This lab on Support Vector Machines in R is an adapted version of p. 359-366 of "Introduction to Statistical Learning with Applications in R" by Gareth James, Daniela Witten, Trevor Hastie and Robert Tibshirani. It was re-implemented in Fall 2016 in tidyverse format by Amelia McNamara and R. Jordan Crouser at Smith College.
Want to follow along on your own machine? Download the .Rmd or Jupyter Notebook version.
In this lab, we'll use the e1071 library in R to demonstrate the support vector classifier
and the SVM. Another option is the LiblineaR library, which is particularly useful for
very large linear problems.
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 of the equations we saw in lecture to build 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 can 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. Let's start by generating a set of observations, which belong
to two classes:
set.seed(1)
x = matrix(rnorm(20*2), ncol=2)
class = c(rep(-1,10), rep(1,10))
x[class == 1,] = x[class == 1,] + 1
Let's plot the data to see whether the classes are linearly separable:
library(ggplot2)
ggplot(data.frame(x), aes(X1, X2, colour = factor(class))) +
geom_point()
Nope; not linear. 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:
training_data = data.frame(x = x, class = as.factor(class))
library(e1071)
svmfit = svm(class~.,
data = training_data,
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,
we might prefer to use scale = TRUE. We can now plot the support vector classifier by calling the plot() function on the output of the call to svm(), as well as the data used in the call to svm():
plot(svmfit, training_data)
The region of feature space that will be assigned to the −1 class is shown in
light blue, and the region that will be assigned to the +1 class is shown in
purple. 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. We see that in this case only one observation
is misclassified. (Note also that 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$index
We can obtain some basic information about the support vector classifier
fit using the summary() command:
summary(svmfit)
This tells us, for instance, that a linear kernel was used with cost = 10, and
that there were seven support vectors, four in one class and three in the
other. What if we instead used a smaller value of the cost parameter?
svmfit = svm(class~.,
data = training_data,
kernel = "linear",
cost = 0.1,
scale = FALSE)
plot(svmfit, training_data)
svmfit$index
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,
class~.,
data = training_data,
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)
The tune() function stores the best model obtained, which can be accessed as
follows:
bestmod = tune_out$best.model
summary(bestmod)
As usual, the predict() function can be used to predict the class label on a set of
test observations, at any given value of the cost parameter. Let's
generate a test data set:
xtest = matrix(rnorm(20*2), ncol = 2)
ytest = sample(c(-1,1), 20, rep = TRUE)
xtest[ytest == 1,] = xtest[ytest == 1,] + 1
test_data = data.frame(x = xtest, class = as.factor(ytest))
Now we predict the class labels of these test observations. Here we use the best model obtained through cross-validation in order to make predictions:
class_pred = predict(bestmod, test_data)
table(predicted = class_pred, true = test_data$class)
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(class~., data = training_data, kernel = "linear", cost = .01, scale = FALSE)
class_pred = predict(svmfit, test_data)
table(predicted = class_pred, true = test_data$class)
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. First we'll give our simulated data a little nudge so that they are linearly separable:
x[class == 1,] = x[class == 1,] + 0.5
ggplot(data.frame(x), aes(X1, X2, colour = factor(class))) +
geom_point()
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.
training_data2 = data.frame(x = x, class = as.factor(class))
svmfit = svm(class~., data = training_data2, kernel = "linear", cost = 1e5)
summary(svmfit)
plot(svmfit, training_data2)
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. Let's try a smaller value of cost:
svmfit = svm(class~., data = training_data2, kernel = "linear", cost = 1)
summary(svmfit)
plot(svmfit, training_data2)
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.
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, and in the latter case we use gamma to specify a
value of $\gamma$ for the radial basis kernel.
Let's generate some data with a non-linear class boundary:
set.seed(1)
x = matrix(rnorm(200*2), ncol = 2)
x[1:100,] = x[1:100,]+2
x[101:150,] = x[101:150,]-2
class = c(rep(1,150),rep(2,50))
nonlinear_data = data.frame(x = x, class = as.factor(class))
ggplot(nonlinear_data, aes(x.1, x.2, colour = factor(class))) +
geom_point()
See how one class is kind of stuck in the middle of another class? This suggests that we might want to use a radial kernel in our SVM. Now let's randomly split this data into training and testing groups, and then fit
the training data using the svm() function with a radial kernel and $\gamma = 1$:
library(dplyr)
nonlinear_train = nonlinear_data %>%
sample_frac(0.5)
nonlinear_test = nonlinear_data %>%
setdiff(nonlinear_train)
svmfit = svm(class~., data = nonlinear_train, kernel = "radial", gamma = 1, cost = 1)
plot(svmfit, nonlinear_train)
Not too shabby! The plot shows that the resulting SVM has a decidedly non-linear boundary. 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:
svmfit = svm(class~., data = nonlinear_train, kernel = "radial", gamma = 1, cost = 1e5)
plot(svmfit, nonlinear_train)
However, this comes at the price of a more irregular decision boundary that seems to be at risk of overfitting the data. 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, class~., data = nonlinear_train, kernel = "radial",
ranges = list(cost = c(0.1,1,10,100,1000), gamma = c(0.5,1,2,3,4)))
bestmod = tune_out$best.model
summary(bestmod)
Therefore, the best choice of parameters involves cost = 1 and gamma = 2. We
can plot the resulting fit using the plot() function, and view the test set predictions for this model by applying the predict()
function to the test data.
plot(bestmod, nonlinear_train)
table(true = nonlinear_test$class, pred = predict(tune_out$best.model, newdata = nonlinear_test))
90% of test observations are correctly classified by this SVM. Not bad!
The ROCR package can be used to produce ROC curves such as those we saw in lecture. 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 $X = (X_1,X_2, . . .,X_p)^T$ takes the form $\hat\beta_0 + \hat\beta_1X_1 + \hat\beta_2X_2 + . . . + \hat\beta_pX_p$.
For an SVM with a non-linear kernel, the equation that yields the fitted value is given in (9.23) on p. 352 of the ISLR book. 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 than 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. Let's fit models using the $\gamma$ selected by cross-validation, and a higher value, which will produce a more flexible fit:
svmfit_opt = svm(class~., data = nonlinear_train, kernel = "radial",
gamma = 2, cost = 1, decision.values = TRUE)
svmfit_flex = svm(class~., data = nonlinear_train, kernel = "radial",
gamma = 50, cost = 1, decision.values = TRUE)
Now we can produce the ROC plot to see how the models perform on both the training and the test data:
par(mfrow = c(1,2))
# Plot optimal parameter model's performance on training data
fitted_opt_train = attributes(predict(svmfit_opt, nonlinear_train,
decision.values = TRUE))$decision.values
rocplot(fitted_opt_train, nonlinear_train$class, main = "Training Data")
# Add more flexible model's performance to the plot
fitted_flex_train = attributes(predict(svmfit_flex, nonlinear_train,
decision.values = TRUE))$decision.values
rocplot(fitted_flex_train, nonlinear_train$class, add = TRUE, col = "red")
# Plot optimal parameter model's performance on test data
fitted_opt_test = attributes(predict(svmfit_opt, nonlinear_test,
decision.values = TRUE))$decision.values
rocplot(fitted_opt_test, nonlinear_test$class, main = "Test Data")
# Add more flexible model's performance to the plot
fitted_flex_test = attributes(predict(svmfit_flex, nonlinear_test,
decision.values = TRUE))$decision.values
rocplot(fitted_flex_test, nonlinear_test$class, add = TRUE, col = "red")
If the response is a factor containing more than two levels, then the ${\tt svm()}$ function will perform multi-class classification using the one-versus-one approach. We explore that setting here by generating a third class of observations:
x = rbind(x, matrix(rnorm(50*2), ncol = 2))
class = c(class, rep(0,50))
x[class == 0,2] = x[class == 0,2]+2
data_3_classes = data.frame(x = x, class = as.factor(class))
ggplot(data_3_classes, aes(x.1, x.2, colour = factor(class))) +
geom_point()
Fitting an SVM to multiclass data uses identical syntax to fitting a simple two-class model:
svmfit = svm(class~., data = data_3_classes, kernel = "radial", cost = 10, gamma = 1)
plot(svmfit, data_3_classes)
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.
We now examine Optical Recognition of Handwritten Digits Data Set, which contains 5,620 samples of handwritten digits 0..9. You can use these links to download the training data and test data, and then we'll load them into R:
digits_train = read.csv("optdigits.tra", header = FALSE)
digits_train = data.frame(x = digits_train[,-1], y = as.factor(digits_train[,65]))
digits_test = read.csv("optdigits.tes", header = FALSE)
digits_test = data.frame(x = digits_test[,-1], y = as.factor(digits_test[,65]))
Let's take a look at the dimensions of this dataset:
dim(digits_train)
dim(digits_test)
This data set consists of preprocessed images of handwriting samples gathered from 43 different people. Each image was converted into an 8x8 matrix (64 pixels), which was then flattened into a vector of 64 numeric values. The final column contains the class label for each digit.
The training and test sets consist of 3,823 and 1,797 observations respectively. Let's see what one of these digits looks like:
m = matrix(unlist(digits_train[1,-1]),8,8)
image(m, axes = FALSE, col = grey(seq(0, 1, length = 256)))
That's a pretty messy digit. Let's peek at the true class:
digits_train[1,65]
Phew, looks like our SVM has its work cut out for it! Let's start with a linear kernel to see how we do:
digits_svm = svm(y~., data = digits_train, kernel = "linear", cost = 10, scale = FALSE)
table(digits_svm$fitted, digits_train$y)
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:
pred = predict(digits_svm, newdata = digits_test)
table(pred, digits_test$y)
We see that using cost = 10 yields just 40 test set errors on this data. Now try using the tune() function to select an optimal value for cost, and refit the model using that value. Consider values in the range 0.01 to 100:
# Your code here
To get credit for this lab, report your optimal values and comment on your final model's performance on Moodle: https://moodle.smith.edu/mod/quiz/view.php?id=266457