Problem 9.5

We have seen that we can fit an SVM with a non-linear kernel in order to perform classification using a non-linear decision boundary. We will now see that we can also obtain a non-linear decision boundary by performing logistic regression using non-linear transformations of the features.

  1. Generate a data set with n = 500 and p = 2, such that the observations belong to two classes with a quadratic decision boundary between them. For instance, you can do this as follows:
x1 <- runif (500) - 0.5
x2 <- runif (500) - 0.5
y <- as.factor(1 * (x1^2 - x2^2 > 0))
  1. Plot the observations, colored according to their class labels. Your plot should display X1 on the x-axis, and X2 on the y axis.
library(ggplot2)
## Warning: package 'ggplot2' was built under R version 4.4.3
# data
df <- data.frame(
  x1 = x1,
  x2 = x2,
  y = y
)

# Plot
ggplot(df, aes(x = x1, y = x2, color = y)) +
  geom_point(size = 2) +
  labs(title = "x1 vs x2 by class",
       x = "x1",
       y = "x2") +
  theme_minimal()

  1. Fit a logistic regression model to the data, using X1 and X2 as predictors.
library(glmnet)
## Loading required package: Matrix
## Warning: package 'Matrix' was built under R version 4.4.3
## Loaded glmnet 4.1-8
log <- glm(y ~ x1 + x2, family = 'binomial')
  1. Apply this model to the training data in order to obtain a predicted class label for each training observation. Plot the observations, colored according to the predicted class labels. The decision boundary should be linear.

  1. Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X1^2 , X1 * X2, log(X2), and so forth).

We can repeat the above by using a polynomial of degree 2 for both x1 and x2:

# poly <- glm(y ~ x1 + x2 + poly(x1,3) + poly(x2,3), family = 'binomial')
poly <- glm(y ~ poly(x1, 2) + poly(x2, 2), family = 'binomial')
## Warning: glm.fit: algorithm did not converge
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
poly_preds <- predict(poly, newdata = df[,-3], type='response')
poly_preds <- as.factor(ifelse(poly_preds >= 0.5, 1, 0))
  1. Apply this model to the training data in order to obtain a predicted class label for each training observation. Plot the observations, colored according to the predicted class labels. The decision boundary should be obviously non-linear. If it is not, then repeat (a)-(e) until you come up with an example in which the predicted class labels are obviously non-linear.

We can plot them each as follows:

# predicted data
df <- data.frame(
  x1 = x1,
  x2 = x2,
  y = poly_preds
)

# Plot
ggplot(df, aes(x = x1, y = x2, color = y)) +
  geom_point(size = 2) +
  labs(title = "x1 vs x2 by predicted class for polynomial of degree 2",
       x = "x1",
       y = "x2") +
  theme_minimal()

  1. Fit a support vector classifier to the data with X1 and X2 as predictors. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.
library(e1071)
## Warning: package 'e1071' was built under R version 4.4.3
svm <- svm(y ~ x1 + x2)
svm_preds <- predict(svm, newdata = df[,-3], type = 'response')

# predicted data
df <- data.frame(
  x1 = x1,
  x2 = x2,
  y = svm_preds
)

# Plot
ggplot(df, aes(x = x1, y = x2, color = y)) +
  geom_point(size = 2) +
  labs(title = "x1 vs x2 by predicted class for SVM",
       x = "x1",
       y = "x2") +
  theme_minimal()

  1. Fit a SVM using a non-linear kernel to the data. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.
svm_radial <- svm(y ~ x1 + x2, kernel ='radial')
svm_radial_preds <- predict(svm_radial, newdata = df[,-3], type = 'response')

# predicted data
df <- data.frame(
  x1 = x1,
  x2 = x2,
  y = svm_preds
)

# Plot
ggplot(df, aes(x = x1, y = x2, color = y)) +
  geom_point(size = 2) +
  labs(title = "x1 vs x2 by predicted class for SVM with radial kernel",
       x = "x1",
       y = "x2") +
  theme_minimal()

  1. Comment on your results.

First we will compare the results of each method together:

## SVM (linear kernel) Accuracy: 0.98
## SVM (radial kernel) Accuracy: 0.98
## Logistic Regression Accuracy: 0.394
## Polynomial Logistic Regression Accuracy: 1

We see that the best performance is with logistic regression using polynomial terms, which gives a perfect fit. It is important to note that this is likely overfitting as it is on the training data, however. SVM using either radial or linear kernel gives a near perfect performance without any tuning, showing that they are quite good classifiers out of the box. In practical terms, SVM is the likely winner in this case despite the perfect performance of the polynomial logistic regression.

Problem 9.7

In this problem, you will use support vector approaches in order to predict whether a given car gets high or low gas mileage based on the Auto data set.

  1. Create a binary variable that takes on a 1 for cars with gas mileage above the median, and a 0 for cars with gas mileage below the median.
library(ISLR2)

auto <- Auto
auto$higher_mileage <- as.factor(ifelse(Auto$mpg > median(Auto$mpg), 1, 0))
auto <- auto[,-1] # removing mpg
table(auto$higher_mileage)
## 
##   0   1 
## 196 196
  1. Fit a support vector classifier to the data with various values of cost, in order to predict whether a car gets high or low gas mileage. Report the cross-validation errors associated with different values of this parameter. Comment on your results. Note you will need to fit the classifier without the gas mileage variable to produce sensible results.

We can perform cross validation, and see the results below:

set.seed(123)

tuning_linear <- tune(
  "svm",                                 
  higher_mileage ~ .,                   
  data = auto,                          
  kernel = "linear",                   
  ranges = list(cost = c(0.001, 0.01, 0.1, 1, 5, 10, 100))  # cost parameters
)

summary(tuning_linear)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##  0.01
## 
## - best performance: 0.08910256 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-03 0.13250000 0.06630411
## 2 1e-02 0.08910256 0.03791275
## 3 1e-01 0.09403846 0.04842472
## 4 1e+00 0.09147436 0.05817937
## 5 5e+00 0.10423077 0.06425850
## 6 1e+01 0.10673077 0.06562804
## 7 1e+02 0.12724359 0.06371052

We find the best performance using a cost of 1e-02, with an error of 0.0891 and a dispersion of 0.0379.

  1. Now repeat (b), this time using SVMs with radial and polynomial basis kernels, with different values of gamma and degree and cost. Comment on your results.

We can do tuning on the radial kernel by adding in gamma as a tuning parameter:

set.seed(123)

tuning_radial <- tune(
  "svm",
  higher_mileage ~ .,
  data = auto,
  kernel = "radial",
  ranges = list(
    cost = c(0.001, 0.01, 0.1, 1, 5, 10, 100),
    gamma = c(0.01, 0.1, 1)
  )
)

summary(tuning_radial)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##     5   0.1
## 
## - best performance: 0.07365385 
## 
## - Detailed performance results:
##     cost gamma      error dispersion
## 1  1e-03  0.01 0.58173077 0.04740051
## 2  1e-02  0.01 0.58173077 0.04740051
## 3  1e-01  0.01 0.11211538 0.05486624
## 4  1e+00  0.01 0.08910256 0.03791275
## 5  5e+00  0.01 0.08397436 0.03933133
## 6  1e+01  0.01 0.09403846 0.04842472
## 7  1e+02  0.01 0.09147436 0.05690990
## 8  1e-03  0.10 0.58173077 0.04740051
## 9  1e-02  0.10 0.26000000 0.09153493
## 10 1e-01  0.10 0.08910256 0.03979296
## 11 1e+00  0.10 0.08653846 0.03578151
## 12 5e+00  0.10 0.07365385 0.05975763
## 13 1e+01  0.10 0.07365385 0.05852240
## 14 1e+02  0.10 0.10442308 0.05095556
## 15 1e-03  1.00 0.58173077 0.04740051
## 16 1e-02  1.00 0.58173077 0.04740051
## 17 1e-01  1.00 0.58173077 0.04740051
## 18 1e+00  1.00 0.08660256 0.04466420
## 19 5e+00  1.00 0.08916667 0.05328055
## 20 1e+01  1.00 0.08916667 0.05328055
## 21 1e+02  1.00 0.08916667 0.05328055

Using the radial kernel we see the best results with a cost of 10 and a gamma of 0.10, with an error of 0.0736 and a dispersion of 0.0585. So far this is the best performing model.

We can then do the same with the polynomial kernel, but using degree instead of gamma:

set.seed(123)

tuning_polynomial <- tune(
  "svm",
  higher_mileage ~ .,
  data = auto,
  kernel = "polynomial",
  ranges = list(
    cost = c(0.001, 0.01, 0.1, 1, 5, 10, 100),
    degree = c(2, 3, 4)
  )
)

summary(tuning_polynomial)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##   100      2
## 
## - best performance: 0.3086538 
## 
## - Detailed performance results:
##     cost degree     error dispersion
## 1  1e-03      2 0.5817308 0.04740051
## 2  1e-02      2 0.5817308 0.04740051
## 3  1e-01      2 0.5817308 0.04740051
## 4  1e+00      2 0.5817308 0.04740051
## 5  5e+00      2 0.5817308 0.04740051
## 6  1e+01      2 0.5714744 0.04575370
## 7  1e+02      2 0.3086538 0.10382736
## 8  1e-03      3 0.5817308 0.04740051
## 9  1e-02      3 0.5817308 0.04740051
## 10 1e-01      3 0.5817308 0.04740051
## 11 1e+00      3 0.5817308 0.04740051
## 12 5e+00      3 0.5817308 0.04740051
## 13 1e+01      3 0.5817308 0.04740051
## 14 1e+02      3 0.4159615 0.12008716
## 15 1e-03      4 0.5817308 0.04740051
## 16 1e-02      4 0.5817308 0.04740051
## 17 1e-01      4 0.5817308 0.04740051
## 18 1e+00      4 0.5817308 0.04740051
## 19 5e+00      4 0.5817308 0.04740051
## 20 1e+01      4 0.5817308 0.04740051
## 21 1e+02      4 0.5817308 0.04740051

Interestingly enough we see an almost identical performance across the board, with most models having an error of 0.5817 and dispersion of 0.0474. The best performing model is using a cost of 100 and degree of 2 with an error of 0.3086 and a dispersion of 0.1038.

  1. Make some plots to back up your assertions in (b) and (c).

Hint: In the lab, we used the plot() function for svm objects only in cases with p = 2. When p > 2, you can use the plot() function to create plots displaying pairs of variables at a time. Essentially, instead of typing

plot(svmfit , dat)

where svmfit contains your fitted model and dat is a data frame containing your data, you can type

plot(svmfit , dat , x1 ~ x4)

in order to plot just the first and fourth variables. However, you must replace x1 and x4 with the correct variable names. To find out more, type ?plot.svm.

We can compare the errors of each of these in a cross validation plot each kernel:

## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

We see that as stated above the best model was a radial kernel with a gamma of 0.1 and cost of 10. We will train this best model and then make a few different plots:

We see that we cannot provide an easily visible decision boundary using these variables, due to the high dimensional nature of the data.

Problem 9.8

This problem involves the OJ data set which is part of the ISLR2 package.

  1. Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(123) 

n <- nrow(OJ)
train_idx <- sample(1:n, size = 800)

# Create training and test sets
train <- OJ[train_idx, ]
test  <- OJ[-train_idx, ]
  1. Fit a support vector classifier to the training data using cost = 0.01, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics, and describe the results obtained.
svm_linear <- svm(Purchase ~ ., data = train, cost = 0.01)
summary(svm_linear)
## 
## Call:
## svm(formula = Purchase ~ ., data = train, cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  0.01 
## 
## Number of Support Vectors:  629
## 
##  ( 313 316 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

We see 313 support vectors in class CH, and 316 in class MM.

  1. What are the training and test error rates?
## Training Accuracy: 0.60875
## Test Accuracy: 0.6148148
  1. Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
set.seed(123)
tuning <- tune('svm', Purchase ~ ., data = train, kernel = 'linear', ranges = list(cost = seq(0.01, 0.1, by = 0.01)))
summary(tuning)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##  0.04
## 
## - best performance: 0.17 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.17375 0.04910660
## 2  0.02 0.17250 0.04816061
## 3  0.03 0.17250 0.04887626
## 4  0.04 0.17000 0.04758034
## 5  0.05 0.17125 0.04678927
## 6  0.06 0.17125 0.04678927
## 7  0.07 0.17125 0.04678927
## 8  0.08 0.17250 0.04632314
## 9  0.09 0.17250 0.04669642
## 10 0.10 0.17500 0.04823265

We see the best model is at cost 0.03.

best_model <- tuning$best.model
  1. Compute the training and test error rates using this new value for cost.
## Training Accuracy: 0.60875
## Test Accuracy: 0.6148148
  1. Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
set.seed(123)
tuning_radial <- tune( 'svm', Purchase ~ ., data = train, kernel = 'radial', ranges = list(cost = seq(0.01, 0.1, by = 0.01)))
summary(tuning_radial)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##   0.1
## 
## - best performance: 0.17625 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.39125 0.04411554
## 2  0.02 0.39125 0.04411554
## 3  0.03 0.38625 0.04466309
## 4  0.04 0.26375 0.06079622
## 5  0.05 0.22000 0.06851602
## 6  0.06 0.19500 0.06351946
## 7  0.07 0.18625 0.06108112
## 8  0.08 0.17750 0.06003471
## 9  0.09 0.17750 0.05945353
## 10 0.10 0.17625 0.06108112

Here we see the best model at cost 0.08, which we will continue with and calculate the accuracy for:

## Training Accuracy: 0.84
## Test Accuracy: 0.8037037
  1. Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree = 2.
set.seed(123)
tuning_polynomial <- tune( 'svm', Purchase ~ ., data = train, kernel = 'polynomial', degree = 2, ranges = list(cost = seq(0.01, 0.1, by = 0.01)))
summary(tuning_polynomial)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##   0.1
## 
## - best performance: 0.31 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.39000 0.04281744
## 2  0.02 0.36250 0.04930066
## 3  0.03 0.35750 0.05210833
## 4  0.04 0.35375 0.05272110
## 5  0.05 0.33500 0.04816061
## 6  0.06 0.31875 0.05077524
## 7  0.07 0.31500 0.05296750
## 8  0.08 0.31375 0.05573063
## 9  0.09 0.31375 0.05573063
## 10 0.10 0.31000 0.05361903

Here we see the best model at cost 0.08, which we will continue with and calculate the accuracy for:

## Training Accuracy: 0.71125
## Test Accuracy: 0.6962963
  1. Overall, which approach seems to give the best results on this data?

We can compare the test accuracy for all the models:

## Linear Accuracy: 0.6148148
## Radial Accuracy: 0.8037037
## Polynomial Accuracy: 0.6962963

We see the best performing model to be SVM with a radial kernel, with an accuracy of 80.4%.