. 11.9 Using an Expression Inside a Regression Formula | 285 For the second example we use: > lm(y ~ u + I(u^2)) Here R computes the square of u and then regresses on the sum u + u2. All the basic binary operators (+, -, *, /, ^) have special meanings inside a regression formula. For this reason, you must use the I(...) operator whenever you incorporate calculated values into a regression. A beautiful aspect of these embedded transformations is that R remembers the trans- formations and applies them when you make predictions from the model. Consider the quadratic model described by the second example. It uses u and u^2, but we supply the value of u only and R does the heavy lifting. We don’t need to calculate the square of u ourselves: > m <- lm(y ~ u + I(u^2)) > predict(m, newdata=data.frame(u=13.4)) # R plugs u and u^2 into the calculation 1 11531.59 See Also See Recipe 11.10 for the special case of regression on a polynomial. See Recipe 11.11 for incorporating other data transformations into the regression. 11.10 Regressing on a Polynomial Problem You want to regress y on a polynomial of x. Solution Use the poly(x,n) function in your regression formula to regress on an n-degree poly- nomial of x. This example models y as a cubic function of x: > lm(y ~ poly(x,3,raw=TRUE)) The example’s formula corresponds to the following cubic regression equation: yi = β0 + β1xi + β2xi2 + β3xi3 + εi Discussion When I first used a polynomial model in R, I did something clunky like this: > x_sq <- x^2 > x_cub <- x^3 > m <- lm(y ~ x + x_sq + x_cub) Obviously, this was quite annoying, and it littered my workspace with extra variables. 286 | Chapter 11: Linear Regression and ANOVA It’s much easier to write: > m <- lm(y ~ poly(x,3,raw=TRUE)) The raw=TRUE is necessary. Without it, the poly function computes orthogonal poly- nomials instead of simple polynomials. Beyond the convenience, a huge advantage is that R will calculate all those powers of x when you make predictions from the model (Recipe 11.18). Without that, you are stuck calculating x2 and x3 yourself every time you employ the model. Here is another good reason to use poly. You cannot write your regression formula in this way: > lm(y ~ x + x^2 + x^3) # Does not do what you think! R will interpret x^2 and x^3 as interaction terms, not as powers of x. The resulting model is a one-term linear regression, completely unlike your expectation. You could write the regression formula like this: > lm(y ~ x + I(x^2) + I(x^3)) But that’s getting pretty verbose. Just use poly. See Also See Recipe 11.6 for more about interaction terms. See Recipe 11.11 for other transfor- mations on regression data. 11.11 Regressing on Transformed Data Problem You want to build a regression model for x and y, but they do not have a linear relationship. Solution You can embed the needed transformation inside the regression formula. If, for exam- ple, y must be transformed into log(y), then the regression formula becomes: > lm(log(y) ~ x) Discussion A critical assumption behind the lm function for regression is that the variables have a linear relationship. To the extent this assumption is false, the resulting regression be- comes meaningless. Fortunately, many datasets can be transformed into a linear relationship before apply- ing lm. Figure 11-1 shows an example of exponential decay. The left panel shows the 11.11 Regressing on Transformed Data | 287 original data, z. The dotted line shows a linear regression on the original data; clearly, it’s a lousy fit. If the data is really exponential, then a possible model is: z = exp[β0 + β1t + ε] where t is time and exp[⋅] is the exponential function (ex). This is not linear, of course, but we can linearize it by taking logarithms: log(z) = β0 + β1t + ε In R, that regression is simple because we can embed the log transform directly into the regression formula: > m <- lm(log(z) ~ t) > summary(m) Call: lm(formula = log(z) ~ t) Residuals: Min 1Q Median 3Q Max -0.17366 -0.09070 -0.03761 0.08080 0.36454 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -0.22059 0.01653 -13.35 <2e-16 *** t -1.16110 0.02056 -56.46 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.1275 on 98 degrees of freedom Multiple R-squared: 0.9702, Adjusted R-squared: 0.9699 F-statistic: 3188 on 1 and 98 DF, p-value: < 2.2e-16 The right panel of Figure 11-1 shows the plot of log(z) versus time. Superimposed on that plot is their regression line. The fit appears to be much better; this is confirmed by the R2 = 0.97, compared with 0.76 for the linear regression on the original data. You can embed other functions inside your formula. If you thought the relationship was quadratic, you could use a square-root transformation: > lm(sqrt(y) ~ month) You can apply transformations to variables on both sides of the formula, of course. This formula regresses y on the square root of x: > lm(y ~ sqrt(x)) This regression is for a log-log relationship between x and y: > lm(log(y) ~ log(x)) See Also See Recipe 11.12. 288 | Chapter 11: Linear Regression and ANOVA 11.12 Finding the Best Power Transformation (Box–Cox Procedure) Problem You want to improve your linear model by applying a power transformation to the response variable. Solution Use the Box–Cox procedure, which is implemented by the boxcox function of the MASS package. The procedure will identify a power, λ, such that transforming y into yλ will improve the fit of your model: > library(MASS) > m <- lm(y ~ x) > boxcox(m) Discussion To illustrate the Box–Cox transformation, let’s create some artificial data using the equation y−1.5 = x + ε, where ε is an error term: > x <− 10:100 > eps <- rnorm(length(x), sd=5) > y <- (x + eps)^(−1/1.5) Figure 11-1. Transforming data into a linear relationship 11.12 Finding the Best Power Transformation (Box–Cox Procedure) | 289 Then we will (mistakenly) model the data using a simple linear regression and derive an adjusted R2 of 0.6935: > m <- lm(y ~ x) > summary(m) Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max -0.038866 -0.016128 -0.004948 0.008367 0.083354 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.580e-01 5.710e-03 27.67 <2e-16 *** x -1.340e-03 9.368e-05 -14.30 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.02347 on 89 degrees of freedom Multiple R-squared: 0.6969, Adjusted R-squared: 0.6935 F-statistic: 204.6 on 1 and 89 DF, p-value: < 2.2e-16 When plotting the residuals against the fitted values, we get a clue that something is wrong: > plot(m, which=1) # Plot only the fitted vs residuals The resulting plot, shown in the left panel of Figure 11-2, has a parabolic shape. A possible fix is a power transformation on y, so we run the Box–Cox procedure: > library(MASS) > bc <- boxcox(m) The boxcox function creates a plot, shown in the right panel of Figure 11-2, that plots values of λ against the log-likelihood of the resulting model. We want to maximize that log-likelihood, so the function draws a line at the best value and also draws lines at the limits of its confidence interval. In this case, it looks like the best value is around −1.5, with a confidence interval of about (−1.75, −1.25). Oddly, the boxcox function does not return the best value of λ. Rather, it returns the (x, y) pairs displayed in the plot. It’s pretty easy to find the values of λ that yield the largest log-likelihood y. We use the which.max function: > which.max(bc$y) [1] 13 Then this gives us the position of the corresponding λ: > lambda <- bc$x[which.max(bc$y)] > lambda [1] -1.515152 290 | Chapter 11: Linear Regression and ANOVA The function reports that the best λ is −1.515. In an actual application, I would urge you to interpret this number and choose the power that makes sense to you—rather than blindly accepting this “best” value. Use the graph to assist you in that interpreta- tion. Here, I’ll go with −1.515. We can apply the power transform to y and then fit the revised model; this gives a much better R2 of 0.9595: > z <- y^lambda > m2 <- lm(z ~ x) > summary(m2) Call: lm(formula = z ~ x) Residuals: Min 1Q Median 3Q Max -12.4325 -3.6608 -0.8793 2.3856 15.0977 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 0.61801 1.36496 0.453 0.652 x 1.03382 0.02239 46.164 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 5.612 on 89 degrees of freedom Multiple R-squared: 0.9599, Adjusted R-squared: 0.9595 F-statistic: 2131 on 1 and 89 DF, p-value: < 2.2e-16 Figure 11-2. Box–Cox plots 11.12 Finding the Best Power Transformation (Box–Cox Procedure) | 291 For those who prefer one-liners, the transformation can be embedded right into the revised regression formula: > m2 <- lm(I(y^lambda) ~ x) By default, boxcox searches for values of λ in the range −2 to +2. You can change that via the lambda argument; see the help page for details. I suggest viewing the Box–Cox result as a starting point, not as a definitive answer. If the confidence interval for λ includes 1.0, it may be that no power transformation is actually helpful. As always, inspect the residuals before and after the transformation. Did they really improve? See Also See Recipes 11.11 and 11.15. 11.13 Forming Confidence Intervals for Regression Coefficients Problem You are performing linear regression and you need the confidence intervals for the regression coefficients. Solution Save the regression model in an object; then use the confint function to extract confi- dence intervals: > m <- lm(y ~ x1 + x2) > confint(m) Discussion The Solution uses the model y = β0 + β1(x1)i + β2(x2)i + εi. The confint function returns the confidence intervals for the intercept (β0), the coefficient of x1 (β1), and the coeffi- cient of x2 (β2): > confint(m) 2.5 % 97.5 % (Intercept) -5.332195 12.361525 x1 -3.009250 1.282295 x2 1.892837 7.093099 By default, confint uses a confidence level of 95%. Use the level parameter to select a different level: > confint(m, level=0.99) 0.5 % 99.5 % 292 | Chapter 11: Linear Regression and ANOVA D o wnload from Wow! eBook (Intercept) -8.4316652 15.460995 x1 -3.7610142 2.034059 x2 0.9818897 8.004046 See Also The coefplot function of the arm package can plot confidence intervals for regression coefficients. 11.14 Plotting Regression Residuals Problem You want a visual display of your regression residuals. Solution You can plot the model object by selecting the residuals plot from the available plots: > m <- lm(y ~ x) > plot(m, which=1) Discussion Normally, plotting a regression model object produces several diagnostic plots. You can select just the residuals plot by specifying which=1. Figure 11-3 shows a plot of the residuals from Recipe 11.1. R draws a smoothed line through the residuals as a visual aid to finding significant patterns—for example, a slope or a parabolic shape. See Also See Recipe 11.15, which contains examples of residuals plots and other diagnostic plots. 11.15 Diagnosing a Linear Regression Problem You have performed a linear regression. Now you want to verify the model’s quality by running diagnostic checks. Solution Start by plotting the model object, which will produce several diagnostic plots: > m <- lm(y ~ x) > plot(m) 11.15 Diagnosing a Linear Regression | 293 Next, identify possible outliers either by looking at the diagnostic plot of the residuals or by using the outlier.test function of the car package: > library(car) > outlier.test(m) Finally, identify any overly influential observations (Recipe 11.16). Discussion R fosters the impression that linear regression is easy: just use the lm function. Yet fitting the data is only the beginning. It’s your job to decide whether the fitted model actually works and works well. Figure 11-3. Plotting regression residuals 294 | Chapter 11: Linear Regression and ANOVA Before anything else, you must have a statistically significant model. Check the F sta- tistic from the model summary (Recipe 11.4) and be sure that the p-value is small enough for your purposes. Conventionally, it should be less than 0.05 or else your model is likely meaningless. Simply plotting the model object produces several useful diagnostic plots: > m <- lm(y ~ x) > plot(m) Figure 11-4 shows diagnostic plots for a pretty good regression: • The points in the Residuals vs Fitted plot are randomly scattered with no particular pattern. • The points in the Normal Q–Q plot are more-or-less on the line, indicating that the residuals follow a normal distribution. • In both the Scale–Location plot and the Residuals vs Leverage plots, the points are in a group with none too far from the center. In contrast, Figure 11-5 shows the diagnostics for a not-so-good regression. Observe that the Residuals vs Fitted plot has a definite parabolic shape. This tells us that the model is incomplete: a quadratic factor is missing that could explain more variation in y. Other patterns in residuals are suggestive of additional problems: a cone shape, for example, may indicate nonconstant variance in y. Interpreting those patterns is a bit of an art, so I suggest reviewing a good book on linear regression while evaluating the plot of residuals. There are other problems with the not-so-good diagnostics. The Normal Q–Q plot has more points off the line than it does for the good regression. Both the Scale–Location and Residuals vs Leverage plots show points scattered away from the center, which suggests that some points have excessive leverage. Another pattern is that point number 28 sticks out in every plot. This warns us that something is odd with that observation. The point could be an outlier, for example. We can check that hunch with the outlier.test function of the car package: > outlier.test(m) max|rstudent| = 3.183304, degrees of freedom = 27, unadjusted p = 0.003648903, Bonferroni p = 0.1094671 Observation: 28 The outlier.test identifies the model’s most outlying observation. In this case, it identified observation number 28 and so confirmed that it could be an outlier. 11.15 Diagnosing a Linear Regression | 295 See Also See Recipes 11.4 and 11.16. The car package is not part of the standard distribution of R; see Recipe 3.9. 11.16 Identifying Influential Observations Problem You want to identify the observations that are having the most influence on the re- gression model. This is useful for diagnosing possible problems with the data. Figure 11-4. Diagnostic plots—pretty good fit 296 | Chapter 11: Linear Regression and ANOVA Solution The influence.measures function reports several useful statistics for identifying influential observations, and it flags the significant ones with an asterisk (*). Its main argument is the model object from your regression: > influence.measures(m) Discussion The title of this recipe could be “Identifying Overly Influential Observations”, but that would be redundant. All observations influence the regression model, even if only a little. When a statistician says that an observation is influential, it means that removing Figure 11-5. Diagnostic plots—not-so-good fit 11.16 Identifying Influential Observations | 297 the observation would significantly change the fitted regression model. We want to identify those observations because they might be outliers that distort our model; we owe it to ourselves to investigate them. The influence.measures function reports several statistics: DFBETAS, DFFITS, covariance ratio, Cook’s distance, and hat matrix values. If any of these measures indicate that an observation is influential, the function flags that observation with an asterisk (*) along the righthand side: > influence.measures(m) Influence measures of lm(formula = z ~ x) : dfb.1_ dfb.x dffit cov.r cook.d hat inf 1 0.64663 -0.56267 0.6466 1.04 1.98e-01 0.1373 2 0.40133 -0.33798 0.4020 1.11 8.01e-02 0.1137 3 0.29023 -0.23992 0.2914 1.14 4.29e-02 0.1035 4 0.13917 -0.11036 0.1409 1.16 1.02e-02 0.0862 5 0.40997 -0.33125 0.4133 1.05 8.34e-02 0.0932 . . (etc.) . 25 -0.00985 0.02449 0.0338 1.16 5.93e-04 0.0700 26 -0.02698 0.06341 0.0856 1.15 3.78e-03 0.0740 27 -0.16813 0.31877 0.3888 1.09 7.47e-02 0.1017 28 -0.34708 0.74122 0.9580 0.62 3.46e-01 0.0830 * 29 -0.09644 0.18050 0.2187 1.17 2.44e-02 0.1045 30 -0.37679 0.64112 0.7398 0.97 2.51e-01 0.1339 This is the model from Recipe 11.15, where we suspected that observation 28 was an outlier. An asterisk is flagging that observation, confirming that it’s overly influential. This recipe can identify influential observations, but you shouldn’t reflexively delete them. Some judgment is required here. Are those observations improving your model or damaging it? See Also See Recipe 11.15. Use help(influence.measures) to get a list of influence measures and some related functions. See a regression textbook for interpretations of the various influence measures. 11.17 Testing Residuals for Autocorrelation (Durbin–Watson Test) Problem You have performed a linear regression and want to check the residuals for autocorrelation. 298 | Chapter 11: Linear Regression and ANOVA Solution The Durbin—Watson test can check the residuals for autocorrelation. The test is im- plemented by the dwtest function of the lmtest package: > library(lmtest) > m <- lm(y ~ x) # Create a model object > dwtest(m) # Test the model residuals The output includes a p-value. Conventionally, if p < 0.05 then the residuals are sig- nificantly correlated whereas p > 0.05 provides no evidence of correlation. You can perform a visual check for autocorrelation by graphing the autocorrelation function (ACF) of the residuals: > acf(m) # Plot the ACF of the model residuals Discussion The Durbin–Watson test is often used in time series analysis, but it was originally created for diagnosing autocorrelation in regression residuals. Autocorrelation in the residuals is a scourge because it distorts the regression statistics, such as the F statistic and the t statistics for the regression coefficients. The presence of autocorrelation sug- gests that your model is missing a useful predictor variable or that it should include a time series component, such as a trend or a seasonal indicator. This first example builds a simple regression model and then tests the residuals for autocorrelation. The test returns a p-value of 0.07755, which indicates that there is no significant autocorrelation: > m <- lm(y ~ x) > dwtest(m) Durbin–Watson test data: m DW = 1.5654, p-value = 0.07755 alternative hypothesis: true autocorrelation is greater than 0 This second example exhibits autocorrelation in the residuals. The p-value is only 0.01298, so the autocorrelation is likely positive: > m <- lm(y ~ z) > dwtest(m) Durbin–Watson test data: m DW = 1.2946, p-value = 0.01298 alternative hypothesis: true autocorrelation is greater than 0 11.17 Testing Residuals for Autocorrelation (Durbin–Watson Test) | 299 By default, dwtest performs a one-sided test and answers this question: Is the autocor- relation of the residuals greater than zero? If your model could exhibit negative auto- correlation (yes, that is possible), then you should use the alternative option to perform a two-sided test: > dwtest(m, alternative="two.sided") The Durbin–Watson test is also implemented by the durbin.watson function of the car package. I suggested the dwtest function only because I think the output is easier to read. See Also Neither the lmtestest package nor the car package are included in the standard distri- bution of R; see Recipes 3.6 and 3.9. See Recipes 14.13 and 14.16 for more regarding tests of autocorrelation. 11.18 Predicting New Values Problem You want to predict new values from your regression model. Solution Save the predictor data in a data frame. Use the predict function, setting the newdata parameter to the data frame: > m <- lm(y ~ u + v + w) > preds <- data.frame(u=3.1, v=4.0, w=5.5) > predict(m, newdata=preds) Discussion Once you have a linear model, making predictions is quite easy because the predict function does all the heavy lifting. The only annoyance is arranging for a data frame to contain your data. The predict function returns a vector of predicted values with one prediction for every row in the data. The example in the Solution contains one row, so predict returns one value: > preds <- data.frame(u=3.1, v=4.0, w=5.5) > predict(m, newdata=preds) 1 12.31374 If your predictor data contains several rows, you get one prediction per row: > preds <- data.frame( + u=c(3.0, 3.1, 3.2, 3.3), 300 | Chapter 11: Linear Regression and ANOVA + v=c(3.9, 4.0, 4.1, 4.2), + w=c(5.3, 5.5, 5.7, 5.9) ) > predict(m, newdata=preds) 1 2 3 4 11.97277 12.31374 12.65472 12.99569 In case it’s not obvious: the new data needn’t contain values for response variables, only predictor variables. After all, you are trying to calculate the response, so it would be unreasonable of R to expect you to supply it. See Also These are just the point estimates of the predictions. See Recipe 11.19 for the confidence intervals. 11.19 Forming Prediction Intervals Problem You are making predictions using a linear regression model. You want to know the prediction intervals: the range of the distribution of the prediction. Solution Use the predict function and specify interval="prediction": > predict(m, newdata=preds, interval="prediction") Discussion This is a continuation of Recipe 11.18, which described packaging your data into a data frame for the predict function. We are adding interval="prediction" to obtain pre- diction intervals. Here is the example from Recipe 11.18, now with prediction intervals. The new lwr and upr columns are the lower and upper limits, respectively, for the interval: > predict(m, newdata=preds, interval="prediction") fit lwr upr 1 11.97277 7.999848 15.94569 2 12.31374 8.272327 16.35516 3 12.65472 8.540045 16.76939 4 12.99569 8.803251 17.18813 By default, predict uses a confidence level of 0.95. You can change this via the level argument. A word of caution: these prediction intervals are extremely sensitive to deviations from normality. If you suspect that your response variable is not normally distributed, con- sider a nonparametric technique, such as the bootstrap (Recipe 13.8), for prediction intervals. 11.19 Forming Prediction Intervals | 301 11.20 Performing One-Way ANOVA Problem Your data is divided into groups, and the groups are normally distributed. You want to know if the groups have significantly different means. Solution Use a factor to define the groups. Then apply the oneway.test function: > oneway.test(x ~ f) Here, x is a vector of numeric values and f is a factor that identifies the groups. The output includes a p-value. Conventionally, a p-value of less than 0.05 indicates that two or more groups have significantly different means whereas a value exceeding 0.05 pro- vides no such evidence. Discussion Comparing the means of groups is a common task. One-way ANOVA performs that comparison and computes the probability that they are statistically identical. A small p-value indicates that two or more groups likely have different means. (It does not indicate that all groups have different means.) The basic ANOVA test assumes that your data has a normal distribution or that, at least, it is pretty close to bell-shaped. If not, use the Kruskal–Wallis test instead (Recipe 11.23). We can illustrate ANOVA with stock market historical data. Is the stock market more profitable in some months than in others? For instance, a common folk myth says that October is a bad month for stock market investors.‡ I explored this question by creating a vector and a factor. The vector, r, contains the daily percentage change in the Standard & Poor’s 500 index, a broad measure of stock market performance. The factor, mon, indicates the calendar month in which that change occurred: Jan, Feb, Mar, and so forth. The data covers the period 1950 though 2009. The one-way ANOVA shows a p-value of 0.03347: > oneway.test(r ~ mon) One-way analysis of means (not assuming equal variances) data: r and mon F = 1.9102, num df = 11.000, denom df = 5930.142, p-value = 0.03347 ‡ In the words of Mark Twain, “October: This is one of the peculiarly dangerous months to speculate in stocks in. The others are July, January, September, April, November, May, March, June, December, August and February.” 302 | Chapter 11: Linear Regression and ANOVA We can conclude that stock market changes varied significantly according to the cal- endar month. Before you run to your broker and start flipping your portfolio monthly, however, we should check something: did the pattern change recently? We can limit the analysis to recent data by specifying a subset parameter. This works for oneway.test just as it does for the lm function. The subset contains the indexes of observations to be analyzed; all other observations are ignored. Here, we give the indexes of the 2,500 most recent observations, which is about 10 years of data: > oneway.test(r ~ mon, subset=tail(seq_along(r),2500)) One-way analysis of means (not assuming equal variances) data: r and mon F = 0.7973, num df = 11.000, denom df = 975.963, p-value = 0.643 Uh-oh! Those monthly differences evaporated during the past 10 years. The large p-value, 0.643, indicates that changes have not recently varied according to calendar month. Apparently, those differences are a thing of the past. Notice that the oneway.test output says “(not assuming equal variances)”. If you know the groups have equal variances, you’ll get a less conservative test by specifying var.equal=TRUE: > oneway.test(x ~ f, var.equal=TRUE) You can also perform one-way ANOVA by using the aov function like this: > m <- aov(x ~ f) > summary(m) However, the aov function always assumes equal variances and so is somewhat less flexible than oneway.test. See Also If the means are significantly different, use Recipe 11.22 to see the actual differences. Use Recipe 11.23 if your data is not normally distributed, as required by ANOVA. 11.21 Creating an Interaction Plot Problem You are performing multiway ANOVA: using two or more categorical variables as pre- dictors. You want a visual check of possible interaction between the predictors. 11.21 Creating an Interaction Plot | 303 Solution Use the interaction.plot function: > interaction.plot(pred1, pred2, resp) Here, pred1 and pred2 are two categorical predictors and resp is the response variable. Discussion ANOVA is a form of linear regression, so ideally there is a linear relationship between every predictor and the response variable. One source of nonlinearity is an interac- tion between two predictors: as one predictor changes value, the other predictor changes its relationship to the response variable. Checking for interaction between predictors is a basic diagnostic. The faraway package contains a dataset called rats. In it, treat and poison are catego- rical variables and time is the response variable. When plotting poison against time we are looking for straight, parallel lines, which indicate a linear relationship. However, using the interaction.plot function reveals that something is not right: > library(faraway) > data(rats) > interaction.plot(rats$poison, rats$treat, rats$time) The resulting graph is shown in Figure 11-6. Each line graphs time against poison. The difference between lines is that each line is for a different value of treat. The lines should be parallel, but the top two are not exactly parallel. Evidently, varying the value of treat “warped” the lines, introducing a nonlinearity into the relationship between poison and time. This signals a possible interaction that we should check. For this data it just so happens that yes, there is an interaction but no, it is not statistically significant. The moral is clear: the visual check is useful, but it’s not foolproof. Follow up with a statistical check. See Also See Recipe 11.6. 11.22 Finding Differences Between Means of Groups Problem Your data is divided into groups, and an ANOVA test indicates that the groups have significantly different means. You want to know the differences between those means for all groups. 304 | Chapter 11: Linear Regression and ANOVA Solution Perform the ANOVA test using the aov function, which returns a model object. Then apply the TukeyHSD function to the model object: > m <- aov(x ~ f) > TukeyHSD(m) Here, x is your data and f is the grouping factor. You can plot the TukeyHSD result to obtain a graphical display of the differences: > plot(TukeyHSD(m)) Discussion The ANOVA test is important because it tells you whether or not the groups’ means are different. But the test does not identify which groups are different, and it does not report their differences. The TukeyHSD function can calculate those differences and help you identify the largest ones. It uses the “honest significant differences” method invented by John Tukey. Figure 11-6. Plot of interaction between treat and poison 11.22 Finding Differences Between Means of Groups | 305 I’ll illustrate TukeyHSD by continuing the example from Recipe 11.20, which grouped daily stock market changes by month. Here, I will group them by weekday instead, using a factor called wday that identifies the day of the week (Mon, ..., Fri) on which the change occurred. I’ll use the first 2,500 observations, which roughly cover the pe- riod from 1950 to 1960: > oneway.test(r ~ wday, subset=1:2500) One-way analysis of means (not assuming equal variances) data: r and wday F = 12.6051, num df = 4.000, denom df = 1242.499, p-value = 4.672e-10 The p-value is essentially zero, indicating that average changes varied significantly de- pending on the weekday. To use the TukeyHSD function, I first perform the ANOVA test using the aov function, which returns a model object, and then apply the TukeyHSD function to the object: > m <- aov(r ~ wday, subset=1:2500) > TukeyHSD(m) Tukey multiple comparisons of means 95% family-wise confidence level Fit: aov(formula = r ~ wday, subset = 1:2500) $wday diff lwr upr p adj Tue-Mon 0.13131281 0.007025859 0.2555998 0.0323054 Wed-Mon 0.23846814 0.114846431 0.3620898 0.0000015 Thu-Mon 0.22371916 0.099432205 0.3480061 0.0000094 Fri-Mon 0.31728974 0.192816883 0.4417626 0.0000000 Wed-Tue 0.10715532 -0.015961497 0.2302721 0.1222470 Thu-Tue 0.09240635 -0.031378436 0.2161911 0.2481996 Fri-Tue 0.18597693 0.062005489 0.3099484 0.0004185 Thu-Wed -0.01474898 -0.137865796 0.1083678 0.9975329 Fri-Wed 0.07882161 -0.044482882 0.2021261 0.4064721 Fri-Thu 0.09357058 -0.030400857 0.2175420 0.2378246 Each line in the output table includes the difference between the means of two groups (diff) as well as the lower and upper bounds of the confidence interval (lwr and upr) for the difference. The first line in the table, for example,compares the Tue group and the Wed group: the difference of their means is 0.1313 with a confidence interval of (0.0070, 0.2556). Scanning the table, we see that the Fri–Mon comparison had the largest difference, which was 0.3173. A cool feature of TukeyHSD is that it can display these differences visually, too. Simply plot the function’s return value: > plot(TukeyHSD(m)) 306 | Chapter 11: Linear Regression and ANOVA Figure 11-7 shows the graph for our stock market example. The horizontal lines plot the confidence intervals for each pair. With this visual representation you can quickly see that several confidence intervals cross over zero, indicating that the difference is not necessarily significant. You can also see that the Fri–Mon pair has the largest difference because their confidence interval is farthest to the right. Figure 11-7. Differences between group means See Also See Recipe 11.20. 11.22 Finding Differences Between Means of Groups | 307 11.23 Performing Robust ANOVA (Kruskal–Wallis Test) Problem Your data is divided into groups. The groups are not normally distributed, but their distributions have similar shapes. You want to perform a test similar to ANOVA—you want to know if the group medians are significantly different. Solution Create a factor that defines the groups of your data. Use the kruskal.test function, which implements the Kruskal–Wallis test. Unlike the ANOVA test, this test does not depend upon the normality of the data: > kruskal.test(x ~ f) Here, x is a vector of data and f is a grouping factor. The output includes a p-value. Conventionally, p < 0.05 indicates that there is a significant difference between the medians of two or more groups whereas p > 0.05 provides no such evidence. Discussion Regular ANOVA assumes that your data has a Normal distribution. It can tolerate some deviation from normality, but extreme deviations will produce meaningless p-values. The Kruskal–Wallis test is a nonparametric version of ANOVA, which means that it does not assume normality. However, it does assume same-shaped distributions. You should use the Kruskal–Wallis test whenever your data distribution is nonnormal or simply unknown. The null hypothesis is that all groups have the same median. Rejecting the null hy- pothesis (with p < 0.05) does not indicate that all groups are different, but it does suggest that two or more groups are different. One year, I taught Business Statistics to 94 undergraduate students. The class included a midterm examination, and there were four homework assignments prior to the exam. I wanted to know: What is the relationship between completing the homework and doing well on the exam? If there is no relation, then the homework is irrelevant and needs rethinking. I created a vector of grades, one per student. I also created a parallel factor that captured the number of homework assignments completed by that student: > print(midterm) [1] 0.8182 0.6818 0.5114 0.6705 0.6818 0.9545 0.9091 0.9659 1.0000 0.7273 [11] 0.7727 0.7273 0.6364 0.6250 0.6932 0.6932 0.0000 0.8750 0.8068 0.9432 . . (etc.) . [91] 0.9006 0.7029 0.9171 0.9628 308 | Chapter 11: Linear Regression and ANOVA > print(hw) [1] 4 4 2 3 4 4 4 3 4 4 3 3 1 4 4 3 3 4 3 4 4 4 3 2 4 4 4 4 3 2 3 4 4 0 4 4 4 4 [39] 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 2 4 4 4 4 4 1 4 4 4 4 4 4 4 4 4 2 4 4 [77] 3 4 2 2 4 4 4 4 1 3 0 4 4 0 4 3 4 4 Levels: 0 1 2 3 4 Notice that the hw variable—although it appears to be numeric—is actually a factor. It assigns each midterm grade to one of five groups depending upon how many homework assignments the student completed. The distribution of exam grades is definitely not Normal: the students have a wide range of math skills, so there are an unusual number of A and F grades. Hence regular ANOVA would not be appropriate. Instead I used the Kruskal–Wallis test and obtained a p-value of essentially zero (3.99 × 10−5, or 0.00003669): > kruskal.test(midterm ~ hw) Kruskal–Wallis rank sum test data: midterm by hw Kruskal–Wallis chi-squared = 25.6813, df = 4, p-value = 3.669e-05 Obviously, there is a significant performance difference between students who com- plete their homework and those who do not. But what could I actually conclude? At first, I was pleased that the homework appeared so effective. Then it dawned on me that this was a classic error in statistical reasoning: I assumed that correlation implied causality. It does not, of course. Perhaps strongly motivated students do well on both homework and exams whereas lazy students do not. In that case, the causal factor is degree of motivation, not the brilliance of my homework selection. In the end, I could only conclude something very simple: students who complete the homework will likely do well on the midterm exam, but I don’t really know why. 11.24 Comparing Models by Using ANOVA Problem You have two models of the same data, and you want to know whether they produce different results. Solution The anova function can compare two models and report if they are significantly different: > anova(m1, m2) Here, m1 and m2 are both model objects returned by lm. The output from anova includes a p-value. Conventionally, a p-value of less than 0.05 indicates that the models are significantly different whereas a value exceeding 0.05 provides no such evidence. 11.24 Comparing Models by Using ANOVA | 309 Discussion In Recipe 11.3, we used the anova function to print the ANOVA table for one regression model. Now we are using the two-argument form to compare two models. The anova function has one strong requirement when comparing two models: one model must be contained within the other. That is, all the terms of the smaller model must appear in the larger model. Otherwise, the comparison is impossible. The ANOVA analysis performs an F test that is similar to the F test for a linear regres- sion. The difference is that this test is between two models whereas the regression F test is between using the regression model and using no model. Suppose we build three models of y, adding terms as we go: > m1 <- lm(y ~ u) > m2 <- lm(y ~ u + v) > m3 <- lm(y ~ u + v + w) Is m2 really different from m1? We can use anova to compare them, and the result is a p-value of 0.003587: > anova(m1, m2) Analysis of Variance Table Model 1: y ~ u Model 2: y ~ u + v Res.Df RSS Df Sum of Sq F Pr(>F) 1 28 108.968 2 27 79.138 1 29.83 10.177 0.003587 ** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 The small p-value indicates that the models are significantly different. Comparing m2 and m3, however, yields a p-value of 0.05744: > anova(m2, m3) Analysis of Variance Table Model 1: y ~ u + v Model 2: y ~ u + v + w Res.Df RSS Df Sum of Sq F Pr(>F) 1 27 79.138 2 26 68.696 1 10.442 3.9522 0.05744 . --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 This is right on the edge. Strictly speaking, it does not pass our requirement to be smaller than 0.05; however, it’s close enough that you might judge the models to be “different enough.” 310 | Chapter 11: Linear Regression and ANOVA This example is a bit contrived, so it does not show the larger power of anova. I use anova when, while experimenting with complicated models by adding and deleting multiple terms, I need to know whether or not the new model is really different from the original one. In other words: if I add terms and the new model is essentially un- changed, then the extra terms are not worth the additional complications. 11.24 Comparing Models by Using ANOVA | 311 CHAPTER 12 Useful Tricks Introduction The recipes in this chapter are neither obscure numerical calculations nor deep statis- tical techniques. Yet they are useful functions and idioms that you will likely need at one time or another. 12.1 Peeking at Your Data Problem You have a lot of data—too much to display at once. Nonetheless, you want to see some of the data. Solution Use head to view the first few data or rows: > head(x) Use tail to view the last few data or rows: > tail(x) Discussion Printing a large dataset is pointless because everything just rolls off your screen. Use head to see a little bit of the data: > head(dfrm) x y z 1 0.7533110 0.57562846 -0.1710760 2 2.0143547 0.83312274 0.3698584 3 -0.3551345 0.57471542 2.0132348 4 2.0281678 0.78945319 -0.5378854 313 5 -2.2168745 0.01758024 1.8344879 6 0.7583962 -1.78214755 2.2848990 Use tail to see the last few rows and the number of rows. Here, we see that this data frame has 10,120 rows: > tail(dfrm) x y z 10115 -0.0314354 -0.74988291 -0.2048963 10116 -0.4779001 0.93407510 1.0509977 10117 -1.1314402 1.89308417 1.7207972 10118 0.4891881 -1.20792811 -1.4630227 10119 1.2349013 -0.09615198 -0.9887513 10120 -1.3763834 -2.25309628 0.9296106 See Also See Recipe 12.15 for seeing the structure of your variable’s contents. 12.2 Widen Your Output Problem You are printing wide datasets. R is wrapping the output, making it hard to read. Solution Set the width option to reflect the true number of columns in your output window: > options(width=numcols) Discussion By default, R formats your output to be 80 columns wide. It’s frustrating when you print a wide dataset and R wraps the output lines according to its assumed limitation of 80 columns. Your expensive, wide-screen display can probably accommodate many more columns than that. If your display is 120 characters wide, for example, then ask R to use the entire 120 columns by setting the width option: > options(width=120) Now R will wrap its lines after 120 characters, not 80, letting you comfortably view wider datasets. See Also See Recipe 3.16 for more about setting options. 314 | Chapter 12: Useful Tricks 12.3 Printing the Result of an Assignment Problem You are assigning a value to a variable and you want to see its value. Solution Simply put parentheses around the assignment: > x <- 1/pi # Prints nothing > (x <- 1/pi) # Prints assigned value [1] 0.3183099 Discussion Normally, R inhibits printing when it sees you enter a simple assignment. When you surround the assignment with parentheses, however, it is no longer a simple assignment and so R prints the value. See Also See Recipe 2.1 for more ways to print things. 12.4 Summing Rows and Columns Problem You want to sum the rows or columns of a matrix or data frame. Solution Use rowSums to sum the rows: > rowSums(m) Use colSums to sum the columns: > colSums(m) Discussion This is a mundane recipe, but it’s so common that it deserves mentioning. I use this recipe, for example, when producing reports that include column totals. In this exam- ple, daily.prod is a record of this week’s factory production and we want totals by product and by day: > daily.prod Widgets Gadgets Thingys Mon 179 167 182 12.4 Summing Rows and Columns | 315 D o wnload from Wow! eBook Tue 153 193 166 Wed 183 190 170 Thu 153 161 171 Fri 154 181 186 > colSums(daily.prod) Widgets Gadgets Thingys 822 892 875 > rowSums(daily.prod) Mon Tue Wed Thu Fri 528 512 543 485 521 These functions return a vector. In the case of column sums, we can append the vector to the matrix and thereby neatly print the data and totals together: > rbind(daily.prod, Totals=colSums(daily.prod)) Widgets Gadgets Thingys Mon 179 167 182 Tue 153 193 166 Wed 183 190 170 Thu 153 161 171 Fri 154 181 186 Totals 822 892 875 12.5 Printing Data in Columns Problem You have several parallel data vectors, and you want to print them in columns. Solution Use cbind to form the data into columns; then print the result. Discussion When you have parallel vectors, it’s difficult to see their relationship if you print them separately: > print(x) [1] -0.3142317 2.4033751 -0.7182640 -1.7606110 -1.1252812 -0.7195406 1.3102240 [8] 0.4518985 0.1521774 0.6533708 > print(y) [1] -0.9473355 -1.0710863 -0.1760594 1.8720502 -0.5103384 1.4388904 1.1531710 [8] -0.1548398 -0.2599055 0.4713668 Use the cbind function to form them into columns that, when printed, show the data’s structure: > print(cbind(x,y)) x y [1,] -0.3142317 -0.9473355 [2,] 2.4033751 -1.0710863 [3,] -0.7182640 -0.1760594 316 | Chapter 12: Useful Tricks [4,] -1.7606110 1.8720502 [5,] -1.1252812 -0.5103384 [6,] -0.7195406 1.4388904 [7,] 1.3102240 1.1531710 [8,] 0.4518985 -0.1548398 [9,] 0.1521774 -0.2599055 [10,] 0.6533708 0.4713668 You can include expressions in the output, too. Use a tag to give them a column heading: > print(cbind(x,y,Total=x+y)) x y Total [1,] -0.3142317 -0.9473355 -1.2615672 [2,] 2.4033751 -1.0710863 1.3322888 [3,] -0.7182640 -0.1760594 -0.8943233 [4,] -1.7606110 1.8720502 0.1114392 [5,] -1.1252812 -0.5103384 -1.6356196 [6,] -0.7195406 1.4388904 0.7193499 [7,] 1.3102240 1.1531710 2.4633949 [8,] 0.4518985 -0.1548398 0.2970587 [9,] 0.1521774 -0.2599055 -0.1077281 [10,] 0.6533708 0.4713668 1.1247376 12.6 Binning Your Data Problem You have a vector, and you want to split the data into groups according to intervals. Statisticians call this binning your data. Solution Use the cut function. You must define a vector, say breaks, which gives the ranges of the intervals. The cut function will group your data according to those intervals. It returns a factor whose levels (elements) identify each datum’s group: > f <- cut(x, breaks) Discussion This example generates 1,000 random numbers that have a standard Normal distribu- tion. It breaks them into six groups by defining intervals at ±1, ±2, and ±3 standard deviations: > x <- rnorm(1000) > breaks <- c(-3,-2,-1,0,1,2,3) > f <- cut(x, breaks) The result is a factor, f, that identifies the groups. The summary function shows the number of elements by level. R creates names for each level, using the mathematical notation for an interval: 12.6 Binning Your Data | 317 > summary(f) (-3,-2] (-2,-1] (-1,0] (0,1] (1,2] (2,3] NA's 18 119 348 355 137 21 2 The results are bell-shaped, which is what we expect. There are two NA values, indi- cating that two values in x fell outside the defined intervals. We can use the labels parameter to give nice, predefined names to the six groups instead of the funky synthesized names: > f <- cut(x, breaks, labels=c("Bottom", "Low", "Neg", "Pos", "High", "Top")) Now the summary function uses our names: Bottom Low Neg Pos High Top NA's 18 119 348 355 137 21 2 Don’t Throw Away That Information Binning is useful for summaries such as histograms. But it results in information loss, which can be harmful in modeling. Consider the extreme case of binning a continuous variable into two values, “high” and “low”. The binned data has only two possible values, so you have replaced a rich source of information with one bit of information. Where the continuous variable might be a powerful predictor, the binned variable can distinguish at most two states and so will likely have only a fraction of the original power. Before binning, I suggest exploring other transformations that are less lossy. 12.7 Finding the Position of a Particular Value Problem You have a vector. You know a particular value occurs in the contents, and you want to know its position. Solution The match function will search a vector for a particular value and return the position: > vec <- c(100,90,80,70,60,50,40,30,20,10) > match(80, vec) [1] 3 Here match returns 3, which is the position of 80 within vec. Discussion There are special functions for finding the location of the minimum and maximum values—which.min and which.max, respectively: > vec <- c(100,90,80,70,60,50,40,30,20,10) > which.min(vec) # Position of smallest element 318 | Chapter 12: Useful Tricks [1] 10 > which.max(vec) # Position of largest element [1] 1 See Also This technique is used in Recipe 11.12. 12.8 Selecting Every nth Element of a Vector Problem You want to select every nth element of a vector. Solution Create a logical indexing vector that is TRUE for every nth element. One approach is to find all subscripts that equal zero when taken modulo n: > v[ seq_along(v) %% n == 0 ] Discussion This problem arises in systematic sampling: we want to sample a dataset by selecting every nth element. The seq_along(v) function generates the sequence of integers that can index v; it is equivalent to 1:length(v). We compute each index value modulo n by the expression: seq_along(v) %% n Then we find those values that equal zero: seq_along(v) %% n == 0 The result is a logical vector, the same length as v and with TRUE at every nth element, that can index v to select the desired elements: v[ seq_along(v) %% n == 0 ] If you just want something simple like every second element, you can use the recycling rule in a clever way. Index v with a two-element logical vector, like this: v[ c(FALSE, TRUE) ] If v has more than two elements then the indexing vector is too short. Hence R will invoke the Recycling Rule and expand the index vector to the length of v, recycling its contents. That gives an index vector that is FALSE, TRUE, FALSE, TRUE, FALSE, TRUE, and so forth. Voilà! The final result is every second element of v. See Also See Recipe 5.3 for more about the Recycling Rule. 12.8 Selecting Every nth Element of a Vector | 319 12.9 Finding Pairwise Minimums or Maximums Problem You have two vectors, v and w, and you want to find the minimums or the maximums of pairwise elements. That is, you want to calculate: min(v1, w1), min(v2, w2), min(v3, w3), ... or: max(v1, w1), max(v2, w2), max(v3, w3), ... Solution R calls these the parallel minimum and the parallel maximum. The calculation is per- formed by pmin(v,w) and pmax(v,w), respectively: > pmin(1:5, 5:1) # Find the element-by-element minimum [1] 1 2 3 2 1 > pmax(1:5, 5:1) # Find the element-by-element maximum [1] 5 4 3 4 5 Discussion When an R beginner wants pairwise minimums or maximums, a common mistake is to write min(v,w) or max(v,w). Those are not pairwise operations: min(v,w) returns a single value, the minimum over all v and w. Likewise, max(v,w) returns a single value from all of v and w. The pmin and pmax values compare their arguments in parallel, picking the minimum or maximum for each subscript. They return a vector that matches the length of the inputs. You can combine pmin and pmax with the Recycling Rule to perform useful hacks. Sup- pose the vector v contains both positive and negative values, and you want to reset the negative values to zero. This does the trick: > v <- pmax(v, 0) By the Recycling Rule, R expands the zero-valued scalar into a vector of zeros that is the same length as v. Then pmax does an element-by-element comparison, taking the larger of zero and each element of v: > v <- c(1, -2, 3, -4, 5, -6, 7) > pmax(v,0) [1] 1 0 3 0 5 0 7 Actually, pmin and pmax are more powerful than the Solution indicates. They can take more than two vectors, comparing all vectors in parallel. 320 | Chapter 12: Useful Tricks See Also See Recipe 5.3 for more about the Recycling Rule. 12.10 Generating All Combinations of Several Factors Problem You have two or more factors. You want to generate all combinations of their levels, also known as their Cartesian product. Solution Use the expand.grid function. Here, f and g are two factors: > expand.grid(f, g) Discussion This code snippet creates two factors—sides represents the two sides of a coin, and faces represents the six faces of a die (those little spots on a die are called pips): > sides <- factor(c("Heads", "Tails")) > faces <- factor(c("1 pip", paste(2:6, "pips"))) We can use expand.grid to find all combinations of one roll of the die and one coin toss: > expand.grid(faces, sides) Var1 Var2 1 1 pip Heads 2 2 pips Heads 3 3 pips Heads 4 4 pips Heads 5 5 pips Heads 6 6 pips Heads 7 1 pip Tails 8 2 pips Tails 9 3 pips Tails 10 4 pips Tails 11 5 pips Tails 12 6 pips Tails Similarly, we can find all combinations of two dice: > expand.grid(faces, faces) Var1 Var2 1 1 pip 1 pip 2 2 pips 1 pip 3 3 pips 1 pip 4 4 pips 1 pip 5 5 pips 1 pip 6 6 pips 1 pip 7 1 pip 2 pips 12.10 Generating All Combinations of Several Factors | 321 8 2 pips 2 pips . . (etc.) . 34 4 pips 6 pips 35 5 pips 6 pips 36 6 pips 6 pips The result is a data frame. R automatically provides the row names and column names. The Solution and the example show the Cartesian product of two factors, but expand.grid can handle three or more factors, too. See Also If you’re working with strings and not factors, then you can also use Recipe 7.7 to generate combinations. 12.11 Flatten a Data Frame Problem You have a data frame of numeric values. You want to process all its elements together, not as separate columns—for example, to find the mean across all values. Solution Convert the data frame to a matrix and then process the matrix. This example finds the mean of all elements in the data frame dfrm: > mean(as.matrix(dfrm)) It is sometimes necessary then to convert the matrix to a vector. In that case, use as.vector(as.matrix(dfrm)). Discussion Suppose we have a data frame, such as the factory production data from Recipe 12.4: > daily.prod Widgets Gadgets Thingys Mon 179 167 182 Tue 153 193 166 Wed 183 190 170 Thu 153 161 171 Fri 154 181 186 Suppose also that we want the average daily production across all days and products. This won’t work: 322 | Chapter 12: Useful Tricks > mean(daily.prod) Widgets Gadgets Thingys 164.4 178.4 175. The mean function thinks it should average individual columns, which is usually what you want. But when you want the average across all values, first collapse the data frame down to a matrix: > mean(as.matrix(daily.prod)) [1] 172.6 This recipe works only on data frames with all-numeric data. Recall that converting a data frame with mixed data (numeric columns mixed with character columns or fac- tors) into a matrix forces all columns to be converted to characters. See Also See Recipe 5.33 for more about converting between datatypes. 12.12 Sorting a Data Frame Problem You have a data frame. You want to sort the contents, using one column as the sort key. Solution Use the order function on the sort key, and then rearrange the data frame rows ac- cording to that ordering: > dfrm <- dfrm[order(dfrm$key),] Here dfrm is a data frame and dfrm$key is the sort-key column. Discussion The sort function is great for vectors but is ineffective for data frames. Sorting a data frame requires an extra step. Suppose we have the following data frame and we want to sort by month: > print(dfrm) month day outcome 1 7 11 Win 2 8 10 Lose 3 8 25 Tie 4 6 27 Tie 5 7 22 Win The order function tells us how to rearrange the months into ascending order: > order(dfrm$month) [1] 4 1 5 2 3 12.12 Sorting a Data Frame | 323 It does not return data values; rather, it returns the subscripts of dfrm$month that would sort it. We use those subscripts to rearrange the entire data frame: > dfrm[order(dfrm$month),] month day outcome 4 6 27 Tie 1 7 11 Win 5 7 22 Win 2 8 10 Lose 3 8 25 Tie After rearranging the data frame, the month column is in ascending order—just as we wanted. 12.13 Sorting by Two Columns Problem You want to sort the contents of a data frame, using two columns as sort keys. Solution Sorting by two columns is similar to sorting by one column. The order function accepts multiple arguments, so we can give it two sort keys: > dfrm <- dfrm[order(dfrm$key1,dfrm$key2),] Discussion In Recipe 12.12 we used the order function to rearrange data frame columns. We can give a second sort key to order, and it will use the additional key to break ties in the first key. Continuing our example from Recipe 12.12, we can sort on both month and day by giving both those columns to order: > dfrm[order(dfrm$month,dfrm$day),] month day outcome 4 6 27 Tie 1 7 11 Win 5 7 22 Win 2 8 10 Lose 3 8 25 Tie Within months 7 and 8, the days are now sorted into ascending order. See Also See Recipe 12.12. 324 | Chapter 12: Useful Tricks 12.14 Stripping Attributes from a Variable Problem A variable is carrying around old attributes. You want to remove some or all of them. Solution To remove all attributes, assign NULL to the variable’s attributes property: > attributes(x) <- NULL To remove a single attribute, select the attribute using the attr function, and set it to NULL: > attr(x, "attributeName") <- NULL Discussion Any variable in R can have attributes. An attribute is simply a name/value pair, and the variable can have many of them. A common example is the dimensions of a matrix variable, which are stored in an attribute. The attribute name is dim and the attribute value is a two-element vector giving the number of rows and columns. You can view the attributes of x by printing attributes(x) or str(x). Sometimes you want just a number and R insists on giving it attributes. This can happen when you fit a simple linear model and extract the slope, which is the second regression coefficient: > m <- lm(resp ~ pred) > slope <- coef(m)[2] > slope pred 1.011960 When we print slope, R also prints "pred". That is a name attribute given by lm to the coefficient (because it’s the coefficient for the pred variable). We can see that more clearly by printing the internals of slope, which reveals a "names" attribute: > str(slope) Named num 1.01 - attr(*, "names")= chr "pred" Stripping off all the attributes is easy, after which the slope value becomes simply a number: > attributes(slope) <- NULL # Strip off all attributes > str(slope) # Now the "names" attribute is gone num 1.01 > slope # And the number prints cleanly without a label [1] 1.011960 12.14 Stripping Attributes from a Variable | 325 Alternatively, we could have stripped the single offending attribute this way: > attr(slope, "names") <- NULL Remember that a matrix is a vector (or list) with a dim attribute. If you strip all the attributes from a matrix, that will strip away the dimensions and thereby turn it into a mere vector (or list). Furthermore, stripping the attributes from an object (specifically, an S3 object) can render it useless. So remove attributes with care. See Also See Recipe 12.15 for more about seeing attributes. 12.15 Revealing the Structure of an Object Problem You called a function that returned something. Now you want to look inside that something and learn more about it. Solution Use class to determine the thing’s object class: > class(x) Use mode to strip away the object-oriented features and reveal the underlying structure: > mode(x) Use str to show the internal structure and contents: > str(x) Discussion I am amazed how often I call a function, get something back, and wonder: “What the heck is this thing?” Theoretically, the function documentation should explain the re- turned value, but somehow I feel better when I can see its structure and contents myself. This is especially true for objects with a nested structure: objects within objects. Let’s dissect the value returned by lm (the linear modeling function) in the simplest linear regression recipe, Recipe 11.1: > m <- lm(y ~ x) > print(m) Call: lm(formula = y ~ x) Coefficients: 326 | Chapter 12: Useful Tricks (Intercept) x 17.72 3.25 Always start by checking the thing’s class. The class indicates if it’s a vector, matrix, list, data frame, or object: > class(m) [1] "lm" Hmmm. It seems that m is an object of class lm. That doesn’t mean anything to me. I know, however, that all object classes are built upon the native data structures (vector, matrix, list, or data frame), so I use mode to strip away the object facade and reveal the underlying structure: > mode(m) [1] "list" Ah-ha! It seems that m is built on a list structure. Now I can use list functions and operators to dig into its contents. First, I want to know the names of its list elements: > names(m) [1] "coefficients" "residuals" "effects" "rank" "fitted.values" [6] "assign" "qr" "df.residual" "xlevels" "call" [11] "terms" "model" The first list element is called “coefficients”. I’m guessing those are the regression co- efficients. Let’s have a look: > m$coefficients (Intercept) x 17.722623 3.249677 Yes, that’s what they are. I recognize those values. I could continue digging into the list structure of m, but that would get tedious. The str function does a good job of revealing the internal structure of any variable. The dump of m is shown in Example 12-1. Example 12-1. Dumping the structure of a variable > str(m) List of 12 $ coefficients : Named num [1:2] 17.72 3.25 ..- attr(*, "names")= chr [1:2] "(Intercept)" "x" $ residuals : Named num [1:30] -12.471 -3.983 -2.899 0.192 14.302 ... ..- attr(*, "names")= chr [1:30] "1" "2" "3" "4" ... $ effects : Named num [1:30] -372.443 155.463 -0.614 2.418 16.552 ... ..- attr(*, "names")= chr [1:30] "(Intercept)" "x" "" "" ... $ rank : int 2 $ fitted.values: Named num [1:30] 17.9 23.9 26.8 32.2 30 ... ..- attr(*, "names")= chr [1:30] "1" "2" "3" "4" ... $ assign : int [1:2] 0 1 $ qr :List of 5 ..$ qr : num [1:30, 1:2] -5.477 0.183 0.183 0.183 0.183 ... .. ..- attr(*, "dimnames")=List of 2 .. .. ..$ : chr [1:30] "1" "2" "3" "4" ... 12.15 Revealing the Structure of an Object | 327 .. .. ..$ : chr [1:2] "(Intercept)" "x" .. ..- attr(*, "assign")= int [1:2] 0 1 ..$ qraux: num [1:2] 1.18 1.23 ..$ pivot: int [1:2] 1 2 ..$ tol : num 1e-07 ..$ rank : int 2 ..- attr(*, "class")= chr "qr" $ df.residual : int 28 $ xlevels : list() $ call : language lm(formula = y ~ x) $ terms :Classes 'terms', 'formula' length 3 y ~ x .. ..- attr(*, "variables")= language list(y, x) .. ..- attr(*, "factors")= int [1:2, 1] 0 1 .. .. ..- attr(*, "dimnames")=List of 2 .. .. .. ..$ : chr [1:2] "y" "x" .. .. .. ..$ : chr "x" .. ..- attr(*, "term.labels")= chr "x" .. ..- attr(*, "order")= int 1 .. ..- attr(*, "intercept")= int 1 .. ..- attr(*, "response")= int 1 .. ..- attr(*, ".Environment")= .. ..- attr(*, "predvars")= language list(y, x) .. ..- attr(*, "dataClasses")= Named chr [1:2] "numeric" "numeric" .. .. ..- attr(*, "names")= chr [1:2] "y" "x" $ model :'data.frame': 30 obs. of 2 variables: ..$ y: num [1:30] 5.41 19.94 23.92 32.43 44.26 ... ..$ x: num [1:30] 0.0478 1.9086 2.7999 4.4676 3.7649 ... ..- attr(*, "terms")=Classes 'terms', 'formula' length 3 y ~ x .. .. ..- attr(*, "variables")= language list(y, x) .. .. ..- attr(*, "factors")= int [1:2, 1] 0 1 .. .. .. ..- attr(*, "dimnames")=List of 2 .. .. .. .. ..$ : chr [1:2] "y" "x" .. .. .. .. ..$ : chr "x" .. .. ..- attr(*, "term.labels")= chr "x" .. .. ..- attr(*, "order")= int 1 .. .. ..- attr(*, "intercept")= int 1 .. .. ..- attr(*, "response")= int 1 .. .. ..- attr(*, ".Environment")= .. .. ..- attr(*, "predvars")= language list(y, x) .. .. ..- attr(*, "dataClasses")= Named chr [1:2] "numeric" "numeric" .. .. .. ..- attr(*, "names")= chr [1:2] "y" "x" - attr(*, "class")= chr "lm" Notice that str shows all the elements of m and then recursively dumps each element’s contents and attributes. Long vectors and lists are truncated to keep the output manageable. There is an art to exploring an R object. Use class, mode, and str to dig through the layers. 328 | Chapter 12: Useful Tricks 12.16 Timing Your Code Problem You want to know how much time is required to run your code. This is useful, for example, when you are optimizing your code and need “before” and “after” numbers to measure the improvement. Solution The system.time function will execute your code and then report the execution time: > system.time(aLongRunningExpression) The output is three to five numbers, depending on your platform. The first three are usually the most important: User CPU time Number of CPU seconds spent running R System CPU time Number of CPU seconds spent running the operating system, typically for input/ output Elapsed time Number of seconds on the clock Discussion Suppose we want to know the time required to generate 10,000,000 random normal variates and sum them: > system.time(sum(rnorm(10000000))) [1] 3.06 0.01 3.14 NA NA The output from system.time shows that R used 3.06 seconds of CPU time; the oper- ating system used 0.01 seconds of CPU time; and 3.14 seconds elapsed during the test. (The NA values are subprocess times; in this example, there were no subprocesses.) The third value, elapsed time, is not simply the sum of User CPU time and System CPU time. If your computer is busy running other processes, then more clock time will elapse while R is sharing the CPU with those other processes. 12.17 Suppressing Warnings and Error Messages Problem A function is producing annoying error messages or warning messages. You don’t want to see them. 12.17 Suppressing Warnings and Error Messages | 329 Solution Surround the function call with suppressMessage(...) or suppressWarnings(...): > suppressMessage(annoyingFunction()) > suppressuWarnings(annoyingFunction()) Discussion I use the adf.test function quite a bit. However, it produces an annoying warning message, shown here at the bottom of the output, when the p-value is below 0.01: > library(tseries) > adf.test(x) Augmented Dickey-Fuller Test data: x Dickey-Fuller = -4.3188, Lag order = 4, p-value = 0.01 alternative hypothesis: stationary Warning message: In adf.test(x) : p-value smaller than printed p-value Fortunately, I can muzzle the function by calling it inside suppressWarnings(...): > suppressWarnings(adf.test(x)) Augmented Dickey-Fuller Test data: x Dickey-Fuller = -4.3188, Lag order = 4, p-value = 0.01 alternative hypothesis: stationary Notice that the warning message disappeared. The message is not entirely lost because R retains it internally. I can retrieve the message at my leisure by using the warnings function: > warnings() Warning message: In adf.test(x) : p-value smaller than printed p-value Some functions also produce “messages” (in R terminology), which are even more benign than warnings. Typically, they are merely informative and not signals of problems. If such a message is annoying you, call the function inside suppressMes sages(...), and the message will disappear. See Also See the options function for other ways to control the reporting of errors and warnings. 330 | Chapter 12: Useful Tricks 12.18 Taking Function Arguments from a List Problem Your data is captured in a list structure. You want to pass the data to a function, but the function does not accept a list. Solution In simple cases, convert the list to a vector. For more complex cases, the do.call func- tion can break the list into individual arguments and call your function: > do.call(function, list) Discussion If your data is captured in a vector, life is simple and most R functions work as expected: > vec <- c(1, 3, 5, 7, 9) > mean(vec) [1] 5 If your data is captured in a list, then some functions complain and return a useless result, like this: > numbers <- list(1, 3, 5, 7, 9) > mean(numbers) [1] NA Warning message: In mean.default(numbers) : argument is not numeric or logical: returning NA The numbers list is a simple, one-level list, so we can just convert it to a vector and call the function: > mean(unlist(numbers)) [1] 5 The big headaches come when you have multilevel list structures: lists within lists. These can occur within complex data structures. Here is a list of lists in which each sublist is a column of data: > lists <- list(col1=list(7,8,9), col2=list(70,80,90), col3=list(700,800,900)) Suppose we want to form this data into a matrix. The cbind function is supposed to create data columns, but it gets confused by the list structure and returns something useless: > cbind(lists) lists col1 List,3 col2 List,3 col3 List,3 12.18 Taking Function Arguments from a List | 331 D o wnload from Wow! eBook If we unlist the data then we just get one big, long column: > cbind(unlist(lists)) [,1] col11 7 col12 8 col13 9 col21 70 col22 80 col23 90 col31 700 col32 800 col33 900 The solution is to use do.call, which splits the list into individual items and then calls cbind on those items: > do.call(cbind, lists) col1 col2 col3 [1,] 7 70 700 [2,] 8 80 800 [3,] 9 90 900 Using do.call in that way is functionally identical to calling cbind like this: > cbind(lists[[1]], lists[[2]], lists[[3]]) [,1] [,2] [,3] [1,] 7 70 700 [2,] 8 80 800 [3,] 9 90 900 Be careful if the list elements have names. In that case, do.call interprets the element names as names of parameters to the function, which might cause trouble. This recipe presents the most basic use of do.call. The function is quite powerful and has many other uses. See the help page for more details. See Also See Recipe 5.33 for converting between datatypes. 12.19 Defining Your Own Binary Operators Problem You want to define your own binary operators, making your R code more streamlined and readable. Solution R recognizes any text between percent signs (%...%) as a binary operator. Create and define a new binary operator by assigning a two-argument function to it. 332 | Chapter 12: Useful Tricks Discussion R contains an interesting feature that lets you define your own binary operators. Any text between two percentage signs (%...%) is automatically interpreted by R as a binary operator. R predefines several such operators, such as %/% for integer division and %*% for matrix multiplication. You can create a new binary operator by assigning a function to it. This example creates an operator, %+-%: > '%+-%' <- function(x,margin) x + c(-1,+1)*margin The expression x %+-% m calculates x ± m. Here it calculates 100 ± (1.96 × 15), the two- standard-deviation range of a standard IQ test: > 100 %+-% (1.96*15) [1] 70.6 129.4 Notice that we quote the binary operator when defining it but not when using it. The pleasure of defining your own operators is that you can wrap commonly used operations inside a succinct syntax. If your application frequently concatenates strings without an intervening blank, then you might define a binary concatenation operator for that purpose: > '%+%' <- function(s1,s2) paste(s1,s2,sep="") > "Hello" %+% "World" [1] "HelloWorld" > "limit=" %+% round(qnorm(1-0.05/2), 2) [1] "limit=1.96" A danger of defining your own operators is that the code becomes less portable to other environments. Bring the definitions along with the code in which they are used; otherwise, R will complain about undefined operators. All user-defined operators have the same precedence and are listed collectively in Ta- ble 2-1 as %any%. Their precedence is fairly high: higher than multiplication and division but lower than exponentiation and sequence creation. As a result, it’s easy to misexpress yourself. If we omit parentheses from the "%+-%" example, we get an unexpected result: > 100 %+-% 1.96*15 [1] 1470.6 1529.4 R interpreted the expression as (100 %+-% 1.96) * 15. See Also See Recipe 2.11 for more about operator precedence and Recipe 2.12 for how to define a function. 12.19 Defining Your Own Binary Operators | 333 CHAPTER 13 Beyond Basic Numerics and Statistics Introduction This chapter presents a few advanced techniques such as those you might encounter in the first or second year of a graduate program in applied statistics. Most of these recipes use functions available in the base distribution. Through add-on packages, R provides some of the world’s most advanced statistical techniques. This is because researchers in statistics now use R as their lingua franca, showcasing their newest work. Anyone looking for a cutting-edge statistical technique is urged to search CRAN and the Web for possible implementations. 13.1 Minimizing or Maximizing a Single-Parameter Function Problem Given a single-parameter function f, you want to find the point at which f reaches its minimum or maximum. Solution To minimize a single-parameter function, use optimize. Specify the function to be minimized and the bounds for its domain (x): > optimize(f, lower=lowerBound, upper=upperBound) If you instead want to maximize the function, specify maximum=TRUE: > optimize(f, lower=lowerBound, upper=upperBound, maximum=TRUE) Discussion The optimize function can handle functions of one argument. It requires upper and lower bounds for x that delimit the region to be searched. The following example finds the minimum of a polynomial, 3x4 − 2x3 + 3x2 − 4x + 5: 335 > f <− function(x) 3*x^4 − 2*x^3 + 3*x^2 − 4*x + 5 > optimize(f, lower=-20, upper=20) $minimum [1] 0.5972778 $objective [1] 3.636756 The returned value is a list with two elements: minimum, the x value that minimizes the function; and objective, the value of the function at that point. A tighter range for lower and upper means a smaller region to be searched and hence a faster optimization. However, if you are unsure of the appropriate bounds then use big- but-reasonable values such as lower=-1000 and upper=1000. Just be careful that your function does not have multiple minima within that range! The optimize function will find and return only one such minimum. See Also See Recipe 13.2. 13.2 Minimizing or Maximizing a Multiparameter Function Problem Given a multiparameter function f, you want to find the point at which f reaches its minimum or maximum. Solution To minimize a multiparameter function, use optim. You must specify the starting point, which is a vector of initial arguments for f: > optim(startingPoint, f) To maximize the function instead, specify this control parameter: > optim(startingPoint, f, control=list(fnscale=-1)) Discussion The optim function is more general than optimize (Recipe 13.1) because it handles multiparameter functions. To evaluate your function at a point, optim packs the point’s coordinates into a vector and calls your function with that vector. The function should return a scalar value. optim will begin at your starting point and move through the parameter space, searching for the function’s minimum. 336 | Chapter 13: Beyond Basic Numerics and Statistics Here is an example of using optim to fit a nonlinear model. Suppose you believe that the paired observations z and x are related by zi = (xi + α)β + εi, where α and β are unknown parameters and where the εi are nonnormal noise terms. Let’s fit the model by minimizing a robust metric, the sum of the absolute deviations: ∑|z − (x + a)b| First we define the function to be minimized. Observe that the function has only one formal parameter, a two-element vector. The actual parameters to be evaluated, a and b, are packed into the vector in locations 1 and 2: f <- function(v) { a <- v[1]; b <- v[2] # "Unpack" v, giving a and b sum(abs(z - ((x + a)^b))) } This call to optim starts from (1, 1) and searches for the minimum point of f: > optim(c(1,1), f) $par [1] 10.0072591 0.6997801 $value [1] 1.26311 $counts function gradient 485 NA $convergence [1] 0 $message NULL The returned list includes convergence, the indicator of success or failure. If this indicator is 0 then optim found a minimum; otherwise, it did not. Obviously, the con- vergence indicator is the most important returned value because other values are mean- ingless if the algorithm did not converge. The returned list also includes par, the parameters that minimize our function; and value, the value of f at that point. In this case, optim did converge and found a minimum point at approximately a = 10.01 and b = 0.700. There are no lower and upper bounds for optim, just the starting point that you provide. A better guess for the starting point means a faster minimization. The optim function supports several different minimization algorithms, and you can select among them. If the default algorithm does not work for you, see the help page for alternatives. A typical problem with multidimensional minimization is that the al- gorithm gets stuck at a local minimum and fails to find a deeper, global minimum. Generally speaking, the algorithms that are more powerful are less likely to get stuck. However, there is a trade-off: they also tend to run more slowly. 13.2 Minimizing or Maximizing a Multiparameter Function | 337 See Also The R community has implemented many tools for optimization. On CRAN, see the task view for Optimization and Mathematical Programming for more solutions. 13.3 Calculating Eigenvalues and Eigenvectors Problem You want to calculate the eigenvalues or eigenvectors of a matrix. Solution Use the eigen function. It returns a list with two elements, values and vectors, which contain (respectively) the eigenvalues and eigenvectors. Discussion Suppose we have a matrix such as the Fibonacci matrix: > fibmat [,1] [,2] [1,] 0 1 [2,] 1 1 Given the matrix, the eigen function will return a list of its eigenvalues and eigenvectors: > eigen(fibmat) $values [1] 1.618034 -0.618034 $vectors [,1] [,2] [1,] 0.5257311 -0.8506508 [2,] 0.8506508 0.5257311 Use either eigen(fibmat)$values or eigen(fibmat)$vectors to select the needed value from the list. 13.4 Performing Principal Component Analysis Problem You want to identify the principal components of a multivariable dataset. 338 | Chapter 13: Beyond Basic Numerics and Statistics Solution Use the prcomp function. The first argument is a formula whose righthand side is the set of variables, separated by plus signs (+). The lefthand side is empty: > r <- prcomp( ~ x + y + z) > summary(r) Discussion The base software for R includes two functions for principal components analysis, prcomp and princomp. The documentation mentions that prcomp has better numerical properties, so that’s the function presented here. An important use of principal components analysis (PCA) is to reduce the dimension- ality of your dataset. Suppose your data contains a large number N of variables. Ideally, all the variables are more-or-less independent and contributing equally. But if you sus- pect that some variables are redundant, the PCA can tell you the number of sources of variance in your data. If that number is near N, then all the variables are useful. If the number is less than N, then your data can be reduced to a dataset of smaller dimensionality. The PCA recasts your data into a vector space where the first dimension captures the most variance, the second dimension captures the second most, and so forth. The actual output from prcomp is an object that, when printed, gives the needed vector rotation: > r <- prcomp( ~ x + y) > print(r) Standard deviations: [1] 0.3724471 0.1306414 Rotation: PC1 PC2 x -0.5312726 -0.8472010 y -0.8472010 0.5312726 I find the summary of the PCA much more useful. It shows the proportion of variance that is captured by each component: > summary(r) Importance of components: PC1 PC2 Standard deviation 0.372 0.131 Proportion of Variance 0.890 0.110 Cumulative Proportion 0.890 1.000 In this example, the first component captured 89% of the variance and the second component only 11%, so we know the first component captured most of it. After calling prcomp, use plot(r) to view a bar chart of the variances of the principal components and predict(r) to rotate your data to the principal components. 13.4 Performing Principal Component Analysis | 339 See Also See Recipe 13.9 for an example of using principal components analysis. Further uses of PCA in R are discussed in Modern Applied Statistics with S-Plus (Venables and Ripley, Springer). 13.5 Performing Simple Orthogonal Regression Problem You want to create a linear model using orthogonal regression in which the variances of x and y are treated symmetrically. Solution Use prcomp to perform principal components analysis on x and y. From the resulting rotation, compute the slope and intercept: > r <- prcomp( ~ x + y ) > slope <- r$rotation[2,1] / r$rotation[1,1] > intercept <- r$center[2] - slope*r$center[1] Discussion Orthogonal regression is also known as total least squares (TLS). The ordinary least squares (OLS) algorithm has an odd property: it is asymmetric. That is, calculating lm(y ~ x) is not the mathematical inverse of calculating lm(x ~ y). The reason is that OLS assumes the x values to be constants and the y values to be random variables, so all the variance is attributed to y and none is attributed to x. This creates an asymmetric situation. The asymmetry is illustrated in Figure 13-1, whose upper-left panel illustrates the fit for lm(y ~ x). The OLS algorithm tries the minimize the vertical distances, which are shown as dotted lines. The upper-right panel shows the identical dataset but fit with lm(x ~ y) instead, so the algorithm is minimizing the horizontal dotted lines. Obvi- ously, you’ll get a different result depending upon which distances are minimized. The lower panel of Figure 13-1 is quite different. It uses PCA to implement orthogonal regression. Now the distances being minimized are the orthogonal distances from the data points to the regression line. This is a symmetric situation: reversing the roles of x and y does not change the distances to be minimized. Implementing a basic orthogonal regression in R is quite simple. First, perform the PCA: > r <- prcomp( ~ x + y) Next, use the rotations to compute the slope: > slope <- r$rotation[2,1] / r$rotation[1,1] 340 | Chapter 13: Beyond Basic Numerics and Statistics and then, from the slope, calculate the intercept: > intercept <- r$center[2] - slope*r$center[1] I call this a “basic” regression because it yields only the point estimates for the slope and intercept, not the confidence intervals. Obviously, we’d like to have the regression statistics, too. Recipe 13.8 shows one way to estimate the confidence intervals using a bootstrap algorithm. Figure 13-1. Ordinary least squares versus orthogonal regression See Also Principal components analysis is described in Recipe 13.4. The graphics in Fig- ure 13-1 were inspired by the work of Vincent Zoonekynd and his tutorial on regression at http://zoonek2.free.fr/UNIX/48_R/09.html. 13.5 Performing Simple Orthogonal Regression | 341 13.6 Finding Clusters in Your Data Problem You believe your data contains clusters: groups of points that are “near” each other. You want to identify those clusters. Solution Your dataset, x, can be a vector, data frame, or matrix. Assume that n is the number of clusters you desire: > d <- dist(x) # Compute distances between observations > hc <- hclust(d) # Form hierarchical clusters > clust <- cutree(hc, k=n) # Organize them into the n largest clusters The result, clust, is a vector of numbers between 1 and n, one for each observation in x. Each number classifies its corresponding observation into one of the n clusters. Discussion The dist function computes distances between all the observations. The default is Euclidean distance, which works well for many applications, but other distance measures are also available. The hclust function uses those distances to form the observations into a hierarchical tree of clusters. You can plot the result of hclust to create a visualization of the hier- archy, called a dendrogram, as shown in Figure 13-2. Finally, cutree extracts clusters from that tree. You must specify either how many clus- ters you want or the height at which the tree should be cut. Often the number of clusters is unknown, in which case you will need to explore the dataset for clustering that makes sense in your application. I’ll illustrate clustering of a synthetic dataset. We start by generating 99 normal variates, each with a randomly selected mean of either −3, 0, or +3: > means <- sample(c(-3,0,+3), 99, replace=TRUE) > x <- rnorm(99, mean=means) For our own curiosity, we can compute the true means of the original clusters. (In a real situation, we would not have the means factor and would be unable to perform this computation.) We can confirm that the groups’ means are pretty close to −3, 0, and +3: > tapply(x, factor(means), mean) -3 0 3 -3.0151424 -0.2238356 2.7602507 To discover the clusters, we first compute the distances between all points: > d <- dist(x) 342 | Chapter 13: Beyond Basic Numerics and Statistics Then we create the hierarchical clusters: > hc <- hclust(d) Now we can extract the three largest clusters. Obviously, we have a huge advantage here because we know the true number of clusters. Real life is rarely that easy: > clust <- cutree(hc, k=3) clust is a vector of integers between 1 and 3, one integer for each observation in the sample, that assigns each observation to a cluster. Here are the first 20 cluster assignments: > head(clust,20) [1] 1 2 2 2 1 2 3 3 2 3 1 3 2 3 2 1 2 1 1 3 By treating the cluster number as a factor, we can compute the mean of each statistical cluster (Recipe 6.5): > tapply(x, clust, mean) 1 2 3 3.1899159 -2.6992682 0.2363639 Figure 13-2. Plotting the cluster hierarchy 13.6 Finding Clusters in Your Data | 343 R did a good job of splitting data into clusters: the means appear distinct, with one near −2.7, one near 0.27, and one near +3.2. (The order of the extracted means does not necessarily match the order of the original groups, of course.) The extracted means are similar but not identical to the original means. Side-by-side box plots can show why: > plot(x ~ factor(means), main="Original Clusters", xlab="Cluster Mean") > plot(x ~ factor(clust), main="Identified Clusters", xlab="Cluster Number") The box plots are shown in Figure 13-3. The clustering algorithm perfectly separated the data into nonoverlapping groups. The original clusters overlapped, whereas the identified clusters do not. Figure 13-3. Clusters comparison This illustration used one-dimensional data, but the dist function works equally well on multidimensional data stored in a data frame or matrix. Each row in the data frame or matrix is treated as one observation in a multidimensional space, and dist computes the distances between those observations. See Also This demonstration is based on the clustering features of the base package. There are other packages such as mclust that offer alternative clustering mechanisms. 344 | Chapter 13: Beyond Basic Numerics and Statistics 13.7 Predicting a Binary-Valued Variable (Logistic Regression) Problem You want to perform logistic regression, a regression model that predicts the probability of a binary event occurring. Solution Call the glm function with family=binomial to perform logistic regression. The result is a model object: > m <- glm(b ~ x1 + x2 + x3, family=binomial) Here, b is a factor with two levels (e.g., TRUE and FALSE, 0 and 1) x1, x2, and x3 are predictor variables. Use the model object, m, and the predict function to predict a probability from new data: > dfrm <- data.frame(x1=value, x2=value, x3=value) > predict(m, type="response", newdata=dfrm) Discussion Predicting a binary-valued outcome is a common problem in modeling. Will a treat- ment be effective or not? Will prices rise or fall? Who will win the game, team A or team B? Logistic regression is useful for modeling these situations. In the true spirit of statistics, it does not simply give a “thumbs up” or “thumbs down” answer; rather, it computes a probability for each of the two possible outcomes. In the call to predict, we set type="response" so that predict returns a probability. Otherwise, it returns log-odds, which I don’t find as useful. In his unpublished book entitled Practical Regression and ANOVA Using R, Faraway gives an example of predicting a binary-valued variable: test from the dataset pima is true if the patient tested positive for diabetes. The predictors are diastolic blood pres- sure and body mass index (BMI). Faraway uses linear regression, so let’s try logistic regression instead: > data(pima, package="faraway") > b <- factor(pima$test) > m <- glm(b ~ diastolic + bmi, family=binomial, data=pima) The summary of the resulting model, m, shows that the respective p-values for the diastolic and the bmi variables are 0.805 and (essentially) zero. We can therefore con- clude that only the bmi variable is significant: > summary(m) Call: glm(formula = b ~ diastolic + bmi, family = binomial, data = pima) 13.7 Predicting a Binary-Valued Variable (Logistic Regression) | 345 Deviance Residuals: Min 1Q Median 3Q Max -1.9128 -0.9180 -0.6848 1.2336 2.7417 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -3.629553 0.468176 -7.753 9.01e-15 *** diastolic -0.001096 0.004432 -0.247 0.805 bmi 0.094130 0.012298 7.654 1.95e-14 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 993.48 on 767 degrees of freedom Residual deviance: 920.65 on 765 degrees of freedom AIC: 926.65 Number of Fisher Scoring iterations: 4 Because only the bmi variable is significant, we can create a reduced model like this: > m.red <- glm(b ~ bmi, family=binomial, data=pima) Let’s use the model to calculate the probability that someone with an average BMI (32.0) will test positive for diabetes: > newdata <- data.frame(bmi=32.0) > predict(m.red, type="response", newdata=newdata) 1 0.3332689 According to this model, the probability is about 33.3%. The same calculation for someone in the 90th percentile gives a probability of 54.9%: > newdata <- data.frame(bmi=quantile(pima$bmi, .90)) > predict(m.red, type="response", newdata=newdata) 90% 0.5486215 See Also Using logistic regression involves interpreting the deviance to judge the significance of the model. I suggest reviewing a text on logistic regression before attempting to draw any conclusions from your regression. 13.8 Bootstrapping a Statistic Problem You have a dataset and a function to calculate a statistic from that dataset. You want to estimate a confidence interval for the statistic. 346 | Chapter 13: Beyond Basic Numerics and Statistics Solution Use the boot package. Apply the boot function to calculate bootstrap replicates of the statistic: > library(boot) > bootfun <- function(data, indices) { + # . . . calculate statistic using data[indices] . . . + return(statistic) + } > reps <- boot(data, bootfun, R=999) Here, data is your original data set, which can be stored in either a vector or a data frame. The statistic function (bootfun in this case) should expect two arguments: data is the original dataset, and indices is a vector of integers that selects the bootstrap sample from data. Next, use the boot.ci function to estimate a confidence interval from the replications: > boot.ci(reps, type=c("perc","bca")) Discussion Anybody can calculate a statistic, but that’s just the point estimate. We want to take it to the next level: What is the confidence interval? For some statistics, we can calculate the confidence interval (CI) analytically. The confidence interval for a mean, for in- stance, is calculated by the t.test function. Unfortunately, that is the exception and not the rule. For most statistics, the mathematics is too tortuous or simply unknown, and there is no known closed-form calculation for the confidence interval. The bootstrap algorithm can estimate a confidence interval even when no closed-form calculation is available. It works like this. The algorithm assumes that you have a sample of size N and a function to calculate the statistic: 1. Randomly select N elements from the sample, sampling with replacement. That set of elements is called a bootstrap sample. 2. Apply the function to the bootstrap sample to calculate the statistic. That value is called a bootstrap replication. 3. Repeat steps 1 and 2 many times to yield many (typically thousands) of bootstrap replications. 4. From the bootstrap replications, compute the confidence interval. That last step may seem mysterious, but there are several algorithms for computing the confidence interval. A simple one uses percentiles of the replications, such as taking the 2.5 percentile and the 97.5 percentile to form the 95% confidence interval. I am a huge fan of the bootstrap because I work daily with obscure statistics, it is important that I know their confidence interval, and there is definitely no known for- mula for their CI. The bootstrap gives me a good approximation. 13.8 Bootstrapping a Statistic | 347 Let’s work an example. In Recipe 13.4 we estimated the slope of a line using orthogonal regression. That gave us a point estimate, but how can we find the confidence interval? First, we encapsulate the slope calculation within a function: > stat <- function(data, indices) { + r <- prcomp( ~ x + y, data=data, subset=indices) + slope <- r$rotation[2,1] / r$rotation[1,1] + return(slope) + } Notice that the function is careful to select the subset defined by indices and to compute the slope from that exact subset. Next, we calculate 999 replications of the slope. Recall that we had two vectors, x and y, in the original recipe; here, we combine them into a data frame: > boot.data <- data.frame(x=x, y=y) > reps <- boot(boot.data, stat, R=999) The choice of 999 replications is a good starting point. You can always repeat the bootstrap with more and see if the results change significantly. The boot.ci function can estimate the CI from the replications. It implements several different algorithms, and the type argument selects which algorithms are performed. For each selected algorithm, boot.ci will return the resulting estimate: > boot.ci(reps, type=c("perc","bca")) BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS Based on 999 bootstrap replicates CALL : boot.ci(boot.out = reps, type = c("perc", "bca")) Intervals : Level Percentile BCa 95% ( 1.144, 2.277 ) ( 1.123, 2.270 ) Calculations and Intervals on Original Scale Here I chose two algorithms, percentile and BCa, by setting type=c("perc","bca"). The two resulting estimates appear at the bottom under their names. Other algorithms are available; see the help page. You will note that the two confidence intervals are slightly different: (1.144, 2.277) versus (1.123, 2.270). This is an uncomfortable but inevitable result of using two dif- ferent algorithms. I don’t know any method for deciding which is better. If the selection is a critical issue, you will need to study the reference and understand the differences. In the meantime, my best advice is to be conservative and use the wider interval; in this case, that would be (1.123, 2.270). By default, boot.ci estimates a 95% confidence interval. You can change that via the conf argument, like this: > boot.ci(reps, type=c("perc","bca"), conf=0.90) 348 | Chapter 13: Beyond Basic Numerics and Statistics See Also See Recipe 13.4 for the slope calculation. A good tutorial and reference for the bootstrap algorithm is An Introduction to the Bootstrap by Efron and Tibshirani (Chapman & Hall/CRC). 13.9 Factor Analysis Problem You want to perform factor analysis on your dataset, usually to discover what your variables have in common. Solution Use the factanal function, which requires your dataset and your estimate of the number of factors: > factanal(data, factors=n) The output includes n factors, showing the loadings of each input variable for each factor. The output also includes a p-value. Conventionally, a p-value of less than 0.05 indicates that the number of factors is too small and does not capture the full dimensionality of the dataset; a p-value exceeding 0.05 indicates that there are likely enough (or more than enough) factors. Discussion Factor analysis creates linear combinations of your variables, called factors, that abstract the variable’s underlying commonality. If your N variables are perfectly inde- pendent, then they have nothing in common and N factors are required to describe them. But to the extent that the variables have an underlying commonality, fewer fac- tors capture most of the variance and so fewer than N factors are required. For each factor and variable, we calculate the correlation between them and call it the loading. Variables with a high loading are well explained by the factor. We can square the loading to know what fraction of the variable’s total variance is explained by the factor. Factor analysis is useful when it shows that a few factors capture most of the variance of your variables. Thus, it alerts you to redundancy in your data. In that case you can reduce your dataset by combining closely related variables or by eliminating redundant variables altogether. 13.9 Factor Analysis | 349 A more subtle application of factor analysis is interpreting the factors to find interre- lationships between your variables. If two variables both have large loadings for the same factor, then you know they have something in common. What is it? There is no mechanical answer. You’ll need to study the data and its meaning. There are two tricky aspects of factor analysis. The first is choosing the number of factors. Fortunately, you can use principal components analysis to get a good initial estimate of the number of factors. The second tricky aspect is interpreting the factors themselves. I’ll illustrate factor analysis by using stock prices—or, more precisely, changes in stock prices. The dataset contains 6 months of price changes for the stocks of 12 companies. Every company is involved in the petroleum and gasoline industry. Their stock prices probably move together, since they are subject to the similar economic and market forces. We might ask: How many factors are required to explain their changes? If only one factor is required, then all the stocks are the same and one is as good as another. If many factors are required, we know that owning several of them provides diversification. We start by doing a principal components analysis on diffs, the data frame of price changes. Plotting the PCA results shows the variance captured by the components: > plot(prcomp(diffs)) The plot is shown in Figure 13-4. We can see that the first two components capture much of the variance, but possibly a third is required. So we perform the initial factor analysis while assuming that two factors are required: > factanal(diffs, factors=2) Call: factanal(x = diffs, factors = 2) . . (etc.) . Test of the hypothesis that 2 factors are sufficient. The chi square statistic is 102.25 on 43 degrees of freedom. The p-value is 9.83e-07 We can ignore most of the output because the p-value at the bottom is essentially zero (9.83 × 10−7). The small p-value indicates that two factors are insufficient, so the analysis isn’t good. More are required, so we try again with three factors instead: > factanal(diffs, factors=3) Call: factanal(x = diffs, factors = 3) Uniquenesses APC BP BRY CVX HES MRO NBL OXY SUN TOT VLO XOM 350 | Chapter 13: Beyond Basic Numerics and Statistics D o wnload from Wow! eBook 0.005 0.509 0.299 0.162 0.150 0.207 0.321 0.265 0.098 0.185 0.257 0.155 Loadings: Factor1 Factor2 Factor3 APC 0.286 0.176 0.939 BP 0.212 0.150 0.651 BRY 0.628 0.474 0.286 CVX 0.819 0.320 0.253 HES 0.751 0.415 0.338 MRO 0.721 0.396 0.342 NBL 0.606 0.382 0.406 OXY 0.713 0.436 0.192 SUN 0.394 0.847 0.172 TOT 0.786 0.372 0.241 VLO 0.465 0.672 0.275 XOM 0.833 0.218 0.322 Factor1 Factor2 Factor3 SS loadings 4.835 2.400 2.151 Proportion Var 0.403 0.200 0.179 Cumulative Var 0.403 0.603 0.782 Test of the hypothesis that 3 factors are sufficient. The chi square statistic is 35.74 on 33 degrees of freedom. The p-value is 0.341 Figure 13-4. Principal components of price changes 13.9 Factor Analysis | 351 The large p-value (0.341) confirms that three factors are sufficient, so we can use the analysis. The output includes a table of explained variance, shown here: Factor1 Factor2 Factor3 SS loadings 4.835 2.400 2.151 Proportion Var 0.403 0.200 0.179 Cumulative Var 0.403 0.603 0.782 This table shows that the proportion of variance explained by each factor is 0.403, 0.200, and 0.179, respectively. Cumulatively, they explain 0.782 of the variance, which leaves 1 − 0.782 = 0.218 unexplained. Next we want to interpret the factors, which is more like voodoo than science. Let’s look at the loadings, repeated here: Loadings: Factor1 Factor2 Factor3 APC 0.286 0.176 0.939 BP 0.212 0.150 0.651 BRY 0.628 0.474 0.286 CVX 0.819 0.320 0.253 HES 0.751 0.415 0.338 MRO 0.721 0.396 0.342 NBL 0.606 0.382 0.406 OXY 0.713 0.436 0.192 SUN 0.394 0.847 0.172 TOT 0.786 0.372 0.241 VLO 0.465 0.672 0.275 XOM 0.833 0.218 0.322 Each row is labeled with the variable name (stock symbol): APC, BP, BRY, and so forth. The first factor has many large loadings, indicating that it explains the variance of many stocks. This is a common phenomenon in factor analysis. We are often looking at related variables, and the first factor captures their most basic relationship. In this example, we are dealing with stocks, and most stocks move together in concert with the broad market. That’s probably captured by the first factor. The second factor is more subtle. Notice that the loadings for SUN (0.847) and VLO (0.672) are the dominant ones and that all other stocks have noticeably smaller load- ings. This indicates a connection between SUN and VLO. Perhaps they operate together in a common area (e.g., retail sales) and so tend to move together. The third factor also has two dominant loadings: APC (0.939) and BP (0.651). During the sample period, there was a horrific environmental disaster—caused by an oil rig explosion in the Gulf of Mexico—that cast a dark shadow on the future of oil explo- ration in the Gulf. British Petroleum (BP) owned the rig, and Anadarko Petroleum (APC) is a major supplier of equipment to Gulf operators. Those two stocks moved in unison while the market assessed the disaster’s impact on BP and APC. 352 | Chapter 13: Beyond Basic Numerics and Statistics In summary, it seems there are three groups of stocks here: • SUN and VLO • APC and BP • Everything else Factor analysis is an art and a science. I suggest that you read a good book on multi- variate analysis before employing it. See Also See Recipe 13.4 for more about principal components analysis. 13.9 Factor Analysis | 353 CHAPTER 14 Time Series Analysis Introduction Time series analysis has become a hot topic with the rise of quantitative finance and automated trading of securities. Many of the facilities described in this chapter were invented by practitioners and researchers in finance, securities trading, and portfolio management. Before starting any time series analysis in R, a key decision is your choice of data rep- resentation (object class). This is especially critical in an object-oriented language such as R because the choice affects more than how the data is stored; it also dictates which functions (methods) will be available for loading, processing, analyzing, printing, and plotting your data. When I first started using R, for example, I simply stored my time series data in vectors. That seemed natural. However, I quickly discovered that none of the coolest analytics for time series analysis worked with simple vectors. I switched to using an object class intended for time series analysis, and this was the gateway to important functions and analytics. This chapter’s first recipe recommends using the zoo or xts packages for representing time series data. They are quite general and should meet the needs of most users. Nearly every subsequent recipe assumes you are using one of those two representations. The xts implementation is a superset of zoo, so xts can do everything that zoo can do. In this chapter, whenever a recipe works for a zoo object, you can safely assume (unless stated otherwise) that it also works for an xts object. The base distribution of R includes a time series class called ts. I don’t recommend this representation for general use because the implementation itself is too restrictive and because the associated functions (methods) are limited in scope, usefulness, and power. 355 Date Versus Datetime Every observation in a time series has an associated date or time. The object classes used in this chapter, zoo and xts, give you the choice of using either dates or datetimes for representing the data’s time component. You would use dates to represent daily data, of course, and also for weekly, monthly, or even annual data; in these cases, the date gives the day on which the observation occurred. You would use datetimes for intraday data, where both the date and time of observation are needed. In describing this chapter’s recipes, it’s pretty cumbersome to keep saying “date or datetime.” So I simplified the prose by assuming that your data are daily and thus use whole dates. Please bear in mind, of course, that you are free and able to use timestamps below the resolution of a calendar date. See Also R has many useful functions and packages for time series analysis. You’ll find pointers to them in the task view for Time Series Analysis . 14.1 Representing Time Series Data Problem You want an R data structure that can represent time series data. Solution I recommend the zoo and xts packages. They define a data structure for time series, and they contain many useful functions for working with time series data. Create a zoo object this way, where x is a vector or data frame and dt is a vector of corresponding dates or datetimes: > library(zoo) > ts <- zoo(x, dt) Create an xts object in this way: > library(xts) > ts <- xts(x, dt) Convert between representations of the time series data by using as.zoo and as.xts: as.zoo(ts) Converts ts to a zoo object as.xts(ts) Converts ts to an xts object 356 | Chapter 14: Time Series Analysis Discussion R has at least eight different implementations of data structures for representing time series. I haven’t tried them all, but I can say that the zoo and xts packages are excellent packages for working with time series data and are better than the others that I have tried. These representations assume you have two vectors: a vector of observations (data) and a vector of dates or times of those observations. The zoo function combines them into a zoo object: > ts <- zoo(x, dt) The xts function is similar, returning an xts object: > ts <- xts(x, dt) The data, x, should be numeric. The vector of dates or datetimes, dt, is called the index. Legal indices vary between the packages: zoo The index can be any ordered values, such as Date objects, POSIXct objects, integers, or even floating-point values. xts The index must be a supported date or time class. This includes Date, POSIXct, and chron objects. Those should be sufficient for most applications, but you can also use yearmon, yearqtr, and dateTime objects. The xts package is more restrictive than zoo because it implements powerful operations that require a time-based index. The following example creates a zoo object that contains the price of IBM stock for the first five days of 2010; it uses Date objects for the index: > prices <- c(132.45, 130.85, 130.00, 129.55, 130.85) > dates <- as.Date(c("2010-01-04", "2010-01-05", "2010-01-06", + "2010-01-07","2010-01-08")) > ibm.daily <- zoo(prices, dates) > print(ibm.daily) 2010-01-04 2010-01-05 2010-01-06 2010-01-07 2010-01-08 132.45 130.85 130.00 129.55 130.85 In contrast, the next example captures the price of IBM stock at one-second intervals. It represents time by the number of hours past midnight starting at 9:30 a.m. (1 second ≈ 0.00027778 hours): > prices <- c(131.18, 131.20, 131.17, 131.15, 131.17) > seconds <- c(9.5, 9.500278, 9.500556, 9.500833, 9.501111) > ibm.sec <- zoo(prices, seconds) > print(ibm.sec) 9.5 9.5003 9.5006 9.5008 9.5011 131.18 131.20 131.17 131.15 131.17 14.1 Representing Time Series Data | 357 Those two examples used a single time series, where the data came from a vector. Both zoo and xts can also handle multiple, parallel time series. For this, capture the several time series in a data frame and then create a multivariate time series by calling the zoo (or xts) function: > ts <- zoo(dfrm, dt) # OR: ts <- xts(dfrm, dt) The second argument is a vector of dates (or datetimes) for each observation. There is only one vector of dates for all the time series; in other words, all observations in each row of the data frame must have the same date. See Recipe 14.5 if your data have mismatched dates. Once the data is captured inside a zoo or xts object, you can extract the pure data via coredata, which returns a simple vector (or matrix): > coredata(ibm.daily) [1] 132.45 130.85 130.00 0.00 129.55 You can extract the date or time portion via index: > index(ibm.daily) [1] "2010-01-04" "2010-01-05" "2010-01-06" "2010-01-07" "2010-01-08" The xts package is strongly similar to zoo. It is optimized for speed and so is especially well suited for processing large volumes of data. It is also clever about converting to and from other time series representations. One big advantage of capturing data inside a zoo or xts object is that special-purpose functions become available for printing, plotting, differencing, merging, periodic sam- pling, applying rolling functions, and other useful operations. There is even a function, read.zoo, dedicated to reading time series data from ASCII files. Remember that the xts package can do everything that the zoo package can do, so everywhere that this chapter talks about zoo objects you can also use xts objects. If you are a serious user of time series data, I strongly recommend studying the docu- mentation of these packages in order to learn about the ways they can improve your life. They are rich packages with many useful features. See Also See CRAN for documentation on zoo and xts; this documentation includes reference manuals, vignettes, and quick reference cards. If the packages are already installed on your computer, view their documentation using the vignette function: > vignette("zoo") > vignette("xts") The timeSeries package is another good implementation of a time series object. It is part of the Rmetrics project for quantitative finance. 358 | Chapter 14: Time Series Analysis 14.2 Plotting Time Series Data Problem You want to plot one or more time series. Solution Use plot(x), which works for zoo objects and xts objects containing either single or multiple time series. For a simple vector v of time series observations, you can use either plot(v,type="l") or plot.ts(v). Discussion The generic plot function has a version for zoo objects and xts objects. It can plot objects that contain a single time series or multiple time series. In the latter case, it can plot each series in a separate plot or together in one plot. Suppose that ibm.infl is a zoo object that contains two time series. One shows the quoted price of IBM stock during the 1970s, and the other is that same price adjusted for inflation.* If you plot the object with screens=1, R will plot the two time series together in one plot: > plot(ibm.infl, screens=1) If you specify screens=c(1,2), however, then R will plot both time series separately in two plots: > plot(ibm.infl, screens=c(1,2)) The plot function provides a default label for the x-axis and y-axis, but they are not very informative. It does not provide a default title. So I almost always supply my own xlab, ylab, and main (title) parameters. This R code produces the plots shown in Figures 14-1 and 14-2: xlab="Date" ylab="Relative Price" main="IBM: Historical vs. Inflation-Adjusted" lty=c("dotted", "solid") ylim=range(coredata(ibm.infl)) # Plot the two time series in two plots plot(ibm.infl, screens=c(1,2), lty=lty, main=main, xlab=xlab, ylab=ylab, ylim=ylim) # Plot the two time series in one plot * The data shown do not match the historical stock quotes because they have been adjusted for dividends and stock splits, then normalized to a starting value of 100. 14.2 Plotting Time Series Data | 359 plot(ibm.infl, screens=1, lty=lty, main=main, xlab=xlab, ylab=ylab) # Add a legend legend(as.Date("1970-01-01"), 140, c("Hist","Infl-Adj"), lty=c("dotted", "solid")) The code specifies two line types (lty) so that the two lines are drawn in two different styles, making them easier to distinguish. It also specifies the same y limits (ylim) for the separate plots, making them visually comparable. I included both plots so you can experience their difference yourself. For my money, the two-in-one plot illustrates far more dramatically that inflation in the 1970s took a big bite out of IBM’s stock price: when adjusted for inflation, that price actually fell. See Also For working with financial data, the quantmod package contains special plotting func- tions that produce beautiful, stylized plots. Figure 14-1. Plotting two time series with screen=c(1,2) 360 | Chapter 14: Time Series Analysis Figure 14-2. Plotting two time series with screen=1 14.3 Extracting the Oldest or Newest Observations Problem You want to see only the oldest or newest observations of your time series. Solution Use head to view the oldest observations: > head(ts) Use tail to view the newest observations: > tail(ts) Discussion The head and tail functions are generic, so they will work whether your data is stored in a simple vector, a zoo object, or an xts object. 14.3 Extracting the Oldest or Newest Observations | 361 Suppose you have a zoo object with a multiyear history of the price of IBM stock. You can’t display the whole dataset because it would scroll off your screen. But you can view the initial observations: > head(ibm) 1970-01-02 1970-01-05 1970-01-06 1970-01-07 1970-01-08 1970-01-09 364.75 368.25 368.50 368.75 369.50 369.00 And you can view the final observations: > tail(ibm) 1979-12-21 1979-12-24 1979-12-26 1979-12-27 1979-12-28 1979-12-31 64.00 64.87 64.87 64.75 64.25 64.37 By default, head and tail show (respectively) the six oldest and six newest observations. You can see more observations by providing a second argument—for example, tail(ibm, 20). The xts package also includes first and last functions, which use calendar periods instead of number of observations. We can use first and last to select data by number of days, weeks, months, or even years: > first(as.xts(ibm), "3 weeks") [,1] 1970-01-02 364.75 1970-01-05 368.25 1970-01-06 368.50 1970-01-07 368.75 1970-01-08 369.50 1970-01-09 369.00 1970-01-12 367.75 1970-01-13 374.13 1970-01-14 373.75 1970-01-15 381.50 1970-01-16 369.75 > last(as.xts(ibm), "month") [,1] 1979-12-03 65.00 1979-12-04 65.37 1979-12-05 65.75 . . (etc.) . 1979-12-27 64.75 1979-12-28 64.25 1979-12-31 64.37 Notice that I converted the zoo object to an xts object by calling as.xts(ibm), forcing R to use the xts functions (methods). Also notice that a “week” is defined by the cal- endar, not just by any seven consecutive days. 362 | Chapter 14: Time Series Analysis See Also See help(first.xts) and help(last.xts) for details of the first and last functions, respectively. 14.4 Subsetting a Time Series Problem You want to select one or more elements from a time series. Solution You can index a zoo or xts object by position. Use one or two subscripts, depending upon whether the object contains one time series or multiple time series: ts[i] Selects the ith observation from a single time series ts[j,i] Selects the ith observation of the jth time series of multiple time series You can index the time series by a date object. Use the same type of object as the index of your time series. This example assumes that the index contains Date objects: > ts[as.Date("yyyy-mm-dd")] You can index it by a sequence of dates: > dates <- seq(startdate, enddate, increment) > ts[dates] The window function can select a range by start and end date: > window(ts, start=startdate, end=enddate) Discussion Recall our small sample of IBM stock prices: > ibm.daily 2010-01-04 2010-01-05 2010-01-06 2010-01-07 2010-01-08 132.45 130.85 130.00 129.55 130.85 We can select an observation by position, just like selecting elements from a vector (Recipe 2.9): > ibm.daily[2] 2010-01-05 130.85 14.4 Subsetting a Time Series | 363 We can also select multiple observations by position: > ibm.daily[2:4] 2010-01-05 2010-01-06 2010-01-07 130.85 130.00 129.55 Sometimes it’s more useful to select by date. In this case, our index is built from Date objects, so we subscript the time series using a Date object (if the index were built from POSIXct objects, we’d use a POSIXct object): > ibm.daily[as.Date('2010-01-05')] 2010-01-05 130.85 We can select by a vector of Date objects: > dates <- seq(as.Date('2010-01-04'), as.Date('2010-01-08'), by=2) > ibm.daily[dates] 2010-01-04 2010-01-06 2010-01-08 132.45 130.00 130.85 The window function is easier for selecting a range of consecutive dates: > window(ibm.daily, start=as.Date('2010-01-05'), end=as.Date('2010-01-07')) 2010-01-05 2010-01-06 2010-01-07 130.85 130.00 129.55 See Also The xts package provides many other clever ways to index a time series. See the package documentation. 14.5 Merging Several Time Series Problem You have two or more time series. You want to merge them into a single time series object. Solution Use the zoo object to represent the time series; then use the merge function to combine them: > merge(ts1, ts2) Discussion Merging two time series is an incredible headache when the two series have differing timestamps. Consider these two time series: the daily price of IBM stock from 1970 through 1979; and the monthly Consumer Price Index (CPI) for the same period: 364 | Chapter 14: Time Series Analysis > ibm 1970-01-02 1970-01-05 1970-01-06 1970-01-07 1970-01-08 1970-01-09 1970-01-12 364.75 368.25 368.50 368.75 369.50 369.00 367.75 . . (etc.) . 1979-12-21 1979-12-24 1979-12-26 1979-12-27 1979-12-28 1979-12-31 64.00 64.87 64.87 64.75 64.25 64.37 > cpi 1970-01-01 1970-02-01 1970-03-01 1970-04-01 1970-05-01 1970-06-01 1970-07-01 37.8 38.0 38.2 38.5 38.6 38.8 39.0 . . (etc.) . 1979-05-01 1979-06-01 1979-07-01 1979-08-01 1979-09-01 1979-10-01 1979-11-01 71.5 72.3 73.1 73.8 74.6 75.2 75.9 1979-12-01 76.7 Obviously, the two time series have different timestamps because one is daily data and the other is monthly data. Even worse, the downloaded CPI data is timestamped for the first day of every month, even when that day is a holiday or weekend (e.g., New Year’s Day). Thank goodness for the merge function, which handles the messy details of reconciling the different dates: > merge(ibm,cpi) ibm cpi 1970-01-01 NA 37.8 1970-01-02 364.75 NA 1970-01-05 368.25 NA 1970-01-06 368.50 NA 1970-01-07 368.75 NA 1970-01-08 369.50 NA . . (etc.) . By default, merge finds the union of all dates: the output contains all dates from both inputs, and missing observations are filled with NA values. You can replace those NA values with the most recent observation by using the na.locf function from the zoo package: > na.locf(merge(ibm,cpi))) ibm cpi 1970-01-02 364.75 37.8 1970-01-05 368.25 37.8 1970-01-06 368.50 37.8 1970-01-07 368.75 37.8 1970-01-08 369.50 37.8 1970-01-09 369.00 37.8 . . (etc.) . 14.5 Merging Several Time Series | 365 D o wnload from Wow! eBook (Here locf stands for “last observation carried forward”.) Observe that the NAs were replaced. Also, na.locf eliminated the first observation (1970-01-01) because there was no IBM stock price on that day. You can get the intersection of all dates by setting all=FALSE: > merge(ibm,cpi,all=FALSE) ibm cpi 1970-04-01 337.75 38.5 1970-05-01 296.75 38.6 1970-06-01 287.00 38.8 1970-07-01 254.00 39.0 1970-09-01 263.88 39.2 1970-10-01 297.00 39.4 . . (etc.) . Now the output is limited to observations that are common to both files. Notice, however, that the intersection begins on April 1, not January 1. The reason is that January 1, February 1, and March 1 are all holidays or weekends, so there is no IBM stock price for those dates and hence no intersection with the CPI data. To fix this, see Recipe 14.6. 14.6 Filling or Padding a Time Series Problem Your time series data is missing observations. You want to fill or pad the data with the missing dates/times. Solution Create a zero-width (dataless) zoo object with the missing dates/times. Then merge your data with the zero-width object, taking the union of all dates: > empty <- zoo(,dates) # 'dates' is vector of the missing dates > merge(ts, empty, all=TRUE) Discussion The zoo package includes a handy feature in the constructor for zoo objects: you can omit the data and build a zero-width object. The object contains no data, just dates. We can use these “Frankenstein” objects to perform such operations as filling and padding on other time series objects. Suppose you download monthly CPI data for the 1970s. The data is timestamped with the first day of each month: 366 | Chapter 14: Time Series Analysis > cpi 1970-01-01 1970-02-01 1970-03-01 1970-04-01 1970-05-01 1970-06-01 37.8 38.0 38.2 38.5 38.6 38.8 . . (etc.) . 1979-07-01 1979-08-01 1979-09-01 1979-10-01 1979-11-01 1979-12-01 73.1 73.8 74.6 75.2 75.9 76.7 As far as R knows, we have no observations for the other days of the months. However, we know that each CPI value applies to the subsequent days through month-end. So first we build a zero-width object with every day of the decade, but no data: > dates <- seq(from=as.Date("1970-01-01"), to=as.Date("1979-12-31"), by=1) > empty <- zoo(,dates) Then we take the union the CPI data and the zero-width object, yielding a dataset filled with NA values: > filled.cpi <- merge(cpi, empty, all=TRUE) > filled.cpi 1970-01-01 1970-01-02 1970-01-03 1970-01-04 1970-01-05 1970-01-06 1970-01-07 37.8 NA NA NA NA NA NA 1970-01-08 1970-01-09 1970-01-10 1970-01-11 1970-01-12 1970-01-13 1970-01-14 NA NA NA NA NA NA NA 1970-01-15 1970-01-16 1970-01-17 1970-01-18 1970-01-19 1970-01-20 1970-01-21 NA NA NA NA NA NA NA 1970-01-22 1970-01-23 1970-01-24 1970-01-25 1970-01-26 1970-01-27 1970-01-28 NA NA NA NA NA NA NA 1970-01-29 1970-01-30 1970-01-31 1970-02-01 1970-02-02 1970-02-03 1970-02-04 NA NA NA 38.0 NA NA NA . . (etc.) . The resulting time series contains every calendar day, with NAs where there was no observation. That might be what you need. However, a more common requirement is to replace each NA with the most recent observation as of that date. The na.locf func- tion from the zoo package does exactly that: > filled.cpi <- na.locf(merge(cpi, empty, all=TRUE)) > filled.cpi 1970-01-01 1970-01-02 1970-01-03 1970-01-04 1970-01-05 1970-01-06 1970-01-07 37.8 37.8 37.8 37.8 37.8 37.8 37.8 1970-01-08 1970-01-09 1970-01-10 1970-01-11 1970-01-12 1970-01-13 1970-01-14 37.8 37.8 37.8 37.8 37.8 37.8 37.8 1970-01-15 1970-01-16 1970-01-17 1970-01-18 1970-01-19 1970-01-20 1970-01-21 37.8 37.8 37.8 37.8 37.8 37.8 37.8 1970-01-22 1970-01-23 1970-01-24 1970-01-25 1970-01-26 1970-01-27 1970-01-28 37.8 37.8 37.8 37.8 37.8 37.8 37.8 1970-01-29 1970-01-30 1970-01-31 1970-02-01 1970-02-02 1970-02-03 1970-02-04 37.8 37.8 37.8 38.0 38.0 38.0 38.0 . . (etc.) . 14.6 Filling or Padding a Time Series | 367 January’s value of 37.8 is carried forward until February 1, at which time it is replaced by the February value of 38.0. Now every day has the latest CPI value as of that date. We can use this recipe to fix the problem mentioned in Recipe 14.5. There, the daily price of IBM stock and the monthly CPI data had no intersection on certain days. We can fix that using several different methods. One way is to pad the IBM data to include the CPI dates and then take the intersection (recall that index(cpi) returns all the dates in the CPI time series): > filled.ibm <- na.locf(merge(ibm, zoo(,index(cpi)))) > merge(filled.ibm, cpi, all=FALSE) filled.ibm cpi 1970-02-01 335.19 38.0 1970-03-01 340.25 38.2 1970-04-01 337.75 38.5 1970-05-01 296.75 38.6 1970-06-01 287.00 38.8 1970-07-01 254.00 39.0 . . (etc.) . That gives monthly observations. Another way is to fill out the CPI data (as described previously) and then take the intersection with the IBM data. That gives daily obser- vations, as follows: > merge(ibm, filled.cpi, all=FALSE) ibm filled.cpi 1970-01-02 364.75 37.8 1970-01-05 368.25 37.8 1970-01-06 368.50 37.8 1970-01-07 368.75 37.8 1970-01-08 369.50 37.8 1970-01-09 369.00 37.8 . . (etc.) . 14.7 Lagging a Time Series Problem You want to shift a time series in time, either forward or backward. Solution Use the lag function. The second argument, k, is the number of periods to shift the data: > lag(ts, k) 368 | Chapter 14: Time Series Analysis Use positive k to shift the data forward in time (tomorrow’s data becomes today’s data). Use a negative k to shift the data backward in time (yesterday’s data becomes today’s data). Discussion Recall the five days of IBM stock prices from Recipe 14.1: > ibm.daily 2010-01-04 2010-01-05 2010-01-06 2010-01-07 2010-01-08 132.45 130.85 130.00 129.55 130.85 To shift the data forward one day, we use k=+1: > lag(ibm.daily, k=+1, na.pad=TRUE) 2010-01-04 2010-01-05 2010-01-06 2010-01-07 2010-01-08 130.85 130.00 129.55 130.85 NA We also set na.pad=TRUE to fill the trailing dates with NA. Otherwise, they would simply be dropped, resulting in a shortened time series. To shift the data backward one day, we use k=-1. Again we use na.pad=TRUE to pad the beginning with NAs: > lag(ibm.daily, k=-1, na.pad=TRUE) 2010-01-04 2010-01-05 2010-01-06 2010-01-07 2010-01-08 NA 132.45 130.85 130.00 129.55 If the sign convention for k seems odd to you, you are not alone. The function is called lag, but a positive k actually generates leading data, not lagging data. Use a negative k to get lagging data. Yes, this is bizarre. Perhaps the function should have been called lead. 14.8 Computing Successive Differences Problem Given a time series, x, you want to compute the difference between successive obser- vations: (x2 − x1), (x3 − x2), (x4 − x3), .... Solution Use the diff function: > diff(x) 14.8 Computing Successive Differences | 369 Discussion The diff function is generic, so it works on both simple vectors and zoo objects. The beauty of differencing a zoo object is that the result is also a zoo object and the differences have the correct dates. Here we compute the differences for successive prices of IBM stock: > ibm.daily 2010-01-04 2010-01-05 2010-01-06 2010-01-07 2010-01-08 132.45 130.85 130.00 129.55 130.85 > diff(ibm.daily) 2010-01-05 2010-01-06 2010-01-07 2010-01-08 -1.60 -0.85 -0.45 1.30 The difference labeled 2010-01-05 is the change from the previous day (2010-01-04), which is usually what you want. The differenced series is shorter than the original series by one element because R can’t compute the change as of 2010-01-04, of course. By default, diff computes successive differences. You can compute differences that are more widely spaced by using its lag parameter. Suppose you have monthly CPI data and want to compute the change from the previous 12 months, giving the year-over- year change. Specify a lag of 12: > head(cpi, 24) 1970-01-01 1970-02-01 1970-03-01 1970-04-01 1970-05-01 1970-06-01 1970-07-01 37.8 38.0 38.2 38.5 38.6 38.8 39.0 1970-08-01 1970-09-01 1970-10-01 1970-11-01 1970-12-01 1971-01-01 1971-02-01 39.0 39.2 39.4 39.6 39.8 39.8 39.9 1971-03-01 1971-04-01 1971-05-01 1971-06-01 1971-07-01 1971-08-01 1971-09-01 40.0 40.1 40.3 40.6 40.7 40.8 40.8 1971-10-01 1971-11-01 1971-12-01 40.9 40.9 41.1 > diff(cpi, lag=12) # Compute year-over-year change 1971-01-01 1971-02-01 1971-03-01 1971-04-01 1971-05-01 1971-06-01 1971-07-01 2.0 1.9 1.8 1.6 1.7 1.8 1.7 1971-08-01 1971-09-01 1971-10-01 1971-11-01 1971-12-01 1972-01-01 1972-02-01 1.8 1.6 1.5 1.3 1.3 1.3 1.4 . . (etc.) . 14.9 Performing Calculations on Time Series Problem You want to use arithmetic and common functions on time series data. 370 | Chapter 14: Time Series Analysis Solution No problem. R is pretty clever about operations on zoo objects. You can use arithmetic operators (+, −, *, /, etc.) as well as common functions (sqrt, log, etc) as well as usually get what you expect. Discussion When you perform arithmetic on zoo objects, R aligns the objects according to date so that the results make sense. Suppose we want to compute the percentage change in IBM stock. We need to divide the daily change by the price, but those two time series are not naturally aligned—they have different start times and different lengths: > ibm.daily 2010-01-04 2010-01-05 2010-01-06 2010-01-07 2010-01-08 132.45 130.85 130.00 129.55 130.85 > diff(ibm.daily) 2010-01-05 2010-01-06 2010-01-07 2010-01-08 -1.60 -0.85 -0.45 1.30 Fortunately, when we divide one series by the other, R aligns the series for us and returns a zoo object: > diff(ibm.daily) / ibm.daily 2010-01-05 2010-01-06 2010-01-07 2010-01-08 -0.012227742 -0.006538462 -0.003473562 0.009935040 We can scale the result by 100 to compute the percentage change, and the result is another a zoo object: > 100 * (diff(ibm.daily) / ibm.daily) 2010-01-05 2010-01-06 2010-01-07 2010-01-08 -1.2227742 -0.6538462 -0.3473562 0.9935040 Functions work just as well. If we compute the logarithm or square root of a zoo object, the result is a zoo object with the timestamps preserved: > log(ibm.daily) 2010-01-04 2010-01-05 2010-01-06 2010-01-07 2010-01-08 4.886205 4.874052 4.867534 4.864067 4.874052 In investment management, computing the difference of logarithms of prices is quite common. That’s a piece of cake in R: > diff(log(ibm.daily)) 2010-01-05 2010-01-06 2010-01-07 2010-01-08 -0.012153587 -0.006517179 -0.003467543 0.009984722 See Also See Recipe 14.8 for the special case of computing successive differences. 14.9 Performing Calculations on Time Series | 371 14.10 Computing a Moving Average Problem You want to compute the moving average of a time series. Solution Use the rollmean function of the zoo package to calculate the k-period moving average: > library(zoo) > ma <- rollmean(ts, k) Here ts is the time series data, captured in a zoo object, and k is the number of periods. For most financial applications, you want rollmean to calculate the mean using only historical data; that is, for each day, use only the data available that day. To do that, specify align=right. Otherwise, rollmean will “cheat” and use future data that was actually unavailable at the time: > ma <- rollmean(ts, k, align="right") Discussion Traders are fond of moving averages for smoothing out fluctuations in prices. The formal name is the rolling mean. You could calculate the rolling mean as described in Recipe 14.12, by combining the rollapply function and the mean function, but rollmean is much faster. Beside speed, the beauty of rollmean is that it returns a zoo object. For each element in the object, its date is the “as of” date for a calculated mean. Because it’s a zoo object, you can easily merge the original data and the moving average and then plot them together. The output is normally missing a few initial data points, since rollmean needs a full k observations to compute the mean. Consequently, the output is shorter than the input. If that’s a problem, specify na.pad=TRUE; then rollmean will pad the initial output with NA values. See Also See Recipe 14.12 for more about the align parameter. The moving average described here is a simple moving average. The quantmod, TTR, and fTrading packages contain functions for computing and plotting many kinds of moving averages, including simple ones. 372 | Chapter 14: Time Series Analysis 14.11 Applying a Function by Calendar Period Problem Given a time series, you want to group the contents by a calendar period (e.g., week, month, or year) and then apply a function to each group. Solution The xts package includes functions for processing a time series by day, week, month, quarter, or year: > apply.daily(ts, f) > apply.weekly(ts, f) > apply.monthly(ts, f) > apply.quarterly(ts, f) > apply.yearly(ts, f) Here ts is an xts time series and f is the function to apply to each day, week, month, quarter, or year. If your time series is a zoo object, convert to an xts object first so you can access these functions; for example: > apply.monthly(as.xts(ts), f) Discussion It is common to process time series data according to calendar period. But figuring calendar periods is tedious at best and bizarre at worst. Let these functions do the heavy lifting. Suppose we have a five-year history of IBM stock prices stored in a zoo object: > head(ibm) 1970-01-02 1970-01-05 1970-01-06 1970-01-07 1970-01-08 1970-01-09 364.75 368.25 368.50 368.75 369.50 369.00 We can calculate the average price by month if we use apply.monthly and mean together: > apply.monthly(as.xts(ibm), mean) [,1] 1970-01-30 359.8843 1970-02-27 346.8684 1970-03-31 327.4348 . . (etc.) . 1974-10-31 176.4791 1974-11-29 181.0570 1974-12-31 167.9233 14.11 Applying a Function by Calendar Period | 373 Notice that I first converted the zoo object to an xts object via as.xts(ibm). This would be unnecessary had I stored the data in an xts object in the first place. A more interesting application is calculating volatility by calendar month, where vol- atility is measured as the standard deviation of daily log-returns. Daily log-returns are calculated this way: diff(log(ibm)) We calculate their standard deviation, month by month, like this: apply.monthly(as.xts(diff(log(ibm))), sd) We can scale the daily number to estimate annualized volatility: sqrt(251) * apply.monthly(as.xts(diff(log(ibm))), sd) Plotting the result produces the chart shown in Figure 14-3, which was created by this code snippet: > plot(sqrt(251) * apply.monthly(as.xts(diff(log(ibm))), sd), + main="IBM: Monthly Volatility", + ylab="Std dev, annualized") Figure 14-3. IBM volatility by month 374 | Chapter 14: Time Series Analysis 14.12 Applying a Rolling Function Problem You want to apply a function to a time series in a rolling manner: calculate the function at a data point, move to the next data point, calculate the function at that point, move to the next data point, and so forth. Solution Use the rollapply function in the zoo package. The width parameter defines how many data points from the time series (ts) should be processed by the function (f) at each point: > library(zoo) > rollapply(ts, width, f) For many applications, you will likely set align="right" to avoid computing f with historical data that was unavailable at the time: > rollapply(ts, width, f, align="right") Discussion The rollapply function extracts a “window” of data from your time series, calls your function with that data, saves the result, and moves to the next window—and repeats this pattern for the entire input. As an illustration, consider calling rollapply with width=21: > rollapply(ts, 21, f) rollapply will repeatedly call the function, f, with a sliding window of data, like this: 1. f(ts[1:21]) 2. f(ts[2:22]) 3. f(ts[3:23]) 4. ... etc. ... Observe that the function should expect one argument, which is a vector of values. rollapply will save the returned values before packaging them into a zoo object along with a timestamp for every value. The choice of timestamp depends on the align pa- rameter given to rollapply: align="right" The timestamp is taken from the rightmost value. align="left" The timestamp is taken from the leftmost value. 14.12 Applying a Rolling Function | 375 align="center" (default) The timestamp is taken from the middle value. By default, rollapply will recalculate the function at successive data points. You may instead want to calculate the function at every nth data point. Use the by=n parameter to have rollapply move ahead n points after each function call. When I calculate the rolling standard deviation of a time series, for example, I usually want each window of data to be separate, not overlapping, and so I set the by value equal to the window size: > sds <- rollapply(ts, 30, sd, by=30, align="right") rollapply normally skips over the initial data points because it needs a full window of data before calling your function. The result is that the output is shorter than the input. You can force them to be the same length by setting na.pad=TRUE, in which case the output will contain NA values for that initial segment. 14.13 Plotting the Autocorrelation Function Problem You want to plot the autocorrelation function (ACF) of your time series. Solution Use the acf function: > acf(ts) Discussion The autocorrelation function is an important tool for revealing the interrelationships within a time series. It is a collection of correlations, ρk for k = 1, 2, 3, ..., where ρk is the correlation between all pairs of data points that are exactly k steps apart. Visualizing the autocorrelations is much more useful than listing them, so the acf function plots them for each value of k. Figure 14-4 shows the autocorrelation functions for two time series, one with autocorrelations and one without. The dashed line delimits the significant and insignificant correlations: values above the line are significant. (The height of the dashed line is determined by the amount of data.) The presence of autocorrelations is one indication that an autoregressive integrated moving average (ARIMA) model could model the time series. From the ACF, you can count the number of significant autocorrelations, which is a useful estimate of the number of moving average (MA) coefficients in the model. The top graph in Fig- ure 14-4 shows seven significant autocorrelations, for example, so we estimate that its ARIMA model will require seven MA coefficients (MA(7)). That estimate is just a start- ing point, however, and must be verified by fitting and diagnosing the model. 376 | Chapter 14: Time Series Analysis 14.14 Testing a Time Series for Autocorrelation Problem You want to test your time series for the presence of autocorrelations. Solution Use the Box.test function, which implements the Box–Pierce test for autocorrelation: > Box.test(ts) Figure 14-4. Two autocorrelation functions 14.14 Testing a Time Series for Autocorrelation | 377 The output includes a p-value. Conventionally, a p-value of less than 0.05 indicates that the data contains significant autocorrelations, whereas a p-value exceeding 0.05 provides no such evidence. Discussion Graphing the autocorrelation function is useful for digging into your data. Sometimes, however, we just need to know whether or not the data is autocorrelated. A statistical test such as the Box–Pierce test can provide an answer. We can apply the Box–Pierce test to the data whose autocorrelation function we plotted in Recipe 14.13. The test shows p-values for the two time series that are nearly zero and 0.862, respectively: > Box.test(ts1) Box-Pierce test data: ts1 X-squared = 96.9396, df = 1, p-value < 2.2e-16 > Box.test(ts2) Box-Pierce test data: ts2 X-squared = 0.0303, df = 1, p-value = 0.862 The p-value near zero indicates that the first time series has significant autocorrelations. (We don’t know which autocorrelations are significant; we just know they exist.) The p-value of 0.862 indicates that the test did not detect autocorrelations in second time series has none. The Box.test function can also perform the Ljung–Box test, which is better for small samples. That test calculates a p-value whose interpretation is the same as that for the Box–Pierce p-value: > Box.test(ts, type="Ljung-Box") See Also See Recipe 14.13 to plot the autocorrelation function, a visual autocorrelation check. 14.15 Plotting the Partial Autocorrelation Function Problem You want to plot the partial autocorrelation function (PACF) for your time series. 378 | Chapter 14: Time Series Analysis Solution Use the pacf function: > pacf(ts) Discussion The partial autocorrelation function is another tool for revealing the interrelationships in a time series. However, its interpretation is much less intuitive than that of the autocorrelation function. I’ll leave the mathematical definition of partial correlation to a textbook on statistics. Here, I’ll just say that the partial correlation between two random variables, X and Y, is the correlation that remains after accounting for the correlation shown by X and Y, with all other variables. In the case of time series, the partial autocorrelation at lag k is the correlation between all data points that are exactly k steps apart—after accounting for their correlation with the data between those k steps. The practical value of a PACF is that it helps you to identify the number of autoregres- sion (AR) coefficients in an ARIMA model. Figure 14-5 shows the PACF for two time series, one with partial autocorrelations and one without. Lag values whose lines cross above the dotted line are statistically significant. There are two such values, at k = 1 and k = 2, so our initial ARIMA model will have two AR coefficients (AR(2)). As with autocorrelation, however, that is just an initial estimate and must verified by fitting and diagnosing the model. See Also See Recipe 14.13. 14.16 Finding Lagged Correlations Between Two Time Series Problem You have two time series, and you are wondering if there is a lagged correlation between them. Solution Use the ccf function to plot the cross-correlation function, which will reveal lagged correlations: > ccf(ts1, ts2) 14.16 Finding Lagged Correlations Between Two Time Series | 379 Discussion The cross-correlation function helps you discover lagged correlations between two time series. A lagged correlation occurs when today’s value in one time series is correlated with a future or past value in the other time series. Consider the relationship between commodity prices and bond prices. Some analysts believe those prices are connected because changes in commodity prices are a barom- eter of inflation, one of the key factors in bond pricing. Can we discover a correlation between them? Figure 14-5. Two partial autocorrelation functions 380 | Chapter 14: Time Series Analysis Figure 14-6 shows a cross-correlation function generated from daily changes in bond prices and a commodity price index:† > ccf(bonds, cmdtys, main="Bonds vs. Commodities") Figure 14-6. Cross-correlation function Every vertical line shows the correlation between the two time series at some lag, as indicated along the x-axis. If a correlation extends above or below the dotted lines, it is statistically significant. Notice that the correlation at lag 0 is −0.231, which is the simple correlation between the variables: > cor(bonds, cmdtys) [1] -0.2309180 † Specifically: the bonds variable is the log-returns of the 10-year U.S. Treasury Note futures; and the cmdtys variable is the log-returns of the Rogers International Commodity Index. The data was taken from the period 2007-01-03 through 2009-12-31. 14.16 Finding Lagged Correlations Between Two Time Series | 381 Much more interesting are the correlations at ±1 lag, which are statistically significant. Evidently there is some “ripple effect” in the day-to-day prices of bonds and commod- ities because changes today are correlated with changes tomorrow. Discovering this sort of relationship is useful to short-term forecasters such as market analysts and bond traders. 14.17 Detrending a Time Series Problem Your time series data contains a trend that you want to remove. Solution Use linear regression to identify the trend component; then subtract the trend compo- nent from the original time series. These two lines show how to detrend the zoo object ts and put the result in detr: > m <- lm(coredata(ts) ~ index(ts)) > detr <- zoo(resid(m), index(ts)) Discussion Some time series data contain trends, which means that they gradually slope upward or downward over time. Suppose your time series object, ts, contains a trend as shown in the upper graph of Figure 14-7. We can remove the trend component in two easy steps. First, identify the overall trend by using the linear model function, lm. The model should use the time series index for the x variable and the time series data for the y variable: > m <- lm(coredata(ts) ~ index(ts)) Second, remove the linear trend from the original data by subtracting the straight line found by lm. This is easy because we have access to the linear model’s residuals, which are defined by the difference between the original data and the fitted line: ri = yi − β1xi − β0 where ri is the ith residual and where β1 and β0 are the model’s slope and intercept, respectively. We can extract the residuals from the linear model by using the resid function and then embed the residuals inside a zoo object: > detr <- zoo(resid(m), index(ts)) Notice that we use the same time index as the original data. When we plot detr it is clearly trendless, as evident in the lower graph of Figure 14-7. 382 | Chapter 14: Time Series Analysis 14.18 Fitting an ARIMA Model Problem You want to fit an ARIMA model to your time series data. Solution The auto.arima function in the forecast package can select the correct model order and fit the model to your data: > library(forecast) > auto.arima(x) Figure 14-7. Detrending a time series 14.18 Fitting an ARIMA Model | 383 D o wnload from Wow! eBook If you already know the model order, (p, d, q), then the arima function can fit the model directly: > arima(x, order=c(p,d,q)) Discussion Creating an ARIMA model involves three steps: 1. Identify the model order. 2. Fit the model to the data, giving the coefficients. 3. Apply diagnostic measures to validate the model. The model order is usually denoted by three integers, (p, d, q), where p is the number of autoregressive coefficients; d is the degree of differencing; and q is the number of moving average coefficients. When I build an ARIMA model, I am usually clueless about the appropriate order. Rather than tediously searching for the best combination of p, d, and q, I use auto.arima, which does the searching for me: > auto.arima(x) Series: x ARIMA(2,1,2) Call: auto.arima(x = x) Coefficients: ar1 ar2 ma1 ma2 0.0451 -0.9183 -0.0066 0.6178 s.e. 0.0249 0.0252 0.0496 0.0517 sigma^2 estimated as 0.9877: log likelihood = -708.45 AIC = 1426.9 AICc = 1427.02 BIC = 1447.98 In this case, auto.arima decided the best order was (2, 1, 2), which means that it dif- ferenced the data once (d = 1) before selecting a model with two AR coefficients (p = 2) and two MA coefficients (q = 2). By default, auto.arima limits p and q to the range 0 ≤ p ≤ 5 and 0 ≤ q ≤ 5. If you are confident that your model needs fewer than five coefficients, use the max.p and max.q parameters to limit the search further; this makes it faster. Likewise, if you believe that your model needs more coefficients, use max.p and max.q to expand the search limits. If you already know the order of your ARIMA model, the arima function can quickly fit the model to your data: > arima(x, order=c(2,1,2)) Series: x ARIMA(2,1,2) Call: arima(x = x, order = c(2, 1, 2)) 384 | Chapter 14: Time Series Analysis Coefficients: ar1 ar2 ma1 ma2 0.0451 -0.9183 -0.0066 0.6178 s.e. 0.0249 0.0252 0.0496 0.0517 sigma^2 estimated as 0.9877: log likelihood = -708.45 AIC = 1426.9 AICc = 1427.02 BIC = 1447.98 The output looks identical to auto.arima. What you can’t see here is that arima executes much more quickly. The output from auto.arima and arima includes the fitted coefficients and the standard error (s.e.) for each coefficient: Coefficients: ar1 ar2 ma1 ma2 0.0451 -0.9183 -0.0066 0.6178 s.e. 0.0249 0.0252 0.0496 0.0517 You can find the coefficients’ confidence intervals by capturing the ARIMA model in an object and then using the confint function: > m <- arima(x, order=c(2,1,2)) > confint(m) 2.5 % 97.5 % ar1 -0.003811699 0.09396993 ar2 -0.967741073 -0.86891013 ma1 -0.103861019 0.09059546 ma2 0.516499037 0.71903424 This output illustrates a major headache of ARIMA modeling: not all the coefficients are necessarily significant. The ar1 coefficient has a confidence interval of (−0.0038, +0.0940). Since the interval contains zero, the true coefficient might be zero itself, in which case the whole ar1 term is unnecessary. The same is true for the ma1 coefficient, whose confidence interval is (−0.0066, +0.0496). The ar2 and ma2 coefficients are likely significant, however, since their confidence intervals do not contain zero. If you discover that your model contains insignificant coefficients, use Recipe 14.19 to remove them. The auto.arima and arima functions contain useful features for fitting the best model. For example, you can force them to include or exclude a trend component. See the help pages for details. A final caveat: the danger of auto.arima is that it makes ARIMA modeling look simple. ARIMA modeling is not simple. It is more art than science, and the automatically gen- erated model is just a starting point. I urge the reader to review a good book about ARIMA modeling before settling on a final model. See Also See Recipe 14.20 for performing diagnostic tests on the ARIMA model. 14.18 Fitting an ARIMA Model | 385 14.19 Removing Insignificant ARIMA Coefficients Problem One or more of the coefficients in your ARIMA model are statistically insignificant. You want to remove them. Solution The arima function includes the parameter fixed, which is a vector. The vector should contain one element for every coefficient in the model, including a term for the drift (if any). Each element is either NA or 0. Use NA for the coefficients to be kept and use 0 for the coefficients to be removed. This example shows an ARIMA(2, 1, 2) model with the first AR coefficient and the first MA coefficient forced to be 0: > arima(x, order=c(2,1,2), fixed=c(0,NA,0,NA)) Discussion The example in Recipe 14.18 contained two insignificant coefficients, ar1 and ma1. We decided they were insignificant by looking at the confidence intervals: > m <- arima(x, order=c(2,1,2)) > confint(m) 2.5 % 97.5 % ar1 -0.003811699 0.09396993 ar2 -0.967741073 -0.86891013 ma1 -0.103861019 0.09059546 ma2 0.516499037 0.71903424 Since the confidence intervals for ar1 and ma1 contain zero, the true coefficients could be zero. Those are the first and third coefficients, so we can remove them by letting fixed be a vector whose first and third elements are zero: > m <- arima(x, order=c(2,1,2), fixed=c(0,NA,0,NA)) > m Series: x ARIMA(2,1,2) Call: arima(x = x, order = c(2, 1, 2), fixed = c(0, NA, 0, NA)) Coefficients: ar1 ar2 ma1 ma2 0 -0.9082 0 0.5931 s.e. 0 0.0268 0 0.0522 sigma^2 estimated as 0.999: log likelihood = -711.27 AIC = 1428.54 AICc = 1428.66 BIC = 1449.62 Warning message: In arima(x, order = c(2, 1, 2), fixed = c(0, NA, 0, NA)) : some AR parameters were fixed: setting transform.pars = FALSE 386 | Chapter 14: Time Series Analysis The warning message from arima is harmless. The default value for the transform.pars parameter is TRUE, but arima forces it to FALSE, if necessary, and warns you. Observe that the ar1 and ma1 coefficients are now zero. The remaining coefficients (ar2 and ma2) are still significant, as shown by their confidence intervals, so we have a reasonable model: > confint(m) 2.5 % 97.5 % ar1 NA NA ar2 -0.9606839 -0.8557283 ma1 NA NA ma2 0.4907792 0.6954606 14.20 Running Diagnostics on an ARIMA Model Problem You have built an ARIMA model, and you want to run diagnostic tests to validate the model. Solution Use the tsdiag function. This example saves the ARIMA model in m and then runs diagnostics on the model: > m <- arima(x, order=c(2,1,2), fixed=c(0,NA,0,NA), transform.pars=FALSE) > tsdiag(m) Discussion The result of tsdiag is a set of three graphs. A good model should produce results like the ones shown in Figure 14-8. Here’s what’s good about the graphs: • The standardized residuals don’t show clusters of volatility. • The autocorrelation function (ACF) shows no significant autocorrelation between the residuals. • The p-values for the Ljung–Box statistics are all large, indicating that the residuals are patternless. We have extracted the information and left behind only “noise.” For contrast, Figure 14-9 shows diagnostic charts with problems: • The ACF shows significant autocorrelations between residuals. • The p-values for the Ljung–Box statistics are small, indicating there is some pattern in the residuals. There is still information to be extracted from the data. 14.20 Running Diagnostics on an ARIMA Model | 387 These are basic diagnostics, but they are a good start. Find a good book on ARIMA modeling and perform the recommended diagnostic tests before concluding that your model is sound. Additional checks of the residuals could include: • Tests for normality • Quantile-quantile (Q-Q) plot • Histogram • Scatterplot against the fitted values • Time-series plot Figure 14-8. Good time series diagnostics 388 | Chapter 14: Time Series Analysis 14.21 Making Forecasts from an ARIMA Model Problem You have an ARIMA model for your time series. You want to forecast the next few observations in the series. Solution Save the model in an object; then apply the predict function to the object. This example saves the model from Recipe 14.19 and predicts the next observation: Figure 14-9. Bad time series diagnostics 14.21 Making Forecasts from an ARIMA Model | 389 > m <- arima(x, order=c(2,1,2), fixed=c(0,NA,0,NA), transform.pars=FALSE) > predict(m) Discussion The predict function will calculate both the next observation and its standard error according to the model. It returns a list with two elements: pred, which is the prediction itself; and se, which is the standard error. Continuing the example from the Solution, the predicted next value is −8.545 with a standard error of 0.9995: > predict(m) $pred Time Series: Start = 503 End = 503 Frequency = 1 [1] -8.545194 $se Time Series: Start = 503 End = 503 Frequency = 1 [1] 0.9995073 (Both pred and se are encoded as time series, so R prints their time series attributes in addition to their numeric values.) To predict more than one observation into the future, use the n.ahead parameter: > predict(m, n.ahead=10) $pred Time Series: Start = 503 End = 512 Frequency = 1 [1] -8.545194 -9.365685 -9.365838 -8.620663 -8.620524 -9.297297 -9.297423 [8] -8.682773 -8.682659 -9.240887 $se Time Series: Start = 503 End = 512 Frequency = 1 [1] 0.9995073 1.4135168 1.5705650 1.7132774 1.9691066 2.1953229 2.3075363 [8] 2.4145404 2.5935203 2.7609219 As before, the output includes two time series: one of predictions and one of standard errors. Notice that the standard errors grow larger with each additional prediction be- cause the predictions become increasingly more uncertain. 390 | Chapter 14: Time Series Analysis 14.22 Testing for Mean Reversion Problem You want to know if your time series is mean reverting. Solution A common test for mean reversion is the Augmented Dickey−Fuller (ADF) test, which is implemented by the adf.test function of the tseries package: > library(tseries) > adf.test(coredata(ts)) The output from adf.test includes a p-value. Conventionally, if p < 0.05 then the time series is likely mean reverting whereas a p > 0.05 provides no such evidence. Discussion When a time series is mean reverting, it tends to return to its long-run average. It may wander off, but eventually it wanders back. If a time series is not mean reverting, then it can wander away without ever returning to the mean. The upper time series in Figure 14-10 appears to be wandering downward and not returning. The large p-value from the adf.test confirms that it is not mean reverting: > adf.test(coredata(ts1)) Augmented Dickey-Fuller Test data: ts1 Dickey-Fuller = -2.1217, Lag order = 6, p-value = 0.5246 alternative hypothesis: stationary The lower time series, however, is just bouncing around its average value. The small p-value (0.01) confirms that it is mean reverting: > adf.test(coredata(ts2)) Augmented Dickey-Fuller Test data: ts2 Dickey-Fuller = -7.1752, Lag order = 6, p-value = 0.01 alternative hypothesis: stationary The adf.test function massages your data before performing the ADF test. First it automatically detrends your data; then it recenters the data, giving it a mean of zero. 14.22 Testing for Mean Reversion | 391 If either detrending or recentering is undesirable for your application, use the adfTest function in the fUnitRoots package instead:‡ > library(fUnitRoots) > adfTest(coredata(ts1), type="nc") With type="nc", the function neither detrends nor recenters your data. With type="c", the function recenters your data but does not detrend it. Both the adf.test and the adfTest functions let you specify a lag value that controls the exact statistic they calculate. These functions provide reasonable defaults, but any serious user should study the textbook description of the ADF test to determine the appropriate lag for her application. Figure 14-10. Two time series: wandering and mean reverting ‡ Make careful note of the difference in spelling between adf.test and adfTest. This sort of subtle difference is unfortunately common in R and occurs when several packages implement the same function but with some slight variation from package to package. 392 | Chapter 14: Time Series Analysis See Also The urca and CADFtest packages also implement tests for a unit root, which is the test for mean reversion. Be careful when comparing the tests from several packages. Each package can make slightly different assumptions, which can lead to puzzling differences in the results. 14.23 Smoothing a Time Series Problem You have a noisy time series. You want to smooth the data to eliminate the noise. Solution The KernSmooth package contains functions for smoothing. Use the dpill function to select an initial bandwidth parameter; then use the locploy function to smooth the data: library(KernSmooth) gridsize <- length(y) bw <- dpill(t, y, gridsize=gridsize) lp <- locpoly(x=t, y=y, bandwidth=bw, gridsize=gridsize) smooth <- lp$y Here, t is the time variable and y is the time series. Discussion The KernSmooth package is a standard part of the R distribution. It includes the locpoly function; this function constructs, around each data point, a polynomial that is fitted to the near-by data points. These are called local polynomials. The local poly- nomials are strung together to create a smoothed version of the original data series. The algorithm requires a bandwidth parameter to control the degree of smoothing. A small bandwidth means less smoothing, in which case the result follows the original data more closely. A large bandwidth means more smoothing, so the result contains less noise. The tricky part is choosing just the right bandwidth: not too small, not too large. Fortunately, KernSmooth also includes the function dpill for estimating the appropriate bandwidth, and it works quite well. I recommend that you start with the dpill value and then experiment with values above and below that starting point. There is no magic formula here. You need to decide what level of smoothing works best in your application. 14.23 Smoothing a Time Series | 393 Figure 14-11 shows an example of smoothing. The solid line is a time series that is the sum of a simple sine wave and normally distributed “noise”: t <- seq(from=-10,to=10,length.out=201) noise <- rnorm(201) y <- sin(t) + noise Figure 14-11. Smoothing a time series Both dpill and locpoly require a grid size—in other words, number of points for which a local polynomial is constructed. I often use a grid size equal to the number of data points, which yields a fine resolution. The resulting time series is very smooth. You might use a smaller grid size if you want a coarser resolution or if you have a very large dataset: library(KernSmooth) gridsize <- length(y) bw <- dpill(t, y, gridsize=gridsize) The locpoly function performs the smoothing and returns a list. The y element of that list is the smoothed data: lp <- locpoly(x=t, y=y, bandwidth=bw, gridsize=gridsize) smooth <- lp$y 394 | Chapter 14: Time Series Analysis In Figure 14-11, the smoothed data is shown as a dashed line. The figure demonstrates that locpoly did an excellent job of extracting the original sine wave. See Also The ksmooth, lowess, and HoltWinters functions in the base distribution can also per- form smoothing. The expsmooth package implements exponential smoothing. 14.23 Smoothing a Time Series | 395 Index Symbols ! (logical negation), 40 != (comparison operator), 34, 40 # (comment), 80 #! (Linux/Unix self-contained script), 65 $ (component extraction), 40 %% (modulo operator), 41, 319 %*% (matrix multiplication), 41, 120 %...% (binary operators), 40 %/% (integer division), 41 %in% (contained in operator), 41 & (logical and), 40, 48 && (short-circuit and), 40, 48 () (print assigned value), 315 * (multiplication operator), 40, 120, 280, 286 *, **, *** (significance codes), 277, 297 + (continuation prompt), 47 + (unary plus, addition operator), 40, 286 - (unary minus, subtraction operator), 40, 286 --help option, 6 --no-init option, 64 --no-restore option, 64 --no-save option, 64 --quiet option, 62, 64 --slave option, 64 ->, ->> (rightwards assignment operators), 26 . 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(search shortcut), 13 @ (slot extraction), 40 [ ] (vector indexing), 7, 35, 40 [[ ]] (list indexing), 40, 110 \ (backslash), 46, 76, 168 \n (newline character), 24 ^ (exponentiation, interaction term), 40, 285, 287 | (logical or), 40, 48 || (short-circuit or), 40, 48 ~ (formula), 40, 233, 248, 269, 303 A abline() function, 222, 231, 245, 255 abs() function, 117 abstract type, 97 access variables in name space (::, :::), 40 accessing built-in datasets, 57 data frame column contents, 141 functions in a package, 55 acf() function, 299, 376 active window, choosing, 262 ad.test() function, 210 adding We’d like to hear your suggestions for improving our indexes. Send email to index@oreilly.com. 397 color/shading to bar charts/plots, 239, 256 confidence intervals to bar charts, 237 density estimates to histograms, 250 grids to plots, 226 legends to plots, 229 title/labels to plots, 225 vertical/horizontal lines to plots, 245 addition operator (+), 40 ADF (Augmented Dickey–Fuller test) adf.test() function, 330, 391 Adler, Joseph, xv all() function, 35 alternative hypothesis, 195 An Introduction to R (Venables), xv Anderson–Darling test, 210 annotations, inhibiting, 225 anonymous functions, 42 ANOVA anova() function, 268, 272–275, 309 aov() function, 107, 303, 305 comparing models using, 309 differences between means of groups, 304 interaction between predictors, 303 kruskal.test() function, 268, 308 oneway.test() function, 268, 302 robust, 308 any() function, 35 aov() function, 107, 303, 305 appending append() function, 102, 103 data to vectors, 101 lines to a file, 83 rows to data frames, 125 Applied Linear Regression Models (Kutner, 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function (ACF) plotting for regression residuals, 299 plotting for time series, 376 B backslash (\), 46, 76, 168 Backspace key, 8 backwards stepwise regression, 281, 283 bandwidth, 393 bar charts adding color or shading to, 239 adding confidence intervals to, 237 barplot() function, 236–240 barplot2() function, 237 creating, 236 batch scripts running, 63–65 writing a plot file, 264 Bernoulli trials, 185 beta distributions, 178, 260 binary operators %...%, 40, 332 meanings inside regression formula, 286 user-defined, 332 binary values, 185, 210, 216, 345 binary-formatted data, 92 binning data, 249, 317 binomial coefficients, 179 binomial distributions, 177, 182, 186, 190 Boolean values, 34 boot package, 347 bootstrap procedures boot.ci() function, 347, 348 bootfun() function, 347 for median confidence interval, 207 replication, 347 sample, 347 sampling with replacement, 184 box plots boxplot() function, 222, 246 creating for each factor level, 247 side-by-side, 344 Box.test() function 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126 chi-squared distributions, 178, 188, 190 chisq.test() function, 201 choose() function, 179 chron package, 162 classes abstract type, 97, 153, 326 Index | 399 D o wnload from Wow! eBook class() function, 326 date and time, 161 object, 356 POSIX, 162 unclass() function, 326 clusters, 342 CMD BATCH subcommand, 63 coefficients coef() function, 272 coefficients() function, 272 coefplot() function, 293 determining significance, 268 from regression summary report, 276, 292 of variation, 42 collapse separator, 164 colnames attribute, 134 colon : (multiplication in formula), 280 : (sequence creation operator), 33, 40 ::, ::: (access variables in name space), 40 colors() function, 258 colSums() function, 315 columns applying functions to, 152 data frame, 122, 127–133, 137, 140, 141, 152, 270, 315 header in data files, 80, 81, 83 matrix, 152 width of, 77 combinations, counting, 179 combining combn() function, 180 data frames, 138 multiple vectors into vector and factor, 107 command history, viewing, 53 command line editing, 7 installing packages using, 59 prompt (>), 7 comment lines in data files, 80 comparing locations of two samples, 213 means of two samples, 212 two vectors, 34 comparison operators (==, !=, <, >, <=, >=), 34, 40, 47 complex atomic type, 143 component extraction ($), 40 Comprehensive R Archive Network (see CRAN) concatenating printing objects, 24, 111, 115 strings, 163 conditional execution, 43 conditional tests, 117 conditioning plots, 233 confidence intervals adding to bar charts, 237 using bootstrap algorithm, 347 confint() function, 157, 272, 292 for a mean, 205 for a median, 206 for a proportion, 207 for regression coefficients, 292 connections, 87, 89 contingency tables, 200 continuation prompt (+), 47 continuous distributions, 178, 190 converting atomic values, 143 data frame to vectors, 145 data to z-scores, 203 date to string, 171 list to vector, 115, 145 mixed-mode matrix to data frame, 146 probabilities to quantiles, 189 string to date, 170 structured data types, 144 year, month, day into date, 172 coplot() function, 233 coredata() function, 358 correlations cor() function, 30–32, 153 cor.test() function, 215 testing for significance, 215 cov() function (covariance), 30–32 Cramer–von Mises test, 210 CRAN (Comprehensive R Archive Network), 18 (see also task views) crantastic.org, 18 downloading R from, 2–4 finding functions/packages on, 18 installing packages from, 59 R Language Definition, xvi server and mirror sites, 60–62 creating 400 | Index bar charts, 236 box plots, 246 box plots for each factor level, 247 contingency tables, 200 interaction plot, 303 lists, 108 matrices, 118 non-normal Q-Q plots, 254 normal Q-Q plots, 252 scatter plots, 223, 227 scatter plots for each factor level, 233 sequences, 32, 175 series of repeated values, 33 vectors, 28 cross-correlation, 379 cross-tabulations, 200 CSV files, 80, 82 Ctrl key combinations, 8 cumsum() function, 137 cumulative distribution function, 188 cumulative probability function, 187 curve() function, 222, 258 customizing R, 68–70 cut() function, 317 cutree() function, 342 cvm.test() function, 210 D data appending to vectors, 101 binning, 317 calculating quantiles of, 201 column, initializing data frame from, 122 complex or irregularly structured (in data files), 86–89 data() function, 57 displaying partial, 313, 361 entering from keyboard, 72 inserting into vectors, 103 non-numeric, 81 normalizing, 203 organized by observation, 123 organized by variable, 123 reading from the Web, 83–86 row, initializing data frame from, 123 self-describing (in data files), 80 summarizing, 197 transforming into linear relationship, 287 data editor, 135 data frames accessing column contents, 141 appending rows to, 125 applying functions to columns of, 152 applying functions to groups of rows, 156 with basic statistics functions, 31 changing column names, 133 combining, 138 converted to lists, 100, 148 editing, 135 entering from keyboard, 72 excluding columns by name, 137 flattening, 322 initializing from a list, 122 initializing from column data, 122 initializing from row data, 123 interpretation by functions, 49 merging by common column, 140 normalized values, 203 populating from built-in editor, 72 preallocating, 126 regressions between columns of, 270 removing NAs from, 136 scatter plots from, 224 selecting columns by name, 131 selecting columns by position, 127–130 selecting rows/columns with subset(), 132 sorting, 323 structure of, 100 summaries of, 198 summing rows and columns, 315 temporary copies, 142 data structures, technical aspects of, 95 datasets accessing built-in, 57 dealing with mixed data from, 124 plotting multiple, 243 dates American-style, 170 calendar periods, 161, 362, 373 converting from string, 170 converting into string, 171 converting year, month, day into, 172 creating sequence of, 175 Date class, 161 versus datetimes, 356 extracting year, month, or day from, 174 getting current, 169 dbConnect() function, 89 Index | 401 dbDisconnect() function, 89 dbGetQuery() function, 89, 90 dbinom() function, 186 Debian, installing R on, 3 defaultPackages list, 69 deleting Delete key, 8 variables, 27 dendrograms, 342 density functions for probability distributions, 186, 190 density() function, 250 desktop icon, creating, 5 detach() function, 56, 141 detrending a time series, 382, 392 dev.set() function, 262 deviance() function, 272 dfrm list operators, 127 dgamma() function, 191 dgeom() function, 187 diagnostic checks for regression, 293 dictionary, 96 diff() function, 187, 369 directories home, 67 working, 5, 51 discrete distributions, 177, 186, 190 discrete histograms, 252 dist() function, 342 distribution function, 187 distribution functions for probability distributions, 186, 188 division operator (/), 40 dnorm() function, 177, 191, 258 do.call() function, 124, 331, 332 documentation packages, 3 supplied, 1, 10, 13 double atomic type, 143 down arrow, 8 downloading and installing R, 2–4 dpill() function, 393 dpois() function, 187 dput() function, 92 dump() function, 21, 92 dumping variable structure, 327 dwtest() function (Durbin–Watson test), 299 dynamically typed language, 25 E edit() function, 135 editor window, 43, 135 effects() function, 272 eigen() function, 338 eigenvalues and eigenvectors, 338 elapsed time, 329 End key, 8 environment variables, 66 equals sign = (alternate assignment operator), 26 == (comparison operator), 34, 40, 47 errors “cannot open the connection”, 76 common syntax, 46–49 “could not find function”, 48, 56 “has not been called yet”, 222 “not in a standard unambiguous format”, 170 “object . . . not found”, 47 suppressing messages, 329 escape character (\), 46, 76, 168 Euclid’s algorithm, 42 example() function, 12 Excel worksheet, data frame as, 101 excluding data frame columns by name, 137 expand.grid() function, 321 exponential distributions, 178, 188, 190, 255 exponential smoothing, 395 exponentiation (^), 40, 286, 287 expsmooth package, 395 extracting Julian date, 173 oldest/newest observations, 361 parts of a date, 174 substrings, 164 F F distributions, 178 F statistic, 267, 268, 277 factor analysis/factanal() function, 349 factors categorical variables as, 105 creating, 105 creating box plot for each level, 247 creating scatter plot for each level, 233 defining groups via, 147 factor() function, 105, 126 402 | Index generating all combinations of several, 321 levels of, 100, 106 non-numeric data in data files as, 81 summaries of, 198 tabulating, 200 testing categorical variables for independence, 201 faraway package, 304 Faraway, Julian, 269, 345 Fedora, installing R on, 3 files, listing, 75 filling a time series, 366 finding best power transformation, 289 data clusters, 342 differences between means of groups, 304 lagged correlations between time series, 379 pairwise minimums or maximums, 320 position of a value, 318 relevant functions and packages, 18 and removing NULL elements, 116 .First(), .First.sys() functions, 70 first() function, 362 fitted() function, 272 fitting the model, 267, 294, 383 fix() function, 135 fixed-width records (in data files), 77 flattening data frames, 322 list into vector, 115 forecast package, 383 format() function, 73, 171 forming, prediction intervals, 301 forward stepwise regression, 281 full model, 281 function keyword, 41 functions anonymous, 42 applying by calendar period, 373 applying rolling, 375 applying to groups of data, 154 applying to list elements, 149 applying to matrix columns, 152 applying to matrix rows, 151 arguments to, 12, 331 defining, 41 examples/demonstrations of, 12 finding on CRAN, 18 getting help on, 11–13 graphing, 258 searching packages for, 13 fUnitRoots package, 392 G gamma distributions, 178, 188, 190 generating all combinations of several factors, 321 combinations, 180 random numbers, 180 random permutations of vectors, 185 random samples, 183 random sequences, 184 reproducible random numbers, 182 generic functions (methods), 223 geometric distributions, 177, 187, 190 getting current date, 169 environment variables, 66 Julian date, 173 length of a string, 163 regression statistics, 272 working directory, 51 ggplot2 package, 223 ggplot2: Elegant Graphics for Data Analysis (Wickham), xv, 223 glm() function, 345 global assignment operator (<<-), 26, 43 global graphics parameters, 264 global variables, 43 Golden Ratio graph dimensions, 263 Goodman, Sasha, xv gplots library, 237 Grammar of Graphics paradigm, 223 graphics functions changing software parameters, 264 low-level versus high-level, 222 graphics package, 221 graphics window, 260, 262 graphing a function, 258 gray() function, 239 grid package, 262 grid() function, 226 Grothendieck, Gabor, 162 grouping factors, 100, 147, 154–158 groups multiple, creating scatter plots of, 227 pairwise comparisons between means, 218 Index | 403 testing for equal proportions, 216 gsub() function, 166 H hashes, 96 hclust() function, 342 head() function, 313, 361 header lines, 80, 81, 83, 120 help ? shortcut, 12, 40 for functions, 11–13 help() function, 12–13, 57 help.search() function, 13 help.start() function, 10 on packages, 14 resources, 1 websites, 16–18 heterogeneity of lists, 96, 99 hidden files, 76 high-level graphics functions, 222 histograms hist() function, 222, 248 testing for normality, 210 history() function, 53 Home key, 8 honest significant differences (Tukey), 305 Hornik, Kurt, 163 HTML tables, reading data from, 84 hypergeometric distributions, 177 I I() operator, 285 if statements, 48 independence, testing for, 196, 201 indexing by position, 363 date vectors, 363 index() function, 358 to select rows/columns from data frame, 132 vectors, 35–38, 357 influence.measures() function, 297 influential observations, 296 information loss, avoiding, 318 initializing data frames from column data, 122 matrices, 118 insignificant coefficients, removing, 386 install.packages() function, 59 installed.packages() function, 58 installing R, 2–4 integers as.integer() function, 143 compute greatest common divisor of, 42 converting from/to atomic type, 143 division (%/%), 41 interaction term (^), 285, 287 interaction.plot() function, 304 intercept, computing, 340 interquartile range, 247 interval probability, 187 inverting quantiles, 202 is.null() function, 116 ISOdate() function, 172 J join operation, 141 JPEG files, 263 julian() function (Julian date), 173 K KernSmooth package, 393 keyboard, entering data from, 72 keywords, documentation help for, 10 Kolmogorov-Smirnov test, 219 kruskal.test() function (Kruskal–Wallis Test), 268, 308 ks.test() function, 219 ksmooth() function, 395 Kutner, Michael, 269 L labels, adding to plot, 225 lag() function, 368 lagged correlations, 379 lagging a time series, 368 lapply() function, 117, 149, 152 last() function, 362 .Last.value variable, 53 lattice package, 223, 262 Lattice: Multivariate Data Visualization with R (Sarkar), xv, 223 layout() function, 261 leading data, 369 left arrow, 8 404 | Index leftwards assignment operators (<-, <<-), 25, 40, 46 legend() function, 229 length() function, 150, 155, 163 level of a factor, 100, 106, 147 library() function, 48, 56, 58 lillie.test() function (Lilliefors test), 210 linear regression and ANOVA, 268 diagnosing, 293 getting statistics from, 272 multiple, 270 overview, 267 simple, 269 understanding the summary, 275 with interaction terms, 279 without an intercept, 278 lines adding to histograms, 250 adding to plots, 245 lines() function, 222, 226, 250 plot() function, 241 type, width, color of, 242, 256 Linux/Unix design philosophy used in R, 71 environment variables, 66 exiting from R, 9 installing packages, 60 installing/building R on, 3 interrupting R, 9 location of .Rprofile, 68 root privileges, 60 self-contained script (#!), 65 starting R in, 4, 6 X Windows system, 66 list.files() function, 76 listing variables/functions, 26 lists applying functions to, 149 arguments to, 331 creating, 108 data frames as, 100, 148 flattening into vector, 115, 145 list() function, 108, 112 matrices made from, 99 and modes, 97 name/value association, 112 populating, 109 properties of, 96 removing elements from, 114, 116, 117 selecting elements by name, 111 selecting elements by position, 109 summaries of, 198 Ljung–Box test, 378, 387 lm() function and model object, 273 fitting the data, 294 getting regression statistics from, 272 multiple linear regression with, 270 regressing on data subset, 284 regressing on transformed data, 287 simple linear regression with, 269 lmtest package, 299 load() function, 92 loading (factor analysis), 349 local polynomials, 393 local variables, 43 locf (last observation carried forward), 366 locploy() function, 393 log() function, 39 log-likelihood, 290 log-normal distributions, 178 logical and (&), 40, 48 logical atomic type, 143 logical negation (!), 40 logical or (|), 40, 48 logical values, 34 logical vectors, 117 logistic distributions, 178 logistic regression, 345 lookup tables, 96 loops, 43, 147 low-correlation variables, removing, 153 low-level graphics function, 222 lower.tri() function, 169 lowess() function, 395 ls() function, 26, 28 ls.str() function, 26 lubridate package, 162 M MA (moving average), 376 mailing lists, R, 1, 19–21, 92 Map() function, 125 mapply() function, 158 MASS package, 57, 179 match() function, 318 Index | 405 Mathematical Statistics with Applications (Wackerly), 197 matrices applying functions to, 151 dim attribute and, 325, 326 graphics window, 260 initializing, 118 matrix() function, 118 multiplication (%*%), 41, 120 naming rows/columns, 120 normalized values, 203 operations, 119 reshaped from vectors, 98, 119 selecting rows/columns, 121 summaries of, 198 summing rows/columns, 315 max() function (maximum values), 48, 318, 320 maximizing multiparameter function, 336 single-parameter function, 335 means forming confidence interval for, 205 group, finding differences between, 304 group, pairwise comparisons between, 218 mean() function, 30–32, 48, 151, 156, 199, 202, 322 reversion to, 391 of two samples, comparing, 212 testing for zero, 205 median confidence interval for, 206 median() function, 30–32, 49 memory management, 102 merge() function, 140, 364, 366 merging data frames by common column, 140 methods, 97 min() function, 48, 320 minimal self-contained examples, 20 minimizing multiparameter function, 336 single-parameter function, 335 minimum values, 318, 320 mirror sites, CRAN, 60 mode() function, 96 model formula, 269 model objects abline() function, 231 ANOVA comparisons of, 309 diagnostic plots from, 295 glm() function, 345 identifying influential observations from, 296 lm() function, 267, 273 from residuals, 293 TukeyHSD() function and, 305 Modern Applied Statistics with S (Venables and Ripley), xv modular tools, 71 modulo operator (%%), 41, 319 Moen, Rick, 21 mondate package, 162 mouse, 8 moving average (MA), 376 multifigure plots, 260 multilevel list structures, 331 multiparameter functions, 336 multiple linear regression, 270 multiplication operator (*), 40, 120, 280 multiplication operator in model formula (:), 280 multiway ANOVA, 303 Murrell, Paul, xv, 221 my.cnf file, 90 MySQL client software, 90 N NA values filling in, 367 initializing matrix with, 118 na.locf() function, 365 na.omit() function, 136 na.strings parameter, 79 removing from data frames, 136 replacing, 365 in vectors, 31, 102 Nachtsheim, Christopher, 269 name/value association lists, 112 nchar() function, 163 negative binomial distributions, 178 negative indexes, 36 Neter, John, 269 newline character (\n), 24 noise, eliminating, 393 nonparametric statistics, 214 normal density, 177 normal distributions functions, 188 406 | Index getting help for, 178 names of, 177 qnorm() function, 190 testing for, 209, 252 normal quantiles, 177, 189 normalizing data, 203 null hypothesis, 195, 210, 308 NULL value, 114, 116, 325 numeric atomic type, 144 numeric() function, 126 O objects class of, 97, 355 mode of, 96 saving and transporting, 92 structure of, 326 observations, 123 computing successive differences between, 369 extracting oldest/newest, 313, 361 filling in dates/times for missing, 366 forecasting with ARIMA, 389 identifying influential, 296 paired, 212, 337 oneway.test() function (one-way ANOVA), 268, 302 operator precedence, 40, 333 operators, binary, 332 optim() function, 336 optimize() function, 335 options() function, 69, 74, 314 order() function, 323, 324 orthogonal regression, 340 OS X exiting from R, 9 installing packages, 59 installing R, 2 interrupting R, 9 location of .Rprofile, 68 saving graphics files, 263 starting R, 4, 6 outer() function, 168 outlier.test() function, 294, 295 output, redirecting to file, 74 overwriting errors involving plots, 259 P p-values, 195, 201 p.adjust() function, 218 pacf() function (partial autocorrelation function), 378 packages accessing functions in, 55 attaching to search list, 58 datasets and, 57 for dates and times, 162 documentation for, 1, 3 graphical managers, 3 graphics packages, 221, 223 help on, 14 installing and loading, 2, 48, 56, 59, 69 list of, 10, 19 preloading, 69 and search path, 54 unloading, 56 viewing list of installed, 58 padding a time series, 366 paired observations, 212 pairwise comparison between group means, 218 pairwise.t.test() function, 218 par() function, 261, 264 parallel minimum/maximum, 320 parallel vectors, 158 parentheses, use of, 46, 47 partial autocorrelation function (PACF), 378 partial correlation, 379 password security, 90 paste() function, 164, 168 pattern searching, 13, 14 pausing between plots, 259 pbinom() function, 187 PCA (principal component analysis), 338, 350 pchisq() function, 188 Pearson chi-square test, 210 Pearson method, 215 pearson.test() function, 210 performing calculations on time series, 370 linear regression with interaction terms, 279 linear regression without intercept, 278 matrix operations, 119 multiple linear regression, 270 one-way ANOVA, 302 Index | 407 pairwise comparisons between group means, 218 principal component analysis, 338 robust ANOVA, 308 simple linear regression, 269 simple orthogonal regression, 340 vector arithmetic, 38 Petzoldt, Thomas, 162 pexp() function, 188 pgamma() function, 188 pgeom() function, 187 physical type, 96 plots acf() function (autocorrelation), 376 adding title/labels to, 225 conditioning, 233 of density functions, 190 displaying several on one page, 260 of lines from x and y points, 241 multifigure, 260 in multiple colors, 229 of multiple datasets, 243 of all pairs of variables, 233 pacf() function (partial autocorrelation), 378 pausing between, 259 plot() function, 44, 221–232, 243, 252, 256, 359 of regression residuals, 293 scatter, 223, 231–236 time-series, 359 writing to a file, 263 plyr() package, 140 pmax() function, 48, 320 pmin() function, 48, 320 png() function, 263 pnorm() function, 177, 188 points() function, 222, 226 Poisson distributions, 178, 187, 190 poly() function, 286 polygon() function, 191, 222 polymorphic functions, 223 polynomials, 286 POSIXct datetime class, 161, 172 POSIXlt datetime class, 162, 174 Posting Guide for mailing list, 20 Postscript files, 263 pound sign (#), 80 power transformation, best, 289–292 ppoints() function, 255 ppois() function, 187 Practical Regression and ANOVA Using R (Faraway), 269, 345 prcomp() function, 339, 340 preallocating data frames, 126 predicting binary-valued variable, 345 predict() function, 300, 301, 345, 389 from regression model, 300, 389 prediction intervals, forming, 301 predictor variables, 267, 269 pressure dataset, 57 principal component analysis (PCA), 338, 350 princomp() function, 339 print() function, 167 printing assigned variable value, 315 avoiding wrapping output, 314 cat() function, 24, 73, 74, 111, 115, 167 data in columns, 316 number of digits printed, 69, 73 print() function, 23, 73, 75, 97, 156 special characters, 168 probabilities calculating for continuous distributions, 188 calculating for discrete distributions, 186 common functions, 177 converting to quantiles, 189 mass function, 186 probability function, 186 productivity tips, 43–44, 75, 126 profile scripts, 68 proportions confidence intervals for, 207 prop.test() function, 207, 208, 216 testing groups for equal, 216 pseudomedian, 207 pt() function, 188 Q Q&A websites, 2 q() function, 9, 64 quantile-quantile (Q-Q) plots creating, 252 not normally distributed, 254 qqline() function, 253 qqnorm() function, 222, 253 408 | Index testing for normality, 210, 252 quantiles converting probabilities to, 189 interquartile range, 247 inverting, 202 qbinom() function, 190 qexp() function, 255 qgeom() function, 190 qnorm() function, 177 qpois() function, 190 qt() function, 190, 255 quantile() function, 201 quartiles, 201 R R console version, 6 design philosophy, 71 overview of start-up sequence, 70 R console, 6 R Language Definition, xvi R_HOME environment variable, 67 technical aspects of data structures, 95 website, 10 R² (coefficient of determination), 277 R Data Import/Export, 71 R Graphics (Murrell), xv, 221 R in a Nutshell (Adler), xv rainbow() function, 239 random numbers, generating, 180–183 random permutations of vectors, 185 random samples, generating, 183 random sequences, generating, 184 random variates, 177 range() function, 150, 152 rank() function, 154 Raymond, Eric, 21 rbind() function, 124, 125, 138 .RData file, 5, 9, 52, 64, 70 reading CSV files, 80 files with complex structure, 86–89 fixed-width records, 77 from MySQL databases, 89 read.csv() function, 81, 83, 86 read.fwf() function, 77 read.table() function, 78, 83, 86 read.zoo() function, 358 readHTMLTable() function, 84 readLines() function, 86 tabular data files, 78 from the Web, 83–86 Recycling Rule, 48, 103–105, 118, 139, 173 Red Hat, installing R on, 3 reduced model, 282 regression on data subset, 284 diagnosing a linear, 293 formula, using expression inside, 285 logistic, 345 orthogonal, 340 on polynomials, 286 summary, understanding the, 275–278 on transformed data, 287 regression coefficients, 292 regression line of a scatter plot, 231 regression residuals, 293 relative frequencies, 199 removing elements from lists, 114, 116, 117 insignificant ARIMA coefficients, 386 NA values from data frames, 136 rep() function (repeated values), 33 replacing substrings, 166 reproducible random numbers, 182 require() function, 48, 56, 69 reshape2() package, 140 residuals resid() function, 272, 382 residuals() function, 272 statistics, 276, 277 testing for autocorrelation, 298 resources R books, xv statistics books, xvi on the Web, xv response variables, 267, 269 return value, 42 reversion to mean, 391 .Rhistory file, 53 RHOME subcommand, 67 right arrow, 8 rightwards assignment operators (->, ->>), 26, 40 Ripley, Brian, xv, 163 ripple effect, 382 rm() function, 27 RMySQL package, 90 Index | 409 rnorm() function, 177 robust ANOVA, 308 rollapply() function, 375 rollmean() function, 372 row labels, 83, 120 rowSums() function, 315 .Rprofile files, 70 .Rprofile files, 62, 64, 68–70 Rscript program, 64 rseek.org, 16, 18 RSiteSearch() function, 16, 19 Rterm.exe (console version), 6 runif() function, 181 runs.test() function, 211 S S Programming (Venables and Ripley), xvi S statistical package, 71 samples comparing means of two, 212 proportion-testing, 207 sample() function, 184, 185, 186 size of, 205 testing for same distribution, 219 testing mean of, 203 with/without replacement, 184, 185, 347 sapply() function, 116, 149, 152 Sarkar, Deepayan, xv, 223 SAS dataset, data frame as, 101 saving to a graphics file, 263 objects, 92 save() function, 92 save.image() function, 5, 52 savePlot() function, 263 without exiting R, 52 scalars, 97, 158 scale() function, 203 scan() function, 83, 86 scatter plots, 223, 231–236 scripts, running, 62–65 sd() function, 30–32, 49 searching ?? shortcut, 13 with built-in search engine, 10 using search path, 54 search() function, 54 segments() function, 222, 258 selecting best regression variables, 281 data frame columns by name, 131 data frame columns by position, 127–130 every nth element of vector, 319 matrix row or column, 121 rows and columns with subset(), 132 vector elements, 35 self-describing data, 80 sep parameter, 79 separator, string, 164 sequences creation operator (:), 33, 40 seq() function, 33, 175 seq_along() function, 319 set.seed() function, 183 setting environment variables, 66 working directory, 51 sf.test() function, 210 shading, density plot with, 191 shapiro.test() function, 209 Shapiro–Francia test, 210 Shapiro–Wilk test, 210 shifting a time series, 368 short-circuit and (&&), 40, 48 short-circuit or (||), 40, 48 shortcuts ? help, 12 creating, 5 embedding --quiet option with, 62 starting R from desktop, 5 side-by-side display, 260 significance codes, 277 significance level, 196 simple linear regression, 267, 269 simplifying results, 149 single-parameter function, 335 sink() function, 75 slope, computing, 340 slot extraction (@), 40 smoothing, 393 snippets, using, 44 sorting data frames by two columns, 324 sos package, 19 source() function, 63 Spearman method, 215 special characters, 168 special interest group (SIG) mailing lists, 21, 49 410 | Index split() function, 148 split.screen() function, 261 splitting strings by delimiter, 165 splitting vectors into groups, 148 SQL, executing, 89 sqrt() function, 39 Stack Exchange website, 16, 18 Stack Overflow website, 16, 49 stack() function, 107 stacking of rows, 139 standard deviation, 30–32 starting R, 4–6, 62, 70 statistical analysis computing basic statistics, 30–32 help on Web, 16, 18 testing correlation for significance, 215 statistical significance/F statistic, 267 StatLib, 88 step() function (stepwise regression), 281–284 stftime() function, 170 strings concatenating, 163 converting date into, 171 converting to date, 170 generating all pairwise combinations of, 168 getting length of, 163 read.table stringsAsFactors parameter, 79 splitting by delimiter, 165 str() function, 27, 326 strftime() function, 171 unprintable characters in, 167 stripping attributes from variables, 325 strsplit() function, 165 structure of an object, 326 structured data types, 144 Student’s t distributions, 178, 188, 190, 255 sub() function, 166 submitting mailing list questions, 20, 92 subscribing to R-help list, 20 subset() function, 132, 137 subset, regressing on, 284 subsetting a time series, 363 substrings extracting, 164 replacing, 166 substr() function, 164 subtraction operator (-), 40 summarizing data, 197 summary() function, 156, 197, 272–278, 317 summing rows and columns, 315 SuppDists package, 179 suppressMessages() function, 69, 330 suppressWarnings() function, 330 survival function, 187 syntax errors, 46–49 synthetic names, 134 Sys.Date() function, 169 Sys.getenv() function, 66, 67 Sys.putenv() function, 66 system CPU time/system.time() function, 329 T t statistics, 268 t.test() function, 150, 203, 205, 212, 237 Tab key, 8 table() function, 106, 200, 252 tables in data files header lines in, 80 reading, 78, 84, 86 tabulating factors, 200 tail() function, 313, 361 tapply() function, 154 task views, 1, 18 Optimization and Mathematical Programming, 338 Probability Distributions, 179 Time Series Analysis, 356 temporary copies, data frame, 142 testing categorical variables for independence, 201 correlation for significance, 215 groups for equal proportions, 216 mean of a sample, 203 for mean reversion, 391 for normality, 209 residuals for autocorrelation, 298 for runs, 210 time series for autocorrelation, 377 two samples for same distribution, 219 text() function, 222 tilde (~), 40, 233, 248, 269, 303 time series applying a function by calendar period, 373 ARIMA model, 383–390 autocorrelation testing, 377 Index | 411 D o wnload from Wow! eBook computing successive differences between, 369 detrending, 382, 392 finding lagged correlations between, 379 lagging, 368 mean reversion testing, 391 merging, 364 overview, 355 performing calculations on, 370 plotting data, 359 representing data from, 356 rolling mean, 372 smoothing, 393 subsetting, 363 timeDate package, 162 timeSeries package, 358 timestamps, 364, 366, 371, 375 timing your code, 329 title() function, 225 TLS (total least squares), 340 tokens in data files, 86 total least squares (TLS), 340 transformed data Box–Cox procedure, 289 regressing on, 287 transporting objects, 92 trends, removing, 382 ts class, 355 tsdiag() function, 387 tseries package, 211, 391 Tukey, John, 305 TukeyHSD() function, 305 U Ubuntu R install, 3 unary minus operator (-), 40 unary plus operator (+), 40 unclass() function, 326 undefined operators, 333 uniform distributions, 178 Unix (see Linux/Unix) unlist() function, 115, 117 unprintable characters, 167 unstack() function, 148 up arrow, 8 urca package, 393 user CPU time, 329 Using R for Introductory Statistics (Verzani), xvi V values missing, 79 obtaining last, 53 predicting from regression model, 300 var() function, 30–32 variables categorical (see factors) data organized by, 123 deleting, 27 environment, 66 global, 43 .Last.value, 53 listing, 26 local, 43 low-correlation, removing, 153 setting (assigning), 25 stripping attributes from, 325 vcov() function (variance-covariance), 272 vectors appending data to, 101 c() constructor for, 72, 101 calculating z-scores of, 39 comparing, 34 creating, 28 finding position of value, 318 flattening lists into, 115, 145 and grouping factors, 147 indexing of ([ ]), 35, 40 inserting data into, 103 logical, 117 versus loops, 147 and matrices, 98, 119 and modes, 97 multiple, combining into vector and factor, 107 of normalized values, 203 observations stored in, 124 parallel, 112 performing arithmetic on, 38–39 printing simple, 24 properties of, 95 recentering, 39 rotation, 339 and scalars, 97 selecting elements of, 35 selecting every nth element, 319 summaries of lists of, 197, 198 of unequal length, 48, 103–105 412 | Index Venables, William, xv vertical lines, adding to plots, 245 Verzani, John, xvi vignette() function, 16, 358 W Wackerly, Dennis, 197 warning messages “argument is not numeric or logical”, 331 “In arima . . . some AR parameters were fixed”, 386 suppressing, 69, 329 unrecognized escape in a character string, 76 warnings() function, 330 web data, reading, 83–86 week, defined, 362 Weibull distributions, 178 which.max() function, 318 which.min() function, 318 while statements, 48 Wickham, Hadley, xv, 223 wilcox.test() function, 214 Wilcoxon distribution, 178 Wilcoxon–Mann–Whitney test, 214 win.graph() function, 262 window() function, 363 windows, graphic, 260, 262 Windows, Microsoft “cannot open” file error in, 76 exiting from R, 8 installing packages on, 59 installing R on, 2 interrupting R in, 9 location of .Rprofile, 68 “Save as” for graphics files, 263 starting R, 4–6 with() function, 141 Witmer, Jeff, 88 working directory, 5, 51 workspace data saved outside of, 92 .RData file, 5, 9, 52, 64, 70 restore options, 64 saving, 9, 52, 64 variables held in, 25 World Series dataset, 88 write.csv() function, 82 X X Windows system, 66 XML package, 84 xtabs() function, 200 xts package, 355, 356, 373 Z z-scores, 39, 203 zero intercept, 278 zero-width object, 366 zoo objects, performing calculations on, 371 zoo package, 223, 355, 356 Zoonekynd, Vincent, 341 Index | 413 About the Author Paul Teetor is a quantitative developer with master’s degrees in statistics and computer science. He specializes in analytics and software engineering for investment manage- ment, securities trading, and risk management. He works with hedge funds, market makers, and portfolio managers in the greater Chicago area. Colophon The animal on the cover of R Cookbook is a harpy eagle (Harpia harpyja). One of the fifty species of eagle in the world, the harpy eagle is native to the tropical rain forests of Central and South America, and prefers to nest in the upper canopy layer thereof. Both its genus and species names refer to the harpies of ancient Greek mythology— vicious creatures with the face of a woman and the body of an eagle or vulture. On average, harpy eagles weigh about 18 lbs., are 36 to 40 inches long, and have a wingspan of 6 to 7 feet, though females are consistently larger than males. The plumage of both sexes is identical, however: slate black feathers dominate the animal’s top half, while the underside is white or light gray. Light gray-colored heads are accentuated with a double crest of large feathers, which specimens can raise when showing hostility. Harpy eagles are monogamous, and pairs raise only one chick every two to three years. Females will usually lay two eggs at a time, and after the first hatches, the other is neglected and does not hatch. Though the chick will fledge within 6 months, both parents continue to care for and feed the chick for at least a year. Because of this low rate of population growth, the harpy eagle is particularly susceptible to encroachments on its habitat and losses from human hunting. Throughout its range, the animal’s con- servation status ranges from threatened to critically endangered. The cover image is from J. G. Wood’s Animate Creation. The cover font is Adobe ITC Garamond. The text font is Linotype Birka; the heading font is Adobe Myriad Con- densed; and the code font is LucasFont’s TheSansMonoCondensed.