library(ISLR2)
library(caret)
## Loading required package: ggplot2
## Loading required package: lattice
library(ggplot2)
library(lattice)

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.

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

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 X 1 on the x-axis, and X 2 on the y-axis.

plot(x1[y==0],x2[y==0], col="blue", xlab= "x1", ylab= "x2", pch = '+')
points(x1[y==1],x2[y==1], col="green", pch = 4)

(c) Fit a logistic regression model to the data, using X 1 and X 2 as predictors.

lr.fit= glm(y~ x1+x2, family = binomial)
summary(lr.fit)
## 
## Call:
## glm(formula = y ~ x1 + x2, family = binomial)
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.297  -1.212   1.072   1.132   1.226  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)
## (Intercept)  0.08660    0.08968   0.966    0.334
## x1          -0.29732    0.31023  -0.958    0.338
## x2          -0.11405    0.30730  -0.371    0.711
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 692.18  on 499  degrees of freedom
## Residual deviance: 691.14  on 497  degrees of freedom
## AIC: 697.14
## 
## 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)
lm.prob = predict(lr.fit,data, type = "response")
lm.pred = ifelse(lm.prob > 0.52, 1, 0)
data.pos = data[lm.pred == 1, ]
data.neg = data[lm.pred == 0, ]
plot(data.pos$x1, data.pos$x2, col = "blue", xlab = "X1", ylab = "X2", pch = "+")
points(data.neg$x1, data.neg$x2, col = "red", pch = 4)

(e) Now fit a logistic regression model to the data using non-linear functions of X 1 and X 2 as predictors (e.g. X 21 , X 1 ×X 2 , log(X 2 ),and so forth).

lm.fit = glm(y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), data = data, family = binomial)
## Warning: glm.fit: algorithm did not converge
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred

(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.

lm.prob = predict(lm.fit, data, type = "response")
lm.pred = ifelse(lm.prob > 0.5, 1, 0)
data.pos = data[lm.pred == 1, ]
data.neg = data[lm.pred == 0, ]
plot(data.pos$x1, data.pos$x2, col = "green", xlab = "X1", ylab = "X2", pch = "+")
points(data.neg$x1, data.neg$x2, col = "red", pch = 4)

(g) 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.

library(e1071)
svm.fit = svm(as.factor(y) ~ x1 + x2, data, kernel = "linear", cost = 0.1)
plot(data.pos$x1,data.pos$x2,col = "orange", xlab = "X1", ylab = "X2", pch = "+")
points(data.neg$x1, data.neg$x2, col = "blue", pch = 4)

(h) Fit a SVM using a non-linear kernel to the data. Obtain a classprediction for each training observation. Plot the observations, colored according to the predicted class labels. (i) Comment on your results.

svm.fit = svm(as.factor(y) ~ x1 + x2, data, gamma = 1)
plot(data.pos$x1, data.pos$x2, col = "green", xlab = "X1", ylab = "X2", pch = "+")
points(data.neg$x1, data.neg$x2, col = "red", pch = 4)

#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.

(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.

library(ISLR2)
attach(Auto)
## The following object is masked from package:ggplot2:
## 
##     mpg
mpg.med = median(Auto$mpg)
above.below = ifelse(Auto$mpg > mpg.med, 1, 0)
mpglevel = as.factor(above.below)

(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.

library(e1071)
set.seed(12)
tune.out = tune(svm, mpg ~ ., 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
##   0.1
## 
## - best performance: 11.85867 
## 
## - Detailed performance results:
##    cost    error dispersion
## 1 1e-02 12.83584   4.954577
## 2 1e-01 11.85867   4.269184
## 3 1e+00 12.30706   3.171524
## 4 5e+00 13.02714   3.825037
## 5 1e+01 13.44875   4.060655
## 6 1e+02 14.59619   4.439880

(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(12)
tune.out = tune(svm, mpg ~ ., 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: 57.74487 
## 
## - Detailed performance results:
##    cost degree    error dispersion
## 1   0.1      2 61.62628   11.23107
## 2   1.0      2 60.90499   11.22059
## 3   5.0      2 58.89548   11.55725
## 4  10.0      2 57.74487   11.90850
## 5   0.1      3 61.67504   11.23179
## 6   1.0      3 61.37209   11.18713
## 7   5.0      3 60.04905   10.99319
## 8  10.0      3 58.44108   10.77738
## 9   0.1      4 61.70830   11.23679
## 10  1.0      4 61.70358   11.23658
## 11  5.0      4 61.68262   11.23565
## 12 10.0      4 61.65648   11.23452

(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, 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 
## function(fit) {
##     for (name in names(Auto)[!(names(Auto) %in% c("mpg", "mpglevel", "name"))]) {
##         plot(fit, Auto, as.formula(paste("mpg~", name, sep = "")))
##     }
## }

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

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

library(ISLR2)
set.seed(1)
train = sample(dim(OJ)[1], 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 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:  435
## 
##  ( 219 216 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

(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 420  65
##   MM  75 240
1-((420+240)/800)
## [1] 0.175

the test error rate is 17.5%

test.pred = predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)  
##     test.pred
##       CH  MM
##   CH 153  15
##   MM  33  69
1-((153+69)/270)
## [1] 0.1777778

test error rate is 17.778% (d) Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.

set.seed(1)
tune.out=tune(svm ,Purchase~.,data=OJ ,kernel ="linear",ranges=list(cost=c (0.001, 0.01, 0.1, 1,5,10) ))
summary(tune.out)  
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.1626168 
## 
## - Detailed performance results:
##    cost     error dispersion
## 1 1e-03 0.2373832 0.04561497
## 2 1e-02 0.1691589 0.04024604
## 3 1e-01 0.1663551 0.03984617
## 4 1e+00 0.1626168 0.03945456
## 5 5e+00 0.1654206 0.03917066
## 6 1e+01 0.1682243 0.03865942

(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$best.parameters$cost)
train.pred = predict(svm.linear, OJ.train)
table(OJ.train$Purchase, train.pred)  
##     train.pred
##       CH  MM
##   CH 424  61
##   MM  70 245
1-((424+245)/800)
## [1] 0.16375

training error rate is 16.375%

test.pred =predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 155  13
##   MM  29  73
1-((155+73)/270)
## [1] 0.1555556

training error is 15.556% (f) Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.

set.seed(1)
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:  373
## 
##  ( 188 185 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

there are 373 support vectors.

train.pred = predict(svm.radial, OJ.train)
table(OJ.train$Purchase,train.pred)
##     train.pred
##       CH  MM
##   CH 441  44
##   MM  77 238
1-((441+238)/800)
## [1] 0.15125
test.pred = predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)  
##     test.pred
##       CH  MM
##   CH 151  17
##   MM  33  69
1-((151+69)/270)
## [1] 0.1851852
set.seed(1)
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: 0.16875 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39375 0.04007372
## 2   0.01778279 0.39375 0.04007372
## 3   0.03162278 0.35750 0.05927806
## 4   0.05623413 0.19500 0.02443813
## 5   0.10000000 0.18625 0.02853482
## 6   0.17782794 0.18250 0.03291403
## 7   0.31622777 0.17875 0.03230175
## 8   0.56234133 0.16875 0.02651650
## 9   1.00000000 0.17125 0.02128673
## 10  1.77827941 0.17625 0.02079162
## 11  3.16227766 0.17750 0.02266912
## 12  5.62341325 0.18000 0.02220485
## 13 10.00000000 0.18625 0.02853482
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$Purchase, train.pred)  
##     train.pred
##       CH  MM
##   CH 437  48
##   MM  71 244
1-((437+244)/800)  
## [1] 0.14875
test.pred = predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)  
##     test.pred
##       CH  MM
##   CH 150  18
##   MM  30  72
1-((150+72)/270)  
## [1] 0.1777778

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

set.seed(47)
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:  447
## 
##  ( 225 222 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.pred = predict(svm.poly, OJ.train)
table(OJ.train$Purchase, train.pred)  
##     train.pred
##       CH  MM
##   CH 449  36
##   MM 110 205
1-((446+216)/800)  
## [1] 0.1725
test.pred = predict(svm.poly, OJ.test)
table(OJ.test$Purchase, test.pred)  
##     test.pred
##       CH  MM
##   CH 153  15
##   MM  45  57
1-((160+54)/270)  
## [1] 0.2074074
set.seed(1)
tune.out = tune(svm, Purchase ~ ., data = OJ.train, kernel = "poly", 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: 0.185 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.37125 0.03537988
## 2   0.01778279 0.37125 0.03488573
## 3   0.03162278 0.34500 0.04377975
## 4   0.05623413 0.32750 0.04669642
## 5   0.10000000 0.28750 0.05068969
## 6   0.17782794 0.22625 0.03408018
## 7   0.31622777 0.19625 0.03438447
## 8   0.56234133 0.19125 0.03634805
## 9   1.00000000 0.18500 0.02415229
## 10  1.77827941 0.18875 0.02079162
## 11  3.16227766 0.19125 0.02503470
## 12  5.62341325 0.19000 0.03106892
## 13 10.00000000 0.19500 0.03184162
svm.poly = svm(Purchase ~ ., data = OJ.train, kernel = "poly", cost = tune.out$best.parameters$cost)
train.pred = predict(svm.poly, OJ.train)
table(OJ.train$Purchase, train.pred)  
##     train.pred
##       CH  MM
##   CH 453  32
##   MM  91 224
1-((454+215)/800)  
## [1] 0.16375
test.pred = predict(svm.poly, OJ.test)
table(OJ.test$Purchase, test.pred)  
##     test.pred
##       CH  MM
##   CH 155  13
##   MM  47  55
1-((153+55)/270)  
## [1] 0.2296296

(h) Overall, which approach seems to give the best results on this data? The radial kernel seems to give the lowest error rates.