Linear SVMs on radially separable data

Support Vector Machines in R

Kailash Awati

Instructor

Linear SVM, cost = 1

Partition radially separable dataset into training/test (seed = 10)

# Build default cost linear SVM on training set

svm_model <- svm(y ~ ., data = trainset, type = "C-classification", kernel = "linear", cost = 1)
svm_model

Number of Support Vectors:  126

# Calculate accuracy on test set

pred_test <- predict(svm_model, testset)
mean(pred_test == testset$y)

0.6129032

plot(svm_model, trainset)

Chapter 3.2 - radially separable dataset, default cost linear kernel, plot using svm.plot

Linear SVM, cost = 100

svm_model <- svm(y ~ ., data = trainset, type = "C-classification", kernel = "linear", cost = 100)
svm_model

Number of Support Vectors:  136

# Accuracy
pred_test <- predict(svm_model, testset)
mean(pred_test == testset$y)

0.6129032

plot(svm_model, trainset)

Chapter 3.2 - radially separable dataset, cost=100 linear kernel, plot using svm.plot

A better estimate of accuracy

Calculate average accuracy over a number of independent train/test splits.
Check standard deviation of result to get an idea of variability.

Average accuracy for default cost SVM

accuracy <- rep(NA, 100)
set.seed(10)
for (i in 1:100) {
  sample_size <- floor(0.8 * nrow(df))
  train <- sample(seq_len(nrow(df)), size = sample_size)
  trainset <- df[train, ]
  testset <- df[-train, ]
  svm_model<- svm(y ~ ., data = trainset, type = "C-classification", cost = 1, kernel = "linear")
  pred_test <- predict(svm_model, testset)
  accuracy[i] <- mean(pred_test == testset$y)}
mean(accuracy)
sd(accuracy)

0.544
0.04273184

How well does a linear SVM perform?

Marginally better than a coin toss!
We can use our knowledge of the boundary to do much better.

Time to practice!

Support Vector Machines in R