References

Chapter 9.6 [ISLR2] An Introduction to Statistical Learning - with Applications in R (2nd Edition). Free access to download the book: https://www.statlearning.com/

To see the help file of a function funcname, type ?funcname.

We use the e1071 library in R to demonstrate the support vector classifier and the SVM. Another option is the LiblineaR library, which is useful for very large linear problems.

Support Vector Classifier

The e1071 library contains implementations for a number of statistical learning methods. In particular, the svm() function can be used to fit a support vector classifier when the argument kernel = "linear" is used. This function uses a slightly different formulation from lecture slides for the support vector classifier.A cost argument allows us to specify the cost of a violation to the margin. When the cost argument is small, then the margins will be wide and many support vectors will be on the margin or will violate the margin. When the cost argument is large, then the margins will be narrow and there will be few support vectors on the margin or violating the margin.

We now use the svm() function to fit the support vector classifier for a given value of the cost parameter. Here we demonstrate the use of this function on a two-dimensional example so that we can plot the resulting decision boundary. We begin by generating the observations, which belong to two classes, and checking whether the classes are linearly separable.

set.seed(1)
x <- matrix(rnorm(20 * 2), ncol = 2)
y <- c(rep(-1, 10), rep(1, 10))
x[y == 1, ] <- x[y == 1, ] + 1
plot(x, col = (3 - y))

They are not. Next, we fit the support vector classifier. Note that in order for the svm() function to perform classification (as opposed to SVM-based regression), we must encode the response as a factor variable. We now create a data frame with the response coded as a factor.

dat <- data.frame(x = x, y = as.factor(y))
library(e1071)
svmfit <- svm(y ~ ., data = dat, kernel = "linear", 
    cost = 10, scale = FALSE)

The argument scale = FALSE tells the svm() function not to scale each feature to have mean zero or standard deviation one; depending on the application, one might prefer to use scale = TRUE.

We can now plot the support vector classifier obtained:

plot(svmfit, dat)

Note that the two arguments to the SVM plot() function are the output of the call to svm(), as well as the data used in the call to svm(). The region of feature space that will be assigned to the -1 class is shown in light yellow, and the region that will be assigned to the +1 class is shown in red. The decision boundary between the two classes is linear (because we used the argument kernel = "linear"), though due to the way in which the plotting function is implemented in this library the decision boundary looks somewhat jagged in the plot. (Note that here the second feature is plotted on the \(x\)-axis and the first feature is plotted on the \(y\)-axis, in contrast to the behavior of the usual plot() function in R.) The support vectors are plotted as crosses and the remaining observations are plotted as circles; we see here that there are seven support vectors. We can determine their identities as follows:

svmfit$index
## [1]  1  2  5  7 14 16 17

We can obtain some basic information about the support vector classifier fit using the summary() command:

summary(svmfit)
## 
## Call:
## svm(formula = y ~ ., data = dat, kernel = "linear", cost = 10, scale = FALSE)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  10 
## 
## Number of Support Vectors:  7
## 
##  ( 4 3 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  -1 1

This tells us, for instance, that a linear kernel was used with cost = 10, and that there were seven support vectors, four in one class and three in the other.

What if we instead used a smaller value of the cost parameter?

svmfit <- svm(y ~ ., data = dat, kernel = "linear", 
    cost = 0.1, scale = FALSE)
plot(svmfit, dat)

svmfit$index
##  [1]  1  2  3  4  5  7  9 10 12 13 14 15 16 17 18 20

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-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: 0.05 
## 
## - Detailed performance results:
##    cost error dispersion
## 1 1e-03  0.55  0.4377975
## 2 1e-02  0.55  0.4377975
## 3 1e-01  0.05  0.1581139
## 4 1e+00  0.15  0.2415229
## 5 5e+00  0.15  0.2415229
## 6 1e+01  0.15  0.2415229
## 7 1e+02  0.15  0.2415229

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)
## 
## Call:
## best.tune(METHOD = svm, train.x = y ~ ., data = dat, ranges = list(cost = c(0.001, 
##     0.01, 0.1, 1, 5, 10, 100)), kernel = "linear")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.1 
## 
## Number of Support Vectors:  16
## 
##  ( 8 8 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  -1 1

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.

ypred <- predict(bestmod, testdat)
table(predict = ypred, truth = testdat$y)
##        truth
## predict -1 1
##      -1  9 1
##      1   2 8

Thus, with this value of cost, 17 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  6
##      1   0  3

In this case three additional observations are 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 
## 
## 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 close to the decision boundary). It seems likely that this model will perform poorly on test data. We now try a smaller value of cost:

svmfit <- svm(y ~ ., data = dat, kernel = "linear", cost = 1)
summary(svmfit)
## 
## Call:
## svm(formula = y ~ ., data = dat, kernel = "linear", cost = 1)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  1 
## 
## Number of Support Vectors:  7
## 
##  ( 4 3 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  -1 1
plot(svmfit, dat)

Using cost = 1, we misclassify a training observation, but we also obtain a much wider margin and make use of seven support vectors. It seems likely that this model will perform better on test data than the model with cost = 1e5.

Support Vector Machine

In order to fit an SVM using a non-linear kernel, we once again use the svm() function. However, now we use a different value of the parameter kernel. To fit an SVM with a polynomial kernel we use kernel = "polynomial", and to fit an SVM with a radial kernel we use kernel = "radial". In the former case we also use the degree argument to specify a degree for the polynomial kernel (this is \(d\) in lecture slides), and in the latter case we use gamma to specify a value of \(\gamma\) for the radial basis kernel.

We first generate some data with a non-linear class boundary, as follows:

set.seed(1)
x <- matrix(rnorm(200 * 2), ncol = 2)
x[1:100, ] <- x[1:100, ] + 2
x[101:150, ] <- x[101:150, ] - 2
y <- c(rep(1, 150), rep(2, 50))
dat <- data.frame(x = x, y = as.factor(y))

Plotting the data makes it clear that the class boundary is indeed non-linear:

plot(x, col = y)

The data is randomly split into training and testing groups. We then fit the training data using the svm() function with a radial kernel and \(\gamma=1\):

train <- sample(200, 100)
svmfit <- svm(y ~ ., data = dat[train, ], kernel = "radial",  
    gamma = 1, cost = 1)
plot(svmfit, dat[train, ])

The plot shows that the resulting SVM has a decidedly non-linear boundary. The summary() function can be used to obtain some information about the SVM fit:

summary(svmfit)
## 
## Call:
## svm(formula = y ~ ., data = dat[train, ], kernel = "radial", gamma = 1, 
##     cost = 1)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  31
## 
##  ( 16 15 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  1 2

We can see from the figure that there are a fair number of training errors in this SVM fit. If we increase the value of cost, we can reduce the number of training errors. However, this comes at the price of a more irregular decision boundary that seems to be at risk of overfitting the data.

svmfit <- svm(y ~ ., data = dat[train, ], kernel = "radial", 
    gamma = 1, cost = 1e5)
plot(svmfit, dat[train, ])

We can perform cross-validation using tune() to select the best choice of \(\gamma\) and cost for an SVM with a radial kernel:

set.seed(1)
tune.out <- tune(svm, y ~ ., data = dat[train, ], 
    kernel = "radial", 
    ranges = list(
      cost = c(0.1, 1, 10, 100, 1000),
      gamma = c(0.5, 1, 2, 3, 4)
    )
  )
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##     1   0.5
## 
## - best performance: 0.07 
## 
## - Detailed performance results:
##     cost gamma error dispersion
## 1  1e-01   0.5  0.26 0.15776213
## 2  1e+00   0.5  0.07 0.08232726
## 3  1e+01   0.5  0.07 0.08232726
## 4  1e+02   0.5  0.14 0.15055453
## 5  1e+03   0.5  0.11 0.07378648
## 6  1e-01   1.0  0.22 0.16193277
## 7  1e+00   1.0  0.07 0.08232726
## 8  1e+01   1.0  0.09 0.07378648
## 9  1e+02   1.0  0.12 0.12292726
## 10 1e+03   1.0  0.11 0.11005049
## 11 1e-01   2.0  0.27 0.15670212
## 12 1e+00   2.0  0.07 0.08232726
## 13 1e+01   2.0  0.11 0.07378648
## 14 1e+02   2.0  0.12 0.13165612
## 15 1e+03   2.0  0.16 0.13498971
## 16 1e-01   3.0  0.27 0.15670212
## 17 1e+00   3.0  0.07 0.08232726
## 18 1e+01   3.0  0.08 0.07888106
## 19 1e+02   3.0  0.13 0.14181365
## 20 1e+03   3.0  0.15 0.13540064
## 21 1e-01   4.0  0.27 0.15670212
## 22 1e+00   4.0  0.07 0.08232726
## 23 1e+01   4.0  0.09 0.07378648
## 24 1e+02   4.0  0.13 0.14181365
## 25 1e+03   4.0  0.15 0.13540064

Therefore, the best choice of parameters involves cost = 1 and gamma = 0.5. We can view the test set predictions for this model by applying the predict() function to the data. Notice that to do this we subset the dataframe dat using -train as an index set.

table(
    true = dat[-train, "y"], 
    pred = predict(
      tune.out$best.model, newdata = dat[-train, ]
      )
    )
##     pred
## true  1  2
##    1 67 10
##    2  2 21

12% of test observations are misclassified by this SVM.

ROC Curves

The ROCR package can be used to produce ROC curves. We first write a short function to plot an ROC curve given a vector containing a numerical score for each observation, pred, and a vector containing the class label for each observation, truth.

library(ROCR)
rocplot <- function(pred, truth, ...) {
  predob <- prediction(pred, truth)
  perf <- performance(predob, "tpr", "fpr")
  plot(perf, ...)
}

SVMs and support vector classifiers output class labels for each observation. However, it is also possible to obtain fitted values for each observation, which are the numerical scores used to obtain the class labels. For instance, in the case of a support vector classifier, the fitted value for an observation \(X= (X_1, X_2, \ldots, X_p)^T\) takes the form \(\hat{\beta}_0 + \hat{\beta}_1 X_1 + \hat{\beta}_2 X_2 + \cdots + \hat{\beta}_p X_p\). In essence, the sign of the fitted value determines on which side of the decision boundary the observation lies. Therefore, the relationship between the fitted value and the class prediction for a given observation is simple: if the fitted value exceeds zero then the observation is assigned to one class, and if it is less than zero then it is assigned to the other. In order to obtain the fitted values for a given SVM model fit, we use decision.values = TRUE when fitting svm(). Then the predict() function will output the fitted values.

svmfit.opt <- svm(y ~ ., data = dat[train, ], 
    kernel = "radial", gamma = 2, cost = 1, 
    decision.values = T)
fitted <- attributes(
    predict(svmfit.opt, dat[train, ], decision.values = TRUE)
  )$decision.values

Now we can produce the ROC plot. Note we use the negative of the fitted values so that negative values correspond to class 1 and positive values to class 2.

par(mfrow = c(1, 2))
rocplot(-fitted, dat[train, "y"], main = "Training Data")

SVM appears to be producing accurate predictions. By increasing \(\gamma\) we can produce a more flexible fit and generate further improvements in accuracy.

rocplot(-fitted, dat[train, "y"], main = "Training Data")
svmfit.flex <- svm(y ~ ., data = dat[train, ], 
    kernel = "radial", gamma = 50, cost = 1, 
    decision.values = T)
fitted <- attributes(
    predict(svmfit.flex, dat[train, ], decision.values = T)
  )$decision.values
rocplot(-fitted, dat[train, "y"], add = T, col = "red")

However, these ROC curves are all on the training data. We are really more interested in the level of prediction accuracy on the test data. When we compute the ROC curves on the test data, the model with \(\gamma=2\) appears to provide the most accurate results.

fitted <- attributes(
    predict(svmfit.opt, dat[-train, ], decision.values = T)
  )$decision.values
rocplot(-fitted, dat[-train, "y"], main = "Test Data")
fitted <- attributes(
    predict(svmfit.flex, dat[-train, ], decision.values = T)
  )$decision.values
rocplot(-fitted, dat[-train, "y"], add = T, col = "red")

SVM with Multiple Classes

If the response is a factor containing more than two levels, then the svm() function will perform multi-class classification using the one-versus-one approach. We explore that setting here by generating a third class of observations.

set.seed(1)
x <- rbind(x, matrix(rnorm(50 * 2), ncol = 2))
y <- c(y, rep(0, 50))
x[y == 0, 2] <- x[y == 0, 2] + 2
dat <- data.frame(x = x, y = as.factor(y))
par(mfrow = c(1, 1))
plot(x, col = (y + 1))

We now fit an SVM to the data:

svmfit <- svm(y ~ ., data = dat, kernel = "radial", 
    cost = 10, gamma = 1)
plot(svmfit, dat)

The e1071 library can also be used to perform support vector regression, if the response vector that is passed in to svm() is numerical rather than a factor.

Application to Gene Expression Data

We now examine the Khan data set, which consists of a number of tissue samples corresponding to four distinct types of small round blue cell tumors. For each tissue sample, gene expression measurements are available. The data set consists of training data, xtrain and ytrain, and testing data, xtest and ytest.

We examine the dimension of the data:

library(ISLR2)
names(Khan)
## [1] "xtrain" "xtest"  "ytrain" "ytest"
dim(Khan$xtrain)
## [1]   63 2308
dim(Khan$xtest)
## [1]   20 2308
length(Khan$ytrain)
## [1] 63
length(Khan$ytest)
## [1] 20

This data set consists of expression measurements for 2,308 genes. The training and test sets consist of 63 and 20 observations respectively.

table(Khan$ytrain)
## 
##  1  2  3  4 
##  8 23 12 20
table(Khan$ytest)
## 
## 1 2 3 4 
## 3 6 6 5

We will use a support vector approach to predict cancer subtype using gene expression measurements. In this data set, there are a very large number of features relative to the number of observations. This suggests that we should use a linear kernel, because the additional flexibility that will result from using a polynomial or radial kernel is unnecessary.

dat <- data.frame(
    x = Khan$xtrain, 
    y = as.factor(Khan$ytrain)
  )
out <- svm(y ~ ., data = dat, kernel = "linear", 
    cost = 10)
summary(out)
## 
## Call:
## svm(formula = y ~ ., data = dat, kernel = "linear", cost = 10)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  10 
## 
## Number of Support Vectors:  58
## 
##  ( 20 20 11 7 )
## 
## 
## Number of Classes:  4 
## 
## Levels: 
##  1 2 3 4
table(out$fitted, dat$y)
##    
##      1  2  3  4
##   1  8  0  0  0
##   2  0 23  0  0
##   3  0  0 12  0
##   4  0  0  0 20

We see that there are no training errors. In fact, this is not surprising, because the large number of variables relative to the number of observations implies that it is easy to find hyperplanes that fully separate the classes. We are most interested not in the support vector classifier’s performance on the training observations, but rather its performance on the test observations.

dat.te <- data.frame(
    x = Khan$xtest, 
    y = as.factor(Khan$ytest))
pred.te <- predict(out, newdata = dat.te)
table(pred.te, dat.te$y)
##        
## pred.te 1 2 3 4
##       1 3 0 0 0
##       2 0 6 2 0
##       3 0 0 4 0
##       4 0 0 0 5

We see that using cost = 10 yields two test set errors on this data.