library(e1071)
library(ggplot2)
library(ISLR2)
library(dplyr)
## 
## 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
  1. 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.
set.seed(99)
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 yaxis.
plot(x1[y==0],x2[y==0],col="red")
points(x1[y==1],x2[y==1],col="purple")

  1. Fit a logistic regression model to the data, using X1 and X2 as predictors.
data <- data.frame(x1 = x1, x2 = x2, y = as.factor(y))
log.fit <- glm( y ~ . ,data = data, family = 'binomial')
log.fit
## 
## Call:  glm(formula = y ~ ., family = "binomial", data = data)
## 
## Coefficients:
## (Intercept)           x1           x2  
##    -0.04996     -0.57916     -0.21671  
## 
## Degrees of Freedom: 499 Total (i.e. Null);  497 Residual
## Null Deviance:       692.9 
## Residual Deviance: 688.9     AIC: 694.9
  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.
prob <- predict( log.fit , newdata=data, type = 'response')
pred <-  ifelse( prob > 0.5 , 1 , 0)
ggplot(data = data, mapping = aes(x1, x2)) +
  geom_point(data = data, mapping = aes(colour = pred))

  1. 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).
log.fit2 <- glm( y ~ poly(x1,3) + poly(x2,3), data = data, family='binomial')
## Warning: glm.fit: algorithm did not converge
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
  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.
prob2 <- predict( log.fit2 , newdata=data, type = 'response')
pred2 <-  ifelse( prob2 > 0.5 , 1 , 0)
ggplot(data = data, mapping = aes(x1, x2)) +
  geom_point(data = data, mapping = aes(colour = pred2))

  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.
svm1 <- svm(y ~ . , data = data , kernel = 'linear', cost=0.01)
plot(svm1, data)

  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.
svm2 <- svm( y ~ . , data = data, kernel = 'radial' , gamma = 1)
plot(svm2, data = data)

  1. Comment on your results.

The Support Vector Machine using a radial kernel and the logistic regression with non-linear functions did a better job at classifying and setting a non-linear boundary.

  1. 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.
attach(Auto)
## The following object is masked from package:ggplot2:
## 
##     mpg
avg_mpg <- mean(Auto$mpg)
mpg_fct <- ifelse(Auto$mpg > avg_mpg, 1, 0)
Auto$mpgfct <-  as.factor(mpg_fct)
  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(99)
Auto1 <- select(Auto, -mpg)
linear <- tune(svm, mpgfct ~ . , data = Auto1, kernel='linear' , ranges = list(cost = c(0.001, 0.01, 0.1, 1, 10, 100)))
summary(linear)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##   0.1
## 
## - best performance: 0.09974359 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-03 0.12012821 0.04698834
## 2 1e-02 0.10224359 0.03837508
## 3 1e-01 0.09974359 0.03933969
## 4 1e+00 0.10230769 0.04865343
## 5 1e+01 0.12275641 0.05806720
## 6 1e+02 0.10467949 0.04281835

For the data set without the mpg variable, the value of cost with the lowest error is cost = 0.1, with an associated error of 0.09974

  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.
radial <- tune(svm ,mpgfct~.,data = Auto1, kernel="radial", ranges=list(cost=c(0.1,1,10,100),gamma=c(0.5,1,2,3,4)))
summary(radial)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##     1   0.5
## 
## - best performance: 0.07679487 
## 
## - Detailed performance results:
##     cost gamma      error dispersion
## 1    0.1   0.5 0.11000000 0.05436949
## 2    1.0   0.5 0.07679487 0.05419167
## 3   10.0   0.5 0.08185897 0.05257109
## 4  100.0   0.5 0.08185897 0.05257109
## 5    0.1   1.0 0.47448718 0.08180429
## 6    1.0   1.0 0.07685897 0.05927741
## 7   10.0   1.0 0.07679487 0.05545833
## 8  100.0   1.0 0.07679487 0.05545833
## 9    0.1   2.0 0.47448718 0.08180429
## 10   1.0   2.0 0.19884615 0.08418738
## 11  10.0   2.0 0.17589744 0.07856928
## 12 100.0   2.0 0.17589744 0.07856928
## 13   0.1   3.0 0.47448718 0.08180429
## 14   1.0   3.0 0.42352564 0.06408550
## 15  10.0   3.0 0.41076923 0.07273863
## 16 100.0   3.0 0.41076923 0.07273863
## 17   0.1   4.0 0.47448718 0.08180429
## 18   1.0   4.0 0.42858974 0.05992300
## 19  10.0   4.0 0.42608974 0.06048786
## 20 100.0   4.0 0.42608974 0.06048786

Lowest error for SVM with radial kernel= 0.07923077; and is associated with cost= 1 and alpha=0.5

poly <- tune(svm ,mpgfct~.,data = Auto1, kernel="polynomial", ranges=list(cost=c(0.1,1,10,100),degree = c(2, 3, 4)))
summary(poly)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##   100      2
## 
## - best performance: 0.390641 
## 
## - Detailed performance results:
##     cost degree     error dispersion
## 1    0.1      2 0.4746795 0.08632651
## 2    1.0      2 0.4746795 0.08632651
## 3   10.0      2 0.4746795 0.08632651
## 4  100.0      2 0.3906410 0.10917453
## 5    0.1      3 0.4746795 0.08632651
## 6    1.0      3 0.4746795 0.08632651
## 7   10.0      3 0.4746795 0.08632651
## 8  100.0      3 0.4288462 0.16307913
## 9    0.1      4 0.4746795 0.08632651
## 10   1.0      4 0.4746795 0.08632651
## 11  10.0      4 0.4746795 0.08632651
## 12 100.0      4 0.4746795 0.08632651

Lowest error for SVM with polynomial kernel= 0.4137; and is associated with cost= 100 and degree=2

SVM with radial kernel outperformed the other models with an associated min error of 0.07923077.

  1. Make some plots to back up your assertions in (b) and (c).
lin <- svm(mpgfct~., data=Auto1, kernel="linear", cost=0.1)
rad <- svm(mpgfct~., data=Auto1, kernel="radial", cost=1, gamma=0.5)
pol <- svm(mpgfct~., data=Auto1, kernel="polynomial", cost=100, degree=2)
par(mfrow=c(2,2))
plot(lin, data=Auto, mpg ~ weight)

plot(rad, data=Auto, mpg ~ weight)

plot(pol, data=Auto, mpg ~ weight)

plot(lin, data=Auto, mpg ~ horsepower)

plot(rad, data=Auto, mpg ~ horsepower)

plot(pol, data=Auto, mpg ~ horsepower)

detach(Auto)
  1. 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.
attach(OJ)
set.seed(98)
training <- sample(nrow(OJ), 800)
train <- OJ[training,]
test <-  OJ[-training,]
  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.
lin1 <- svm( Purchase ~ . ,data = train, kernel = "linear" , cost = 0.01)
summary(lin1)
## 
## Call:
## svm(formula = Purchase ~ ., data = train, kernel = "linear", cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  437
## 
##  ( 219 218 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

There are 436 support vectors for the classifier, with two classes: CH and MM. Of Those 436 vectors, 219 belong to the CH class and 217 to the MM class.

  1. What are the training and test error rates?
pred_train <-  predict( lin1, train)
matrix <-table(train$Purchase, pred_train)
matrix
##     pred_train
##       CH  MM
##   CH 447  51
##   MM  84 218
err <- 1-round(mean(train$Purchase==pred_train),5)
err
## [1] 0.16875

The training error is 0.16875

pred_test <-  predict( lin1, test)
matrix2 <-table(test$Purchase, pred_test)
matrix2
##     pred_test
##       CH  MM
##   CH 138  17
##   MM  25  90
err <- 1-round(mean(test$Purchase==pred_test),4)
err
## [1] 0.1556

The test error is 0.1556

  1. Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
set.seed(99)
tuneOJ <- tune(svm, Purchase ~ ., data = train, kernel = "linear", ranges = list(cost = c(0.01,0.1,1,10)))
summary(tuneOJ)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##   0.1
## 
## - best performance: 0.1725 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.17875 0.03729108
## 2  0.10 0.17250 0.03374743
## 3  1.00 0.17750 0.03425801
## 4 10.00 0.18250 0.03184162

Smallest error: 0.17250, occurs with a cost value of 0.10

  1. Compute the training and test error rates using this new value for cost.
lin2 <- svm(Purchase ~ ., kernel = "linear", data = train, cost = 0.10)
pred2 <-  predict(lin2, train)
table(train$Purchase, pred2)
##     pred2
##       CH  MM
##   CH 445  53
##   MM  79 223
err <- 1-round(mean(train$Purchase==pred2),4)
err
## [1] 0.165
lin3 <- svm(Purchase ~ ., kernel = "linear", data = test, cost = 0.10)
pred3 <-  predict(lin3, test)
table(test$Purchase, pred3)
##     pred3
##       CH  MM
##   CH 136  19
##   MM  17  98
err <- 1-round(mean(test$Purchase==pred3),3)
err
## [1] 0.133
  1. Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
set.seed(99)
rad1 <-  svm(Purchase ~ ., data = train, kernel = "radial")
summary(rad)
## 
## Call:
## svm(formula = mpgfct ~ ., data = Auto1, kernel = "radial", cost = 1, 
##     gamma = 0.5)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  278
## 
##  ( 137 141 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  0 1
train.rad <-  predict(rad1, train)
matrix <- table(train$Purchase, train.rad)
matrix
##     train.rad
##       CH  MM
##   CH 455  43
##   MM  85 217
err <- 1- round(mean(train$Purchase==train.rad),3)
err
## [1] 0.16
test.rad <-  predict(rad1, test)
matrix <- table(test$Purchase, test.rad)
matrix
##     test.rad
##       CH  MM
##   CH 140  15
##   MM  26  89
err <- 1- round(mean(test$Purchase==test.rad),3)
err
## [1] 0.152
svm.rad <-  svm(Purchase ~ ., data = train, kernel = "radial", cost = 0.10)
pred.train <-  predict(svm.rad, train)
table(train$Purchase, pred.train)
##     pred.train
##       CH  MM
##   CH 455  43
##   MM  96 206
err <- 1- round(mean(train$Purchase==pred.train),3)
err
## [1] 0.174
svm.rad2 <-  svm(Purchase ~ ., data = test, kernel = "radial", cost = 0.10)
pred.test <-  predict(svm.rad2, test)
table(test$Purchase, pred.test)
##     pred.test
##       CH  MM
##   CH 138  17
##   MM  41  74
err <- 1- round(mean(test$Purchase==pred.test),3)
err
## [1] 0.215
  1. Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree = 2.
set.seed(99)
poly1 <-  svm(Purchase ~ ., data = train, kernel = "radial", degree=2)
summary(poly1)
## 
## Call:
## svm(formula = Purchase ~ ., data = train, kernel = "radial", degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  374
## 
##  ( 184 190 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.poly <-  predict(poly1, train)
matrix <- table(train$Purchase, train.poly)
matrix
##     train.poly
##       CH  MM
##   CH 455  43
##   MM  85 217
err <- 1- round(mean(train$Purchase==train.poly),3)
err
## [1] 0.16
test.poly <-  predict(poly1, test)
matrix <- table(test$Purchase, test.poly)
matrix
##     test.poly
##       CH  MM
##   CH 140  15
##   MM  26  89
err <- 1- round(mean(test$Purchase==test.poly),3)
err
## [1] 0.152
set.seed(99)
tune2 <- tune(svm, Purchase ~ ., data = train, kernel = "poly", degree = 2, ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune2)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##    10
## 
## - best performance: 0.18125 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.37750 0.06286007
## 2   0.01778279 0.35625 0.06270799
## 3   0.03162278 0.35000 0.05921946
## 4   0.05623413 0.33250 0.05986095
## 5   0.10000000 0.31750 0.06157651
## 6   0.17782794 0.25750 0.05927806
## 7   0.31622777 0.21125 0.03839216
## 8   0.56234133 0.20000 0.03679900
## 9   1.00000000 0.20000 0.04409586
## 10  1.77827941 0.19750 0.04362084
## 11  3.16227766 0.18875 0.03793727
## 12  5.62341325 0.18500 0.03809710
## 13 10.00000000 0.18125 0.03830162
  1. Overall, which approach seems to give the best results on this data?

With a test error of 0.1556, the SVM with linear kernel had the best results for this data.