Chapter 9: 5, 7, 8

Question 5

We have seen that we can fit an SVM with a non-linear kernel in order to perfrom 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 dataset with n=500 and p=2, such that the observations belong to two classes with a quadratic decision boudnary between them. For instance, you can do as this as follows:

set.seed(10)
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.

plot(x1, x2, xlab = "X1", ylab = "X2", col=(4-y), pch = (3-y))

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

logit.fit = glm(y~ x1 + x2, family = "binomial")
summary(logit.fit)
## 
## Call:
## glm(formula = y ~ x1 + x2, family = "binomial")
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.347  -1.152   1.024   1.166   1.297  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)  
## (Intercept)  0.02449    0.08988   0.272   0.7853  
## x1           0.23573    0.31281   0.754   0.4511  
## x2          -0.52372    0.30367  -1.725   0.0846 .
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 693.12  on 499  degrees of freedom
## Residual deviance: 689.57  on 497  degrees of freedom
## AIC: 695.57
## 
## 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

data = data.frame(x1 = x1, x2=x2, y=y)
probs = predict(logit.fit, data, type = "response")
preds = rep(0, 500)
preds[probs > 0.47] = 1
plot(data[preds == 1, ]$x1, data[preds ==1, ]$x2, col = (4-1), pch = (3-1), xlab = "X1", ylab = "X2")
points(data[preds == 0, ]$x1, data[preds == 0, ]$x2, col = (4-0), pch = (3-0))

e) Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors.

logitnl.fit = glm(y~ poly(x1,2), + poly(x2, 2) + I(x1*x2), family = "binomial")
summary(logitnl.fit)
## 
## Call:
## glm(formula = y ~ poly(x1, 2), family = "binomial", data = +poly(x2, 
##     2) + I(x1 * x2))
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.4422  -0.7503   0.2098   0.7913   1.8645  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)    
## (Intercept)    0.1438     0.1113   1.292    0.197    
## poly(x1, 2)1   2.9568     2.6618   1.111    0.267    
## poly(x1, 2)2  33.1878     3.1146  10.656   <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: 693.12  on 499  degrees of freedom
## Residual deviance: 518.87  on 497  degrees of freedom
## AIC: 524.87
## 
## Number of Fisher Scoring iterations: 4

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.

probs = predict(logitnl.fit, data, type = "response")
preds = rep(0, 500)
preds[probs > 0.47] = 1
plot(data[preds ==1, ]$x1, data[preds == 1, ]$x2, col = (4-1), pch = (3-1), xlab = "X1", ylab = "X2")
points(data[preds == 0, ]$x1, data[preds == 0, ]$x2, col = (4-0), pch = (3-0))

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.

library(e1071)
## Warning: package 'e1071' was built under R version 4.0.5
data$y = as.factor(data$y)
svm.fit = svm(y~ x1 + x2, data, kernal="linear", cost = 0.01)
preds = predict(svm.fit, data)
plot(data[preds == 1, ]$x1, data[preds == 1, ]$x2, col = (4 - 1), pch = (3 - 1), xlab = "X1", ylab = "X2")
points(data[preds == 0, ]$x1, data[preds == 0, ]$x2, col = (4 - 0), pch = (3 - 0))

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.

data$y = as.factor(data$y)
svmnl.fit = svm(y~ x1+ x2, data, kernel = "radial", gamma = 1)
preds = predict(svmnl.fit, data)
plot(data[preds ==1, ]$x1, data[preds ==1, ]$x2, col = (4 - 1), pch = (3 - 1), xlab = "X1", ylab = "X2")
points(data[preds ==0, ]$x1, data[preds ==0, ]$x2, col = (4 - 0), pch = (3 - 0), xlab = "X1", ylab = "X2")

i) Comment on your results

  • SVM with non-linear kernel and logistic regression with interaction terms are equally powerful for finding non-linear decision boundaries. Using linear SVM and logistic regression without interaction terms did not produce a good model in terms of non-linear decision boundaries.

Question 7

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

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

library(ISLR)
## Warning: package 'ISLR' was built under R version 4.0.5
var = ifelse(Auto$mpg > median(Auto$mpg), 1, 0)
Auto$mpglevel = as.factor(var)

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.

set.seed(10)
tune.out = tune(svm, mpglevel~., data = Auto, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100, 1000)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.007628205 
## 
## - Detailed performance results:
##    cost       error dispersion
## 1 1e-02 0.073974359 0.03067293
## 2 1e-01 0.050897436 0.03153169
## 3 1e+00 0.007628205 0.01228382
## 4 5e+00 0.015256410 0.01784872
## 5 1e+01 0.020384615 0.02019157
## 6 1e+02 0.033205128 0.01737168
## 7 1e+03 0.033205128 0.01737168

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

tune.out = tune(svm, mpglevel ~., data = Auto, kernel = "polynomial", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100), degree = c(2, 3, 4)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##   100      2
## 
## - best performance: 0.2983974 
## 
## - Detailed performance results:
##     cost degree     error dispersion
## 1  1e-02      2 0.5560256 0.03078585
## 2  1e-01      2 0.5560256 0.03078585
## 3  1e+00      2 0.5560256 0.03078585
## 4  5e+00      2 0.5560256 0.03078585
## 5  1e+01      2 0.5205769 0.07526613
## 6  1e+02      2 0.2983974 0.04882958
## 7  1e-02      3 0.5560256 0.03078585
## 8  1e-01      3 0.5560256 0.03078585
## 9  1e+00      3 0.5560256 0.03078585
## 10 5e+00      3 0.5560256 0.03078585
## 11 1e+01      3 0.5560256 0.03078585
## 12 1e+02      3 0.3339744 0.07186352
## 13 1e-02      4 0.5560256 0.03078585
## 14 1e-01      4 0.5560256 0.03078585
## 15 1e+00      4 0.5560256 0.03078585
## 16 5e+00      4 0.5560256 0.03078585
## 17 1e+01      4 0.5560256 0.03078585
## 18 1e+02      4 0.5560256 0.03078585
  • For polynomial kernel, the lowest cross-validation error is obtained for a degree of 2 and cost of 100.
tune.out = 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, 100)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##   100  0.01
## 
## - best performance: 0.01282051 
## 
## - Detailed performance results:
##     cost gamma      error dispersion
## 1  1e-02 1e-02 0.56128205 0.04043851
## 2  1e-01 1e-02 0.09192308 0.02504871
## 3  1e+00 1e-02 0.07397436 0.02546808
## 4  5e+00 1e-02 0.04583333 0.03565912
## 5  1e+01 1e-02 0.01788462 0.01727588
## 6  1e+02 1e-02 0.01282051 0.01351401
## 7  1e-02 1e-01 0.21955128 0.07414576
## 8  1e-01 1e-01 0.07653846 0.02704017
## 9  1e+00 1e-01 0.05858974 0.03175102
## 10 5e+00 1e-01 0.03051282 0.02878864
## 11 1e+01 1e-01 0.02801282 0.02509397
## 12 1e+02 1e-01 0.03314103 0.02067257
## 13 1e-02 1e+00 0.56128205 0.04043851
## 14 1e-01 1e+00 0.56128205 0.04043851
## 15 1e+00 1e+00 0.06628205 0.03439069
## 16 5e+00 1e+00 0.06378205 0.03019594
## 17 1e+01 1e+00 0.06378205 0.03019594
## 18 1e+02 1e+00 0.06378205 0.03019594
## 19 1e-02 5e+00 0.56128205 0.04043851
## 20 1e-01 5e+00 0.56128205 0.04043851
## 21 1e+00 5e+00 0.50782051 0.07016412
## 22 5e+00 5e+00 0.50525641 0.07145882
## 23 1e+01 5e+00 0.50525641 0.07145882
## 24 1e+02 5e+00 0.50525641 0.07145882
## 25 1e-02 1e+01 0.56128205 0.04043851
## 26 1e-01 1e+01 0.56128205 0.04043851
## 27 1e+00 1e+01 0.53326923 0.06812082
## 28 5e+00 1e+01 0.52301282 0.05688936
## 29 1e+01 1e+01 0.52301282 0.05688936
## 30 1e+02 1e+01 0.52301282 0.05688936
## 31 1e-02 1e+02 0.56128205 0.04043851
## 32 1e-01 1e+02 0.56128205 0.04043851
## 33 1e+00 1e+02 0.56128205 0.04043851
## 34 5e+00 1e+02 0.56128205 0.04043851
## 35 1e+01 1e+02 0.56128205 0.04043851
## 36 1e+02 1e+02 0.56128205 0.04043851
  • For the radial kernel, the lowest cross-validation error is obtained for a gamma of 0.01 and cost of 100.

d) 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, isntead 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 = 100, degree = 2)
svm.radial = svm(mpglevel~., data = Auto, kernel = "radial", cost = 100, 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)

Question 8

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

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

set.seed(10)
train = sample(nrow(OJ), 800)
OJ.train = OJ[train,]
OJ.test = OJ[-train,]

b) Fit a support vector classifier to the training data using cost-0.01, with Purchase as the response and other variables as predictors. Use the summary() function to produce summary statistics, and describe the results obtained.

svm.linear = svm(Purchase~., data = OJ.train, kernel = "linear", 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:  451
## 
##  ( 226 225 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
  • The SVM model creates 451 support vectors. Out of these 226 belong to the level MM and the remaining 225 belong to the level CH.

c) What are the training and test error rates?

train.pred = predict(svm.linear, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 431  60
##   MM  84 225
(431+225)/(431+60+84+225)
## [1] 0.82
test.pred = predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 143  19
##   MM  23  85
(143+85)/(143+19+23+85)
## [1] 0.8444444
  • The training error rate is 0.18.
  • The test error rate is slightly better at 0.16.

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

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
##     1
## 
## - best performance: 0.18125 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.18375 0.05402224
## 2   0.01778279 0.18250 0.05374838
## 3   0.03162278 0.18500 0.05230785
## 4   0.05623413 0.18250 0.05109903
## 5   0.10000000 0.18625 0.05350558
## 6   0.17782794 0.18500 0.05230785
## 7   0.31622777 0.18750 0.04823265
## 8   0.56234133 0.18125 0.04832256
## 9   1.00000000 0.18125 0.05043216
## 10  1.77827941 0.18250 0.05277047
## 11  3.16227766 0.18375 0.05239076
## 12  5.62341325 0.18250 0.05109903
## 13 10.00000000 0.18875 0.05015601

e) 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$best.parameter$cost)
train.pred = predict(svm.linear, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 431  60
##   MM  77 232
(431+232)/(431+232+60+77)
## [1] 0.82875
test.pred = predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 145  17
##   MM  21  87
(145+87)/(145+87+17+21)
## [1] 0.8592593
  • The training error rate is 0.17.
  • The test error rate is slightly better at 0.14.
  • Both the training and test error rates improve after adding the cost parameter.

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

svm.radial = svm(Purchase~., kernel = "radial", data = OJ.train)
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:  389
## 
##  ( 199 190 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.pred = predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 448  43
##   MM  86 223
(448+223)/(448+43+86+223)
## [1] 0.83875
test.pred = predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 147  15
##   MM  23  85
(147+85)/(147+15+23+85)
## [1] 0.8592593
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
##  0.1778279
## 
## - best performance: 0.185 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.38625 0.04308019
## 2   0.01778279 0.38625 0.04308019
## 3   0.03162278 0.38375 0.04372023
## 4   0.05623413 0.22250 0.03899786
## 5   0.10000000 0.19500 0.03593976
## 6   0.17782794 0.18500 0.03106892
## 7   0.31622777 0.18750 0.03486083
## 8   0.56234133 0.18500 0.03374743
## 9   1.00000000 0.18625 0.03304563
## 10  1.77827941 0.18500 0.03987829
## 11  3.16227766 0.18875 0.03701070
## 12  5.62341325 0.19000 0.03525699
## 13 10.00000000 0.20250 0.03106892
svm.radial = svm(Purchase~., data = OJ.train, cost = tune.out$best.parameter$cost)
summary(svm.radial)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, cost = tune.out$best.parameter$cost)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  0.1778279 
## 
## Number of Support Vectors:  497
## 
##  ( 250 247 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.pred = predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 448  43
##   MM  92 217
(448+217)/(448+43+92+217)
## [1] 0.83125
test.pred = predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 146  16
##   MM  27  81
(146+81)/(146+16+27+81)
## [1] 0.8407407
  • The training and test error rates for the svmRadial model are 0.16 and 0.14, respectively.
  • The training and test error rates after adding the cost paramater are ___ and 0.16, respectively.

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

svm.poly = svm(Purchase~., kernel = "polynomial", data = OJ.train, degree = 2)
summary(svm.poly)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "polynomial", 
##     degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  1 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  461
## 
##  ( 234 227 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.pred = predict(svm.poly, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 455  36
##   MM 119 190
(455+190)/(455+36+119+190)
## [1] 0.80625
test.pred = predict(svm.poly, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 149  13
##   MM  34  74
(149+74)/(149+13+34+74)
## [1] 0.8259259
tune.out = tune(svm, Purchase~., data = OJ.train, kernel = "polynomial", 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
##    10
## 
## - best performance: 0.1875 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.38625 0.05382908
## 2   0.01778279 0.36625 0.04251225
## 3   0.03162278 0.36000 0.04362084
## 4   0.05623413 0.33500 0.05130248
## 5   0.10000000 0.31375 0.04767147
## 6   0.17782794 0.26875 0.04868051
## 7   0.31622777 0.21000 0.03476109
## 8   0.56234133 0.20875 0.03775377
## 9   1.00000000 0.20625 0.03019037
## 10  1.77827941 0.19500 0.03446012
## 11  3.16227766 0.19250 0.03238227
## 12  5.62341325 0.18875 0.03087272
## 13 10.00000000 0.18750 0.03385016
svm.poly = svm(Purchase~., kernel = "polynomial", degree = 2, data = OJ.train, cost = tune.out$best.parameter$cost)
summary(svm.poly)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "polynomial", 
##     degree = 2, cost = tune.out$best.parameter$cost)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  10 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  360
## 
##  ( 185 175 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.pred = predict(svm.poly, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 451  40
##   MM  85 224
(451+224)/(451+40+85+224)
## [1] 0.84375
test.pred = predict(svm.poly, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 148  14
##   MM  27  81
(148+81)/(148+14+27+81)
## [1] 0.8481481
  • The training and test error rates for the svm polynomial model are 0.19 and 0.17, respectively.
  • The training and test error rates after adding cost parameter are 0.16 and 0.16, respectively.

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

  • The approach that gave the lowest test error rates were the svm Radial without the cost parameter and the svm Linear model with the cost parameter. In most of these models, the training error rates were less than the test error rates.