library(ggplot2)
library(tidyverse)
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library(e1071)
library(ISLR2)
## Warning: package 'ISLR2' was built under R version 4.4.2
attach(Auto)
## The following object is masked from package:lubridate:
## 
##     origin
## 
## The following object is masked from package:ggplot2:
## 
##     mpg
attach(OJ)

5. Wehaveseen that we can fit an SVM with a non-linear kernel in orde 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.

(a) 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(38)
x1 <-runif(500)- 0.5
x2 <-runif(500)- 0.5
y <- 1 * (x1^2- x2^2 > 0)

(b) Plot the observations, colored according to their class labels. Your plot should display X1 on the x-axis, and X2 on the y axis.

df <- data.frame(x1 = x1, x2 = x2, y = as.factor(y))

ggplot(df, aes(x = x1, y = x2, color = y)) +
  geom_point() +
  labs(x = "X1", y = "X2") +
  theme_minimal()

(c) Fit a logistic regression model to the data, using X1 and X2 as predictors.

df_glm <- glm(y ~ x1 + x2, data = df, family = "binomial")
summary(df_glm)
## 
## Call:
## glm(formula = y ~ x1 + x2, family = "binomial", data = df)
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)
## (Intercept)  0.02863    0.08964   0.319    0.749
## x1          -0.36562    0.31347  -1.166    0.243
## x2           0.03616    0.32143   0.112    0.910
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 693.02  on 499  degrees of freedom
## Residual deviance: 691.64  on 497  degrees of freedom
## AIC: 697.64
## 
## Number of Fisher Scoring iterations: 3

(d) 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$pred_prob <- predict(df_glm, type = "response")
df$pred_class <- as.factor(ifelse(df$pred_prob > 0.5, 1, 0))

ggplot(df, aes(x = x1, y = x2, color = pred_class)) +
  geom_point() +
  labs(x = "X1", y = "X2", color = "Predicted Class") +
  theme_minimal()

(e) Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X21, X1×X2, log(X2), and so forth).

df_glm2 <- glm(y ~ x1 + x2 + I(x1^2) + I(x2^2) + I(x1 * x2), data = df, family = "binomial")
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
summary(df_glm2)
## 
## Call:
## glm(formula = y ~ x1 + x2 + I(x1^2) + I(x2^2) + I(x1 * x2), family = "binomial", 
##     data = df)
## 
## Coefficients:
##               Estimate Std. Error    z value Pr(>|z|)    
## (Intercept) -3.935e+14  5.494e+06  -71617558   <2e-16 ***
## x1          -1.893e+14  1.050e+07  -18023155   <2e-16 ***
## x2          -2.086e+13  1.079e+07   -1933585   <2e-16 ***
## I(x1^2)      3.555e+16  4.001e+07  888517154   <2e-16 ***
## I(x2^2)     -3.165e+16  4.149e+07 -762811342   <2e-16 ***
## I(x1 * x2)   6.597e+14  3.765e+07   17523197   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance:  693.02  on 499  degrees of freedom
## Residual deviance: 1441.75  on 494  degrees of freedom
## AIC: 1453.7
## 
## Number of Fisher Scoring iterations: 23

(f) 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.

df$pred_prob2 <- predict(df_glm2, type = "response")
df$pred_2 <- as.factor(ifelse(df$pred_prob2 > 0.5, 1, 0))

ggplot(df, aes(x = x1, y = x2, color = pred_2)) +
  geom_point() +
  labs(x = "X1", y = "X2", color = "Predicted Class") +
  theme_minimal()

(g) 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.

svc_model <- svm(y ~ x1 + x2, data = df, kernel = "linear", cost = 1)
summary(svc_model)
## 
## Call:
## svm(formula = y ~ x1 + x2, data = df, kernel = "linear", cost = 1)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  1 
## 
## Number of Support Vectors:  486
## 
##  ( 243 243 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  0 1
df$svc_pred <- predict(svc_model, newdata = df)

ggplot(df, aes(x = x1, y = x2, color = svc_pred)) +
  geom_point() +
  labs(x = "X1", y = "X2", color = "Predicted Class") +
  theme_minimal()

(h) 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.

df_svm <- svm(y ~ x1 + x2, data = df, kernel = 'radial', gamma = 1)
summary(df_svm)
## 
## Call:
## svm(formula = y ~ x1 + x2, data = df, kernel = "radial", gamma = 1)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  157
## 
##  ( 78 79 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  0 1
df$svm_pred <- predict(df_svm, newdata = df)

ggplot(df, aes(x = x1, y = x2, color = svm_pred)) +
  geom_point() +
  labs(x = "X1", y = "X2", color = "Predicted Class") +
  theme_minimal()

(i) Comment on your results.

The basic logistic model and the SVC with a linear kernel didn’t do well because their boundaries were straight, but the actual pattern in the data was curved. When we added curved terms to the logistic model or used an SVM with a radial kernel, the models did much better. The radial SVM gave the best results because it could follow the curved shape of the data.

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

str(Auto)
## 'data.frame':    392 obs. of  9 variables:
##  $ mpg         : num  18 15 18 16 17 15 14 14 14 15 ...
##  $ cylinders   : int  8 8 8 8 8 8 8 8 8 8 ...
##  $ displacement: num  307 350 318 304 302 429 454 440 455 390 ...
##  $ horsepower  : int  130 165 150 150 140 198 220 215 225 190 ...
##  $ weight      : int  3504 3693 3436 3433 3449 4341 4354 4312 4425 3850 ...
##  $ acceleration: num  12 11.5 11 12 10.5 10 9 8.5 10 8.5 ...
##  $ year        : int  70 70 70 70 70 70 70 70 70 70 ...
##  $ origin      : int  1 1 1 1 1 1 1 1 1 1 ...
##  $ name        : Factor w/ 304 levels "amc ambassador brougham",..: 49 36 231 14 161 141 54 223 241 2 ...
##  - attr(*, "na.action")= 'omit' Named int [1:5] 33 127 331 337 355
##   ..- attr(*, "names")= chr [1:5] "33" "127" "331" "337" ...

(a) 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.

median_mpg <- median(Auto$mpg)
Auto$high_mpg <- as.factor(ifelse(Auto$mpg > median_mpg, 1, 0))

(b) 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.

set.seed(38)
Auto2 <- Auto[, !(names(Auto) %in% c("mpg"))]

car_svm <- tune(svm, high_mpg ~ ., data = Auto2,
                 kernel = "linear",
                 ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100)))

summary(car_svm)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##  0.01
## 
## - best performance: 0.08916667 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-02 0.08916667 0.04669520
## 2 1e-01 0.09429487 0.04485190
## 3 1e+00 0.08923077 0.03648006
## 4 5e+00 0.10448718 0.05011219
## 5 1e+01 0.09942308 0.04867149
## 6 1e+02 0.10948718 0.04100538

After performing 10-fold cross-validation, the best SVM model used a cost of 1 and achieved the lowest error rate of 0.0891.

(c) 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(38)
car_radial <- tune(svm, high_mpg ~ ., data = Auto2,
                    kernel = "radial",
                    ranges = list(cost = c(0.1, 1, 10),
                                  gamma = c(0.1, 1, 10)))
summary(car_radial)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##    10   0.1
## 
## - best performance: 0.06365385 
## 
## - Detailed performance results:
##   cost gamma      error dispersion
## 1  0.1   0.1 0.08910256 0.04632762
## 2  1.0   0.1 0.08666667 0.04364606
## 3 10.0   0.1 0.06365385 0.03433469
## 4  0.1   1.0 0.50089744 0.16333516
## 5  1.0   1.0 0.07653846 0.04491370
## 6 10.0   1.0 0.08166667 0.05076232
## 7  0.1  10.0 0.54589744 0.04287924
## 8  1.0  10.0 0.51006410 0.06199973
## 9 10.0  10.0 0.49987179 0.05890991
set.seed(38)
car_poly <- tune(svm, high_mpg ~ ., data = Auto2,
                  kernel = "polynomial",
                  ranges = list(cost = c(0.1, 1, 10),
                                degree = c(2, 3, 4)))
summary(car_poly)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##    10      2
## 
## - best performance: 0.530641 
## 
## - Detailed performance results:
##   cost degree     error dispersion
## 1  0.1      2 0.5408974 0.05097648
## 2  1.0      2 0.5408974 0.05097648
## 3 10.0      2 0.5306410 0.07289163
## 4  0.1      3 0.5408974 0.05097648
## 5  1.0      3 0.5408974 0.05097648
## 6 10.0      3 0.5408974 0.05097648
## 7  0.1      4 0.5408974 0.05097648
## 8  1.0      4 0.5408974 0.05097648
## 9 10.0      4 0.5408974 0.05097648

The radial SVM performed nearly as well as the linear SVM, with a lowest error of 0.0636. In contrast, the polynomial SVM failed to improve predictions, showing high error rates across all settings. This suggests that a linear or radial kernel works best for this classification task.

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

Auto2$high_mpg <- as.factor(Auto2$high_mpg)

car_svm <- svm(high_mpg ~ ., data = Auto2, kernel = "radial", cost = 10, gamma = 1)

plot(car_svm, Auto2, horsepower ~ weight)

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

str(OJ)
## 'data.frame':    1070 obs. of  18 variables:
##  $ Purchase      : Factor w/ 2 levels "CH","MM": 1 1 1 2 1 1 1 1 1 1 ...
##  $ WeekofPurchase: num  237 239 245 227 228 230 232 234 235 238 ...
##  $ StoreID       : num  1 1 1 1 7 7 7 7 7 7 ...
##  $ PriceCH       : num  1.75 1.75 1.86 1.69 1.69 1.69 1.69 1.75 1.75 1.75 ...
##  $ PriceMM       : num  1.99 1.99 2.09 1.69 1.69 1.99 1.99 1.99 1.99 1.99 ...
##  $ DiscCH        : num  0 0 0.17 0 0 0 0 0 0 0 ...
##  $ DiscMM        : num  0 0.3 0 0 0 0 0.4 0.4 0.4 0.4 ...
##  $ SpecialCH     : num  0 0 0 0 0 0 1 1 0 0 ...
##  $ SpecialMM     : num  0 1 0 0 0 1 1 0 0 0 ...
##  $ LoyalCH       : num  0.5 0.6 0.68 0.4 0.957 ...
##  $ SalePriceMM   : num  1.99 1.69 2.09 1.69 1.69 1.99 1.59 1.59 1.59 1.59 ...
##  $ SalePriceCH   : num  1.75 1.75 1.69 1.69 1.69 1.69 1.69 1.75 1.75 1.75 ...
##  $ PriceDiff     : num  0.24 -0.06 0.4 0 0 0.3 -0.1 -0.16 -0.16 -0.16 ...
##  $ Store7        : Factor w/ 2 levels "No","Yes": 1 1 1 1 2 2 2 2 2 2 ...
##  $ PctDiscMM     : num  0 0.151 0 0 0 ...
##  $ PctDiscCH     : num  0 0 0.0914 0 0 ...
##  $ ListPriceDiff : num  0.24 0.24 0.23 0 0 0.3 0.3 0.24 0.24 0.24 ...
##  $ STORE         : num  1 1 1 1 0 0 0 0 0 0 ...

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

set.seed(38)
train_index <- sample(1:nrow(OJ), 800)

oj_train <- OJ[train_index, ]
oj_test <- OJ[-train_index, ]

(b) 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.

oj_svm <- svm(Purchase ~ ., data = oj_train, kernel = "linear", cost = 0.01)

summary(oj_svm)
## 
## Call:
## svm(formula = Purchase ~ ., data = oj_train, kernel = "linear", cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  442
## 
##  ( 220 222 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

The SVM used a low cost of 0.01, so it made a wide margin and allowed some mistakes. It used 442 support vectors, meaning it needed a lot of help to separate the two classes, which could mean it’s not fitting the data very tightly.

(c) What are the training and test error rates?

train_pred <- predict(oj_svm, oj_train)
test_pred <- predict(oj_svm, oj_test)

train_error <- mean(train_pred != oj_train$Purchase)
test_error <- mean(test_pred != oj_test$Purchase)

train_error
## [1] 0.16625
test_error
## [1] 0.1703704

(d) Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.

set.seed(38)

oj_tuned <- tune(svm, Purchase ~ ., data = oj_train,
                  kernel = "linear",
                  ranges = list(cost = c(0.01, 0.1, 1, 5, 10)))

summary(oj_tuned)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##  0.01
## 
## - best performance: 0.1675 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.16750 0.04338138
## 2  0.10 0.17250 0.03670453
## 3  1.00 0.17125 0.03775377
## 4  5.00 0.17125 0.03910900
## 5 10.00 0.17250 0.03717451

(e) Compute the training and test error rates using this new value for cost.

best_oj_svm <- oj_tuned$best.model

train_pred_best <- predict(best_oj_svm, oj_train)
test_pred_best  <- predict(best_oj_svm, oj_test)

train_error_best <- mean(train_pred_best != oj_train$Purchase)
test_error_best  <- mean(test_pred_best != oj_test$Purchase)

train_error_best
## [1] 0.16625
test_error_best
## [1] 0.1703704

(f) Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.

oj_rad <- svm(Purchase ~ ., data = oj_train, kernel = "radial", cost = 0.01)

summary(oj_rad)
## 
## Call:
## svm(formula = Purchase ~ ., data = oj_train, kernel = "radial", cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  0.01 
## 
## Number of Support Vectors:  611
## 
##  ( 308 303 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train_pred2 <- predict(oj_rad, oj_train)
test_pred2  <- predict(oj_rad, oj_test)

train_error2 <- mean(train_pred2 != oj_train$Purchase)
test_error2  <- mean(test_pred2 != oj_test$Purchase)

train_error2
## [1] 0.37875
test_error2
## [1] 0.4222222
set.seed(38)

oj_tuned_rad <- tune(svm, Purchase ~ ., data = oj_train,
                  kernel = "radial",
                  ranges = list(cost = c(0.01, 0.1, 1, 5, 10)))

summary(oj_tuned_rad)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.18 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.37875 0.05497790
## 2  0.10 0.18250 0.03641962
## 3  1.00 0.18000 0.02958040
## 4  5.00 0.18375 0.02766993
## 5 10.00 0.18750 0.02763854
best_oj_rad <- oj_tuned_rad$best.model

train_pred_best2 <- predict(best_oj_rad, oj_train)
test_pred_best2  <- predict(best_oj_rad, oj_test)

train_error_best2 <- mean(train_pred_best2 != oj_train$Purchase)
test_error_best2  <- mean(test_pred_best2 != oj_test$Purchase)

train_error_best2
## [1] 0.14625
test_error_best2
## [1] 0.1703704

(g) Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree = 2.

oj_poly <- svm(Purchase ~ ., data = oj_train,
                kernel = "polynomial", degree = 2, cost = 0.01)

summary(oj_poly)
## 
## Call:
## svm(formula = Purchase ~ ., data = oj_train, kernel = "polynomial", 
##     degree = 2, cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  0.01 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  611
## 
##  ( 308 303 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
mean(predict(oj_poly, oj_train) != oj_train$Purchase)
## [1] 0.37875
mean(predict(oj_poly, oj_test) != oj_test$Purchase)
## [1] 0.4222222
oj_tuned_poly <- tune(svm, Purchase ~ ., data = oj_train,
                   kernel = "polynomial", degree = 2,
                   ranges = list(cost = c(0.01, 0.1, 1, 5, 10)))

summary(oj_tuned_poly)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##    10
## 
## - best performance: 0.1875 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.37875 0.05036326
## 2  0.10 0.31625 0.07121222
## 3  1.00 0.20625 0.03294039
## 4  5.00 0.18875 0.04348132
## 5 10.00 0.18750 0.04409586
best_oj_poly <- oj_tuned_poly$best.model

mean(predict(best_oj_poly, oj_train) != oj_train$Purchase)
## [1] 0.1575
mean(predict(best_oj_poly, oj_test) != oj_test$Purchase)
## [1] 0.1851852

(h) Overall, which approach seems to give the best results on this data?

The radial SVM worked the best with the lowest test error. The polynomial one didn’t do as well. Overall, the radial SVM gave the most accurate results on this data.