Chapter 9: 5,7,8

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<-1*(x12-x22>0)
set.seed(8)
x1 = runif(500) - 0.5
x2 = runif(500) - 0.5
y = as.integer(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 = "green", xlab = "X1", ylab = "X2")
points(x1[y == 1], x2[y == 1], col = "purple")

  1. Fit a logistic regression model to the data, using X1 and X2 as predictors.
dat = data.frame(x1 = x1, x2 = x2, y = as.factor(y))
lr.fit = glm(y ~ ., data = dat, 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.
lr.prob = predict(lr.fit, newdata = dat, type = 'response')
lr.pred = ifelse(lr.prob > 0.5, 1, 0)
plot(dat$x1, dat$x2, col = lr.pred + 2)

  1. Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X2 1, X1×X2, log(X2), and so forth).
lr.nl = glm(y ~ poly(x1, 2) + poly(x2, 2), data = dat, family = 'binomial')
## Warning: glm.fit: algorithm did not converge
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
summary(lr.nl)
## 
## Call:
## glm(formula = y ~ poly(x1, 2) + poly(x2, 2), family = "binomial", 
##     data = dat)
## 
## Deviance Residuals: 
##        Min          1Q      Median          3Q         Max  
## -1.171e-03  -2.000e-08   2.000e-08   2.000e-08   1.201e-03  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)
## (Intercept)      32.11    1699.25   0.019    0.985
## poly(x1, 2)1    461.37   68674.33   0.007    0.995
## poly(x1, 2)2  26770.48  858491.07   0.031    0.975
## poly(x2, 2)1   1207.17   79361.74   0.015    0.988
## poly(x2, 2)2 -27213.11  875178.35  -0.031    0.975
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 6.9179e+02  on 499  degrees of freedom
## Residual deviance: 3.3839e-06  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.
lr.prob.nl = predict(lr.nl, newdata = dat, type = 'response')
lr.pred.nl = ifelse(lr.prob.nl > 0.5, 1, 0)
plot(dat$x1, dat$x2, col = lr.pred.nl + 2)

  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)
svm.lin = svm(y ~ ., data = dat, kernel = 'linear', cost = 0.01)
plot(svm.lin, dat)

  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.nl = svm(y ~ ., data = dat, kernel = 'radial', gamma = 1)
plot(svm.nl, data = dat)

  1. Comment on your results.

SVM with non-linear kernel and logistic regression with interaction terms are both very effective for finding non-linear decision boundaries. SVM with linear kernel and logistic regression without any interaction term are not very useful when it comes to finding non-linear decision boundaries. One argument in favor of SVM is that when using logistic regression, it requires some manual tuning to find the right interaction terms, although when using SVM only the gamma needs to be tuned.

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(ISLR)
## Warning: package 'ISLR' was built under R version 4.1.3
gas.med = median(Auto$mpg)
new.var = ifelse(Auto$mpg > gas.med, 1, 0)
Auto$mpglevel = as.factor(new.var)
  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.
library(e1071)
set.seed(9)
tune.out = tune(svm, mpglevel ~ ., data = Auto, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100)))
summary(tune.out)
## 
## 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 0.07653846 0.05100638
## 2 1e-01 0.05102564 0.03586588
## 3 1e+00 0.01019231 0.01315951
## 4 5e+00 0.02044872 0.01619554
## 5 1e+01 0.02044872 0.01619554
## 6 1e+02 0.03839744 0.03872235
  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(77)
tune.out = tune(svm, mpglevel ~ ., data = Auto, kernel = "polynomial", ranges = list(cost = c(0.1, 1, 5, 10), degree = c(2, 3, 4)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##    10      2
## 
## - best performance: 0.5096795 
## 
## - Detailed performance results:
##    cost degree     error dispersion
## 1   0.1      2 0.5635256 0.05199157
## 2   1.0      2 0.5635256 0.05199157
## 3   5.0      2 0.5635256 0.05199157
## 4  10.0      2 0.5096795 0.11240472
## 5   0.1      3 0.5635256 0.05199157
## 6   1.0      3 0.5635256 0.05199157
## 7   5.0      3 0.5635256 0.05199157
## 8  10.0      3 0.5635256 0.05199157
## 9   0.1      4 0.5635256 0.05199157
## 10  1.0      4 0.5635256 0.05199157
## 11  5.0      4 0.5635256 0.05199157
## 12 10.0      4 0.5635256 0.05199157
tune.out = tune(svm, mpglevel ~ ., data = Auto, kernel = "radial", ranges = list(cost = c(0.1, 1, 5, 10), gamma = c(0.01, 0.1, 1, 5, 10, 100)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##    10  0.01
## 
## - best performance: 0.02557692 
## 
## - Detailed performance results:
##    cost gamma      error dispersion
## 1   0.1 1e-02 0.08955128 0.05590133
## 2   1.0 1e-02 0.07416667 0.05062800
## 3   5.0 1e-02 0.05371795 0.04603578
## 4  10.0 1e-02 0.02557692 0.02093679
## 5   0.1 1e-01 0.07673077 0.05142612
## 6   1.0 1e-01 0.05628205 0.04500063
## 7   5.0 1e-01 0.02814103 0.02246020
## 8  10.0 1e-01 0.03326923 0.03211170
## 9   0.1 1e+00 0.56365385 0.04979325
## 10  1.0 1e+00 0.06391026 0.03485206
## 11  5.0 1e+00 0.06647436 0.03680434
## 12 10.0 1e+00 0.06647436 0.03680434
## 13  0.1 5e+00 0.56365385 0.04979325
## 14  1.0 5e+00 0.51512821 0.05521196
## 15  5.0 5e+00 0.51000000 0.05649372
## 16 10.0 5e+00 0.51000000 0.05649372
## 17  0.1 1e+01 0.56365385 0.04979325
## 18  1.0 1e+01 0.52788462 0.05428129
## 19  5.0 1e+01 0.52532051 0.05013610
## 20 10.0 1e+01 0.52532051 0.05013610
## 21  0.1 1e+02 0.56365385 0.04979325
## 22  1.0 1e+02 0.56365385 0.04979325
## 23  5.0 1e+02 0.56365385 0.04979325
## 24 10.0 1e+02 0.56365385 0.04979325
  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.
svm.linear = svm(mpglevel ~ ., data = Auto, kernel = "linear", cost = 1)
svm.poly = svm(mpglevel ~ ., data = Auto, kernel = "polynomial", cost = 10, 
    degree = 2)
svm.radial = svm(mpglevel ~ ., data = Auto, kernel = "radial", cost = 10, gamma = 0.01)
plotpairs = function(fit) {
    for (name in names(Auto)[!(names(Auto) %in% c("mpg", "mpglevel", "name"))]) {
        plot(fit, Auto, as.formula(paste("mpg~", name, sep = "")))
    }
}
plotpairs(svm.linear)

plotpairs(svm.poly)

plotpairs(svm.radial)

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.
library(ISLR)
set.seed(78)
train = sample(dim(OJ)[1], 800)
OJ.train = OJ[train, ]
OJ.test = OJ[-train, ]
  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.
library(e1071)
svm.linear = svm(Purchase ~ ., kernel = "linear", data = OJ.train, cost = 0.01)
summary(svm.linear)
## 
## 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:  434
## 
##  ( 218 216 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
  1. What are the training and test error rates?
train.pred = predict(svm.linear, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 422  53
##   MM  75 250
(75+53)/(422+53+75+250)
## [1] 0.16
test.pred = predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 150  28
##   MM  22  70
(28+22)/(150+28+22+70)
## [1] 0.1851852

Training test error rate is 16%. Testing test error rate is 18.5%.

  1. Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
set.seed(777)
tune.out = tune(svm, Purchase ~ ., data = OJ.train, kernel = "linear", ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##        cost
##  0.03162278
## 
## - best performance: 0.15875 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.16625 0.03775377
## 2   0.01778279 0.16250 0.03773077
## 3   0.03162278 0.15875 0.03682259
## 4   0.05623413 0.16250 0.03773077
## 5   0.10000000 0.16500 0.03622844
## 6   0.17782794 0.16500 0.03374743
## 7   0.31622777 0.16375 0.03747684
## 8   0.56234133 0.16625 0.03729108
## 9   1.00000000 0.16625 0.03682259
## 10  1.77827941 0.16500 0.03670453
## 11  3.16227766 0.16000 0.03525699
## 12  5.62341325 0.16000 0.03763863
## 13 10.00000000 0.16250 0.03435921
  1. Compute the training and test error rates using this new value for cost.
svm.linear = svm(Purchase ~ ., kernel = "linear", data = OJ.train, cost = tune.out$best.parameters$cost)
train.pred = predict(svm.linear, OJ.train)

table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 417  58
##   MM  68 257
(68+58)/(417+58+68+257)
## [1] 0.1575
test.pred = predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 147  31
##   MM  21  71
(21+31)/(147+31+21+71)
## [1] 0.1925926

Training test error rate is 15.75%. Testing test error rate is 19.26%.

  1. Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
set.seed(1604)
svm.radial = svm(Purchase ~ ., data = OJ.train, kernel = "radial")
summary(svm.radial)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "radial")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  363
## 
##  ( 183 180 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.pred = predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 438  37
##   MM  80 245
(37+80)/(438+37+80+245)
## [1] 0.14625
test.pred = predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 152  26
##   MM  25  67
(25+26)/(152+26+25+67)
## [1] 0.1888889

Training test error rate is 14.6%. Testing test error rate is 18.9%.

set.seed(110)
tune.out = tune(svm, Purchase ~ ., data = OJ.train, kernel = "radial", ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##       cost
##  0.5623413
## 
## - best performance: 0.16375 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.40625 0.06488505
## 2   0.01778279 0.40625 0.06488505
## 3   0.03162278 0.29375 0.06488505
## 4   0.05623413 0.19625 0.04411554
## 5   0.10000000 0.18750 0.04526159
## 6   0.17782794 0.17500 0.05400617
## 7   0.31622777 0.17000 0.05627314
## 8   0.56234133 0.16375 0.04945888
## 9   1.00000000 0.16750 0.04866267
## 10  1.77827941 0.16875 0.04938862
## 11  3.16227766 0.17375 0.05118390
## 12  5.62341325 0.17875 0.05070681
## 13 10.00000000 0.18875 0.05219155
svm.radial = svm(Purchase ~ ., data = OJ.train, kernel = "radial", cost = tune.out$best.parameters$cost)
train.pred = predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 437  38
##   MM  82 243
(38+82)/(437+38+82+243)
## [1] 0.15
test.pred = predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 150  28
##   MM  24  68
(28+24)/(150+28+24+68)
## [1] 0.1925926

Training test error rate is 15%. Testing test error rate is 19.3%.

  1. Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree = 2.
set.seed(2022)
svm.poly = svm(Purchase ~ ., data = OJ.train, kernel = "poly", degree = 2)
summary(svm.poly)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "poly", degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  1 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  441
## 
##  ( 225 216 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.pred = predict(svm.poly, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 448  27
##   MM 112 213
(27+112)/(448+27+112+213)
## [1] 0.17375
test.pred = predict(svm.poly, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 156  22
##   MM  40  52
(22+40)/(156+22+40+52)
## [1] 0.2296296

Training test error rate is 17.4%. Testing test error rate is 23.0%

set.seed(1883)
tune.out = tune(svm, Purchase ~ ., data = OJ.train, kernel = "poly", degree = 2, 
    ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  3.162278
## 
## - best performance: 0.17125 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.37875 0.05304937
## 2   0.01778279 0.37000 0.05407043
## 3   0.03162278 0.34625 0.05529278
## 4   0.05623413 0.32125 0.05894029
## 5   0.10000000 0.30250 0.05552777
## 6   0.17782794 0.22250 0.06118052
## 7   0.31622777 0.19375 0.05008673
## 8   0.56234133 0.19000 0.04594683
## 9   1.00000000 0.18750 0.05527708
## 10  1.77827941 0.18250 0.05439056
## 11  3.16227766 0.17125 0.05894029
## 12  5.62341325 0.17125 0.05560588
## 13 10.00000000 0.17375 0.05252314
svm.poly = svm(Purchase ~ ., data = OJ.train, kernel = "poly", degree = 2, cost = tune.out$best.parameters$cost)
train.pred = predict(svm.poly, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 445  30
##   MM  92 233
(30+92)/(445+30+92+233)
## [1] 0.1525
test.pred = predict(svm.poly, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 156  22
##   MM  30  62
(22+30)/(156+22+30+62)
## [1] 0.1925926

Training test error rate is 15.3%. Testing test error rate is 19.3%.

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

Radial appears to be the best option with lowest error rates.