library(e1071)
## Warning: package 'e1071' was built under R version 4.0.5

\(\textbf{Chapter 9}\)

\(\textbf{Question 5:}\) 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:
set.seed(5)
x1 = runif(500) - 0.5
x2 = runif(500) - 0.5
y = 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.
plot(x1[y == 0], x2[y == 0], xlab = "X1", ylab = "X2", col = "gray", pch = 1)
points(x1[y == 1], x2[y == 1], col = "blue", pch = 3)

  1. Fit a logistic regression model to the data, using \(X_1\) and \(X_2\) as predictors.
fit_1 <- glm(y ~ x1 + x2, family = "binomial")
summary(fit_1)
## 
## Call:
## glm(formula = y ~ x1 + x2, family = "binomial")
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.200  -1.161  -1.131   1.190   1.223  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.03150    0.08949  -0.352    0.725
## x1          -0.06176    0.30506  -0.202    0.840
## x2          -0.11509    0.31086  -0.370    0.711
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 693.02  on 499  degrees of freedom
## Residual deviance: 692.85  on 497  degrees of freedom
## AIC: 698.85
## 
## Number of Fisher Scoring iterations: 3
  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.
df <- data.frame(x1 = x1, x2 = x2, y = y)
prediction_1 <- predict(fit_1, df, type = "response")
prediction_2 <- ifelse(prediction_1 > 0.5, 1, 0)
df_1 <- df[prediction_2 == 1,]
df_0 <- df[prediction_2 == 0,]
plot(df[prediction_2 == 0,]$x1, df[prediction_2 == 0,]$x2, xlab = "X1", ylab = "X2", col = "gray", pch = 1)
points(df[prediction_2 == 1,]$x1, df[prediction_2 == 1,]$x2, col = "blue", pch = 3)

  1. Now fit a logistic regression model to the data using non-linear functions of \(X_1\) and \(X_2\) as predictors (e.g. \(X_1^2\) 1, \(X_1\)×\(X_2\), log(\(X_2\)), and so forth).
fit_2 <- 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
summary(fit_2)
## 
## Call:
## glm(formula = y ~ poly(x1, 2) + poly(x2, 2), family = "binomial")
## 
## Deviance Residuals: 
##       Min         1Q     Median         3Q        Max  
## -0.003739   0.000000   0.000000   0.000000   0.003504  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)
## (Intercept)      252.0     3439.0   0.073    0.942
## poly(x1, 2)1    2792.7    40700.9   0.069    0.945
## poly(x1, 2)2  100456.5  1343613.6   0.075    0.940
## poly(x2, 2)1    -792.9    16782.0  -0.047    0.962
## poly(x2, 2)2  -99176.6  1325929.2  -0.075    0.940
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 6.9302e+02  on 499  degrees of freedom
## Residual deviance: 2.8367e-05  on 495  degrees of freedom
## AIC: 10
## 
## Number of Fisher Scoring iterations: 25
  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.
prediction_3 <- predict(fit_2, df, type = "response")
prediction_4 <- ifelse(prediction_3 > 0.5, 1, 0)
plot(df[prediction_4 == 0,]$x1, df[prediction_4 == 0,]$x2, xlab = "X1", ylab = "X2", col = "gray", pch = 1)
points(df[prediction_4 == 1,]$x1, df[prediction_4 == 1,]$x2, col = "blue", pch = 3)

  1. Fit a support vector classifier to the data with \(X_1\) and \(X_2\) as predictors. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.
fit_svm <- svm(as.factor(df$y) ~ x1 + x2, df, kernel = "linear", cost = .1)
predictions_svm <- predict(fit_svm, df)
plot(df[predictions_svm == 0,]$x1, df[predictions_svm == 0,]$x2, xlab = "X1", ylab = "X2", col = "gray", pch = 1)
points(df[predictions_svm == 1,]$x1, df[predictions_svm == 1,]$x2, col = "blue", pch = 3)

  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.
fit_svm_2 <- svm(as.factor(df$y) ~ x1 + x2, df, kernel = "radial", gamma = 1)
predictions_svm_2 <- predict(fit_svm_2, df)
plot(df[predictions_svm_2 == 0,]$x1, df[predictions_svm_2 == 0,]$x2, xlab = "X1", ylab = "X2", col = "gray", pch = 1)
points(df[predictions_svm_2 == 1,]$x1, df[predictions_svm_2 == 1,]$x2, col = "blue", pch = 3)

\(\textbf{Response:}\) it can be observed that SVM with a non-linear kernel and logistic regression utilizing interaction terms both can offer success in finding non-linear decision boundaries. That being said, the interaction terms I applied to logistic regression appeared to be slightly more effective than the non-linear kernel SVM in this case. On the other hand, both logistic regression without any interaction terms and the linear kernel SVM performed very poorly when tasked with identifying the non-linear decision boundary.

\(\textbf{Question 7:}\) 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.

library(ISLR)
  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
bin_variable <- ifelse(Auto$mpg > median(Auto$mpg), 1, 0)
Auto$mpglevel <- as.factor(bin_variable)
  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.
set.seed(5)
tune_1 <- tune(svm, mpglevel ~ ., data = Auto, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100)))
summary(tune_1)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.01019231 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-02 0.07391026 0.02208631
## 2 1e-01 0.04576923 0.02585417
## 3 1e+00 0.01019231 0.01786828
## 4 5e+00 0.01782051 0.02088734
## 5 1e+01 0.02294872 0.01871043
## 6 1e+02 0.03576923 0.03002696

\(\textbf{Response:}\) it can be observed that the best cost seems to be 1.

  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.
set.seed(5)
tune_2 <- tune(svm, mpglevel ~ ., data = Auto, kernel = "polynomial", ranges = list(cost = c(0.01, 0.1, 1, 5, 10), degree = c(2, 3, 4, 5)))
summary(tune_2)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##    10      2
## 
## - best performance: 0.5738462 
## 
## - Detailed performance results:
##     cost degree     error dispersion
## 1   0.01      2 0.5841026 0.04075521
## 2   0.10      2 0.5841026 0.04075521
## 3   1.00      2 0.5841026 0.04075521
## 4   5.00      2 0.5841026 0.04075521
## 5  10.00      2 0.5738462 0.03762359
## 6   0.01      3 0.5841026 0.04075521
## 7   0.10      3 0.5841026 0.04075521
## 8   1.00      3 0.5841026 0.04075521
## 9   5.00      3 0.5841026 0.04075521
## 10 10.00      3 0.5841026 0.04075521
## 11  0.01      4 0.5841026 0.04075521
## 12  0.10      4 0.5841026 0.04075521
## 13  1.00      4 0.5841026 0.04075521
## 14  5.00      4 0.5841026 0.04075521
## 15 10.00      4 0.5841026 0.04075521
## 16  0.01      5 0.5841026 0.04075521
## 17  0.10      5 0.5841026 0.04075521
## 18  1.00      5 0.5841026 0.04075521
## 19  5.00      5 0.5841026 0.04075521
## 20 10.00      5 0.5841026 0.04075521

\(\textbf{Response:}\) the best performance was given with cost of 10 and degree of 2.

set.seed(5)
tune_2 <- tune(svm, mpglevel ~ ., data = Auto, kernel = "radial", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100), gamma = c(0.01, 0.1, 1, 5, 10, 10)))
summary(tune_2)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##   100  0.01
## 
## - best performance: 0.01012821 
## 
## - Detailed performance results:
##     cost gamma      error dispersion
## 1  1e-02  0.01 0.58410256 0.04075521
## 2  1e-01  0.01 0.08416667 0.02962936
## 3  1e+00  0.01 0.07134615 0.02315997
## 4  5e+00  0.01 0.04833333 0.02175755
## 5  1e+01  0.01 0.02282051 0.02193311
## 6  1e+02  0.01 0.01012821 0.01307720
## 7  1e-02  0.10 0.23211538 0.08391233
## 8  1e-01  0.10 0.07647436 0.02060620
## 9  1e+00  0.10 0.05089744 0.02357836
## 10 5e+00  0.10 0.02044872 0.02354784
## 11 1e+01  0.10 0.01782051 0.01724506
## 12 1e+02  0.10 0.02288462 0.01870128
## 13 1e-02  1.00 0.58410256 0.04075521
## 14 1e-01  1.00 0.58410256 0.04075521
## 15 1e+00  1.00 0.06115385 0.02976796
## 16 5e+00  1.00 0.06365385 0.03620146
## 17 1e+01  1.00 0.06365385 0.03620146
## 18 1e+02  1.00 0.06365385 0.03620146
## 19 1e-02  5.00 0.58410256 0.04075521
## 20 1e-01  5.00 0.58410256 0.04075521
## 21 1e+00  5.00 0.53326923 0.05928665
## 22 5e+00  5.00 0.53070513 0.05708644
## 23 1e+01  5.00 0.53070513 0.05708644
## 24 1e+02  5.00 0.53070513 0.05708644
## 25 1e-02 10.00 0.58410256 0.04075521
## 26 1e-01 10.00 0.58410256 0.04075521
## 27 1e+00 10.00 0.55115385 0.04522834
## 28 5e+00 10.00 0.54608974 0.04883070
## 29 1e+01 10.00 0.54608974 0.04883070
## 30 1e+02 10.00 0.54608974 0.04883070
## 31 1e-02 10.00 0.58410256 0.04075521
## 32 1e-01 10.00 0.58410256 0.04075521
## 33 1e+00 10.00 0.55115385 0.04522834
## 34 5e+00 10.00 0.54608974 0.04883070
## 35 1e+01 10.00 0.54608974 0.04883070
## 36 1e+02 10.00 0.54608974 0.04883070

\(\textbf{Response:}\) in the above case, the lowest cross-validation error was given for a gamma of 0.01 and a cost of 100.

  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.

lin_svm <- svm(mpglevel ~ ., data = Auto, kernel = "linear", cost = 1)
pol_svm <- svm(mpglevel ~ ., data = Auto, kernel = "polynomial", cost = 10, degree = 2)
rad_svm <- svm(mpglevel ~ ., data = Auto, kernel = "radial", cost = 100, gamma = 0.01)
plot_function = function(fit_param) {
    for (n in names(Auto)[!(names(Auto) %in% c("mpg", "mpglevel", "name"))]) {
        plot(fit_param, Auto, as.formula(paste("mpg~", n, sep = "")))
    }
}

\(\textbf{Linear}\)

plot_function(lin_svm)

\(\textbf{Polynomial}\)

plot_function(pol_svm)

\(\textbf{Radial}\)

plot_function(rad_svm)

\(\textbf{Question 8:}\) This problem involves the OJ data set which is part of the ISLR package.

  1. Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(5)
training_sample <- sample(1:nrow(OJ), 800)
training_ds <- OJ[training_sample,]
test_ds <- OJ[-training_sample,]
  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_lin <- svm(Purchase ~ ., data = training_ds, kernel = "linear", cost = 0.01)
summary(svm_lin)
## 
## Call:
## svm(formula = Purchase ~ ., data = training_ds, kernel = "linear", 
##     cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  441
## 
##  ( 219 222 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

\(\textbf{Response:}\) 441 support vectors were created, with 219 belong to level CH and 222 belong to level MM.

  1. What are the training and test error rates?
predictions_training_ds <- predict(svm_lin, training_ds)
conf_mat <- table(training_ds$Purchase, predictions_training_ds)
(conf_mat[2, 1] + conf_mat[1, 2]) / sum(conf_mat)
## [1] 0.16625

\(\textbf{Response:}\) the training error rate is 0.16625.

predictions_test_ds <- predict(svm_lin, test_ds)
conf_mat <- table(test_ds$Purchase, predictions_test_ds)
(conf_mat[2, 1] + conf_mat[1, 2]) / sum(conf_mat)
## [1] 0.1666667

\(\textbf{Response:}\) the test error rate is 0.1666667.

  1. Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
set.seed(5)
tune_3 <- tune(svm, Purchase ~ ., data = training_ds, kernel = "linear", ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune_3)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##    10
## 
## - best performance: 0.16625 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.17625 0.05382908
## 2   0.01778279 0.17500 0.04750731
## 3   0.03162278 0.17375 0.04945888
## 4   0.05623413 0.17375 0.04767147
## 5   0.10000000 0.17375 0.04348132
## 6   0.17782794 0.17250 0.04199868
## 7   0.31622777 0.17375 0.04730589
## 8   0.56234133 0.17250 0.04923018
## 9   1.00000000 0.16750 0.04456581
## 10  1.77827941 0.16750 0.04495368
## 11  3.16227766 0.17000 0.04758034
## 12  5.62341325 0.17250 0.04241004
## 13 10.00000000 0.16625 0.03955042

\(\textbf{Response:}\) the best performance was given for a cost of 10.

  1. Compute the training and test error rates using this new value for cost.
svm_lin_2 <- svm(Purchase ~ ., kernel = "linear", data = training_ds, cost = tune_3$best.parameter$cost)
predictions_training_ds_2 <- predict(svm_lin_2, training_ds)
conf_mat <- table(training_ds$Purchase, predictions_training_ds_2)
(conf_mat[2, 1] + conf_mat[1, 2]) / sum(conf_mat)
## [1] 0.16125

\(\textbf{Response:}\) the training error rate is 0.16125.

predictions_test_ds_2 <- predict(svm_lin_2, test_ds)
conf_mat <- table(test_ds$Purchase, predictions_test_ds_2)
(conf_mat[2, 1] + conf_mat[1, 2]) / sum(conf_mat)
## [1] 0.1740741

\(\textbf{Response:}\) the test error rate is 0.1740741.

  1. Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
svm_rad <- svm(Purchase ~ ., kernel = "radial", data = training_ds)
summary(svm_rad)
## 
## Call:
## svm(formula = Purchase ~ ., data = training_ds, kernel = "radial")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  374
## 
##  ( 183 191 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

\(\textbf{Response:}\) 374 support vectors were created, with 183 belong to level CH and 191 belong to level MM.

predictions_training_ds_3 <- predict(svm_rad, training_ds)
conf_mat <- table(training_ds$Purchase, predictions_training_ds_3)
(conf_mat[2, 1] + conf_mat[1, 2]) / sum(conf_mat)
## [1] 0.15

\(\textbf{Response:}\) the training error rate is 0.15.

predictions_test_ds_3 <- predict(svm_rad, test_ds)
conf_mat <- table(test_ds$Purchase, predictions_test_ds_3)
(conf_mat[2, 1] + conf_mat[1, 2]) / sum(conf_mat)
## [1] 0.1703704

\(\textbf{Response:}\) the test error rate is 0.1703704.

set.seed(5)
tune_4 <- tune(svm, Purchase ~ ., data = training_ds, kernel = "radial", ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune_4)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  3.162278
## 
## - best performance: 0.16625 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.38750 0.04677072
## 2   0.01778279 0.38750 0.04677072
## 3   0.03162278 0.34625 0.06209592
## 4   0.05623413 0.20875 0.04332131
## 5   0.10000000 0.18875 0.04348132
## 6   0.17782794 0.18500 0.04518481
## 7   0.31622777 0.18125 0.03738408
## 8   0.56234133 0.17500 0.03004626
## 9   1.00000000 0.17375 0.02791978
## 10  1.77827941 0.17625 0.03087272
## 11  3.16227766 0.16625 0.03007514
## 12  5.62341325 0.16875 0.03186887
## 13 10.00000000 0.18125 0.03186887

\(\textbf{Response:}\) the best performance was given for a cost of 3.16227766.

svm_rad_2 <- svm(Purchase ~ ., kernel = "radial", data = training_ds, cost = tune_4$best.parameter$cost)
predictions_training_ds_4 <- predict(svm_rad_2, training_ds)
conf_mat <- table(training_ds$Purchase, predictions_training_ds_4)
(conf_mat[2, 1] + conf_mat[1, 2]) / sum(conf_mat)
## [1] 0.14375

\(\textbf{Response:}\) the training error rate is 0.14375.

predictions_test_ds_4 <- predict(svm_rad_2, test_ds)
conf_mat <- table(test_ds$Purchase, predictions_test_ds_4)
(conf_mat[2, 1] + conf_mat[1, 2]) / sum(conf_mat)
## [1] 0.1740741

\(\textbf{Response:}\) the test error rate is 0.1740741.

  1. Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree=2.
svm_pol <- svm(Purchase ~ ., kernel = "polynomial", data = training_ds, degree = 2)
summary(svm_pol)
## 
## Call:
## svm(formula = Purchase ~ ., data = training_ds, kernel = "polynomial", 
##     degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  1 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  439
## 
##  ( 216 223 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

\(\textbf{Response:}\) 439 support vectors were created, with 216 belong to level CH and 223 belong to level MM.

predictions_training_ds_5 <- predict(svm_pol, training_ds)
conf_mat <- table(training_ds$Purchase, predictions_training_ds_5)
(conf_mat[2, 1] + conf_mat[1, 2]) / sum(conf_mat)
## [1] 0.17625

\(\textbf{Response:}\) the training error rate is 0.17625.

predictions_test_ds_5 <- predict(svm_pol, test_ds)
conf_mat <- table(test_ds$Purchase, predictions_test_ds_5)
(conf_mat[2, 1] + conf_mat[1, 2]) / sum(conf_mat)
## [1] 0.2074074

\(\textbf{Response:}\) the test error rate is 0.2074074.

set.seed(5)
tune_5 <- tune(svm, Purchase ~ ., data = training_ds, kernel = "polynomial", degree = 2, ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune_5)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##    10
## 
## - best performance: 0.175 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.36625 0.05205833
## 2   0.01778279 0.35375 0.05744865
## 3   0.03162278 0.34750 0.05645795
## 4   0.05623413 0.32250 0.06061032
## 5   0.10000000 0.30875 0.05070681
## 6   0.17782794 0.25250 0.05857094
## 7   0.31622777 0.20500 0.03689324
## 8   0.56234133 0.20250 0.03809710
## 9   1.00000000 0.20250 0.04199868
## 10  1.77827941 0.19625 0.03682259
## 11  3.16227766 0.19375 0.03596391
## 12  5.62341325 0.18375 0.03387579
## 13 10.00000000 0.17500 0.03280837

\(\textbf{Response:}\) the best performance was given for a cost of 10.

svm_pol_2 <- svm(Purchase ~ ., data = training_ds, kernel = "polynomial", degree = 2, cost = tune_5$best.parameter$cost)
predictions_training_ds_6 <- predict(svm_pol_2, training_ds)
conf_mat <- table(training_ds$Purchase, predictions_training_ds_6)
(conf_mat[2, 1] + conf_mat[1, 2]) / sum(conf_mat)
## [1] 0.14125

\(\textbf{Response:}\) the training error rate is 0.14125.

predictions_test_ds_6 <- predict(svm_pol_2, test_ds)
conf_mat <- table(test_ds$Purchase, predictions_test_ds_6)
(conf_mat[2, 1] + conf_mat[1, 2]) / sum(conf_mat)
## [1] 0.1666667

\(\textbf{Response:}\) the test error rate is 0.1666667.

  1. Overall, which approach seems to give the best results on this data?

\(\textbf{Response:}\) Ultimately, it appears that the polynomial kernel with optimal cost gives the minimum error rate on both training and test data for the experiments I conducted. It should also be noted that, interestingly, the original support vector classifier from (b) using cost=0.01 also gave the same test error rate as the polynomial kernel with optimal cost.