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9.6 Lab: Support Vector Machines 361 Now that a smaller value of the cost parameter is being used, we obtain a larger number of support vectors, because the margin is now wider. Unfortunately, the svm() function does not explicitly output the coefficients of the linear decision boundary obtained when the support vector classifier is fit, nor does it output the width of the margin. The e1071 library includes a built-in function, tune(), to perform cross- tune() validation. By default, tune() performs ten-fold cross-validation on a set of models of interest. In order to use this function, we pass in relevant information about the set of models that are under consideration. The following command indicates that we want to compare SVMs with a linear kernel, using a range of values of the cost parameter. > set.seed(1) > tune.out=tune(svm,y∼.,data=dat,kernel="linear", ranges=list(cost=c(0.001, 0.01, 0.1, 1,5,10,100) )) We can easily access the cross-validation errors for each of these models using the summary() command: > summary(tune.out) Parameter tuning of ’svm’: - sampling method: 10-fold cross validation - best parameters : cost 0.1 - best performance : 0.1 - Detailed performance results: cost error dispersion 1 1e-03 0.70 0.422 2 1e-02 0.70 0.422 3 1e-01 0.10 0.211 4 1e+00 0.15 0.242 5 5e+00 0.15 0.242 6 1e+01 0.15 0.242 7 1e+02 0.15 0.242 We see that cost=0.1 results in the lowest cross-validation error rate. The tune() function stores the best model obtained, which can be accessed as follows: > bestmod=tune.out$best.model > summary(bestmod) The predict() function can be used to predict the class label on a set of test observations, at any given value of the cost parameter. We begin by generating a test data set. > xtest=matrix(rnorm(20*2), ncol=2) > ytest=sample(c(-1,1), 20, rep=TRUE) > xtest[ytest==1,]=xtest[ytest==1,] + 1 > testdat=data.frame(x=xtest , y=as.factor(ytest)) Now we predict the class labels of these test observations. Here we use the best model obtained through cross-validation in order to make predictions.
362 9. Support Vector Machines > ypred=predict(bestmod ,testdat) > table(predict=ypred , truth=testdat$y ) truth predict -1 1 -1 11 1 1 0 8 Thus, with this value of cost, 19 of the test observations are correctly classified. What if we had instead used cost=0.01? > svmfit=svm(y∼., data=dat, kernel="linear", cost=.01, scale=FALSE) > ypred=predict(svmfit ,testdat) > table(predict=ypred , truth=testdat$y ) truth predict -1 1 -1 11 2 1 0 7 In this case one additional observation is misclassified. Now consider a situation in which the two classes are linearly separable. Then we can find a separating hyperplane using the svm() function. We first further separate the two classes in our simulated data so that they are linearly separable: > x[y==1,]=x[y==1,]+0.5 > plot(x, col=(y+5)/2, pch=19) Now the observations are just barely linearly separable. We fit the support vector classifier and plot the resulting hyperplane, using a very large value of cost so that no observations are misclassified. > dat=data.frame(x=x,y=as.factor(y)) > svmfit=svm(y∼., data=dat, kernel="linear", cost=1e5) > summary(svmfit) Call: svm(formula = y ∼ ., data = dat, kernel = "linear", cost = 1e +05) Parameters : SVM-Type: C-classification SVM-Kernel: linear cost: 1e+05 gamma: 0.5 Number of Support Vectors: 3 ( 1 2 ) Number of Classes: 2 Levels: -1 1 > plot(svmfit , dat) No training errors were made and only three support vectors were used. However, we can see from the figure that the margin is very narrow (because the observations that are not support vectors, indicated as circles, are very
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Springer Texts in Statistics Gareth
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Gareth James • Daniela Witten •
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To our parents: Alison and Michael
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viii Preface disciplines who wish t
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x Contents 3 Linear Regression 59 3
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4.2 Why Not Linear Regression? 129
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304 8. Tree-Based Methods Years < 4
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Index C p , 78, 205, 206, 210-213 R
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Index 421 Wage, 1, 2, 9, 10, 14, 26
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Index 423 noise, 22, 228 non-linear
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Index 425 read.table(), 48, 49 regs
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