Chap 9 - Applied

Applied

1. Generate a simulated two-class data set with 100 observations and two features in which there is a visible but non-linear separation between the two classes. Show that in this setting, a support vector machine with a polynomial kernel (with degree greater than 1) or a radial kernel will outperform a support vector classifier on the training data. Which technique performs best on the test data? Make plots and report training and test error rates in order to back up your assertions.

set.seed(30)
x1 <- rnorm(100, mean = 0, sd = 1)
x2 <- x1^2 + rnorm(100, mean = 3, sd = 1)
Seqx <- seq(from = min(x1), to = max(x1),length.out = 100)
Seqy <- Seqx^2 + 3
fc <- rep("Good", 100)
fc[x2<Seqy] <- "Bad"
fc <- as.factor(fc)

Including Plots

You can also embed plots, for example:

Dat <- data.frame(x1,x2, fc)
svmfit <- svm(fc~., data = Dat, kernel = "polynomial", degree = 3,
              cost = 0.1, scale = FALSE)
plot(svmfit, Dat)

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*(x1^2-x2^2 > 0) 
yf <- as.factor(y)
Data.2 <- data.frame(x1,x2,y,yf)

2. 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,x2,col = yf)

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

Logit.model <- glm(formula = y~. -yf, data = Data.2, family = 'binomial')

4. 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.

• Particion the Data

Data.3 <- Data.2[, -4]
library(caret)

## Loading required package: lattice

i.Data.3 <- createDataPartition(Data.3$y, p = 0.8, list = FALSE)
Train.Data.3 <- Data.3[i.Data.3,]
Test.Data.3 <- Data.3[-i.Data.3,]

• Predict the value with Training Data

Logit.model <- glm(formula = y~. , data = Train.Data.3, family = 'binomial')
pred.logit <- predict(Logit.model, type = 'response')
pred.logit.Tb <- rep(0,400)
pred.logit.Tb[pred.logit > 0.5] <- 1
Tb <- table(Pred = pred.logit.Tb, Truth = Train.Data.3$y)
Tb

##     Truth
## Pred   0   1
##    0  63  18
##    1 124 195

(Tb[1,1] + Tb[2,2])/sum(Tb)

## [1] 0.645

plot(Train.Data.3$x1,Train.Data.3$x2, col = as.factor(pred.logit.Tb))

5. Now fit a logistic regression model to the data using non-linear functions of X 1 and X 2 as predictors

Logit.model <- glm(formula = y~ I(x1^2) + I(x2^2), data = Train.Data.3, family = 'binomial')

## Warning: glm.fit: algorithm did not converge

## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred

pred.logit <- predict(Logit.model, type = 'response')
pred.logit.Tb <- rep(0,400)
pred.logit.Tb[pred.logit > 0.5] <- 1
Tb <- table(Pred = pred.logit.Tb, Truth = Train.Data.3$y)
Tb

##     Truth
## Pred   0   1
##    0 187   0
##    1   0 213

Table

(Tb[1,1] + Tb[2,2])/sum(Tb)

## [1] 1

Plot

plot(Train.Data.3$x1,Train.Data.3$x2, col = as.factor(pred.logit.Tb))

7. 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.

• radial

Train.Data.3$y <- as.factor(Train.Data.3$y)
tune.out <- tune(svm, y~., data = Train.Data.3,
                 kernel = "radial", 
                 ranges = list(cost = c(0.1,1,10,100,1000)),
                 gamma = c(0.5,1,2,3,4))
pred.tune.out <- predict(tune.out$best.model,Train.Data.3)
plot(Train.Data.3$x1,Train.Data.3$x2, col = pred.tune.out)

Table

table(Pred = pred.tune.out, Truth = Train.Data.3$y)

##     Truth
## Pred   0   1
##    0 185   0
##    1   2 213

• polynomial

tune.out <- tune(svm, y~., data = Train.Data.3,
                 kernel = "polynomial", 
                 ranges = list(cost = c(0.1,1,10,100,1000)),
                 gamma = c(0.5,1,2,3,4))
pred.tune.out <- predict(tune.out$best.model,Train.Data.3)
plot(Train.Data.3$x1,Train.Data.3$x2, col = pred.tune.out)

Table

table(Pred = pred.tune.out, Truth = Train.Data.3$y)

##     Truth
## Pred   0   1
##    0   0   0
##    1 187 213

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.

auto <- Auto[,-c(9)]
head(auto)

##   mpg cylinders displacement horsepower weight acceleration year origin
## 1  18         8          307        130   3504         12.0   70      1
## 2  15         8          350        165   3693         11.5   70      1
## 3  18         8          318        150   3436         11.0   70      1
## 4  16         8          304        150   3433         12.0   70      1
## 5  17         8          302        140   3449         10.5   70      1
## 6  15         8          429        198   4341         10.0   70      1

auto <- auto %>%
        mutate(mpgbn = ifelse(mpg > median(mpg), 1,0))
auto$mpgbn <- as.factor(auto$mpgbn)
auto <- auto[, -1]

2. 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.

library(caret)
set.seed(30)
i.F.Auto <- createFolds(y = auto$mpgbn, k = 10, list = TRUE)
poly.error.rate <- matrix(NA ,10, 100)
cost <- seq(0.01,10, length.out = 100)
degree <- seq(0.01,10, length.out = 100)
gamma <- seq(0.01,10, length.out = 100)

for(i in 1:10) {
        train.dat <- auto[i.F.Auto[[i]],]
        cv.dat <- auto[-i.F.Auto[[i]],]    
        for(j in 1:100)
        {svm.fit <-svm(mpgbn~., data = train.dat, kernel = "polynomial", degree = degree[j],
              cost = 0.01, scale = TRUE)
        pred <- predict(svm.fit, cv.dat)
        tb <- table(pred, cv.dat$mpgbn)
        poly.error.rate[i,j] <- 1-((tb[1,1]+tb[1,2])/sum(tb))}
}
cv.poly.error <- apply(poly.error.rate, 2, mean)

rd.error.rate <- matrix(NA ,10, 100)

tune.out <- tune(svm, mpgbn~., data = auto,
                 kernel = "radial",
                 range = list(cost = cost),
                 gamma = gamma)

for(i in 1:10) {
        train.dat <- auto[i.F.Auto[[i]],]
        cv.dat <- auto[-i.F.Auto[[i]],]    
        for(j in 1:100)
        {svm.fit <-svm(mpgbn~., data = train.dat, kernel = "radial", gamma = gamma[j],
              cost = 0.6154545, scale = TRUE)
        pred <- predict(svm.fit, cv.dat)
        tb <- table(pred, cv.dat$mpgbn)
        rd.error.rate[i,j] <- 1-((tb[1,1]+tb[1,2])/sum(tb))}
}

rd.cv.Error <- apply(rd.error.rate,2,mean)

round(rd.cv.Error[which.min(rd.cv.Error)],digits = 3); gamma[which.min(rd.cv.Error)]

## [1] 0.524

## [1] 1.523636

table

	Error	Deg/Gam
Radial	0.524	1.523636
Poly	0.625	1.019091

poly.svm.best <- svm(mpgbn~., data = auto, kernel="polynomial", cost = 0.01, degree = 1.019091, scale = TRUE)
rd.svm.best <- svm(mpgbn~., data = auto, kernel="radial", cost = 0.01, gamma = 10, scale = TRUE)
par(mfrow = c(1,2))
plot(rd.svm.best, auto, displacement~weight)

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

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

oj <- OJ
i.Train.o <- createDataPartition(y = oj$Purchase, p = 800/nrow(oj), list = FALSE)   
Train.o <- oj[i.Train.o, ]
Test.o <- oj[-i.Train.o, ]

2. 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.fit <- svm(Purchase ~. , data = Train.o, cost = 0.01,
                  kernel = 'linear')
summary(oj.svm.fit)

## 
## Call:
## svm(formula = Purchase ~ ., data = Train.o, cost = 0.01, kernel = "linear")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
##       gamma:  0.05555556 
## 
## Number of Support Vectors:  448
## 
##  ( 224 224 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

tr.pred <- predict(oj.svm.fit)
ts.pred <- predict(oj.svm.fit, newdata = Test.o)
Tr.table <- table(Predict = tr.pred, Truth = Train.o$Purchase)
Ts.table <- table(Predict = ts.pred, Truth = Test.o$Purchase)

3. What are the training and test error rates?

(Tr.table[1,2]+Tr.table[2,1])/sum(Tr.table); (Ts.table[1,2]+Ts.table[2,1])/sum(Ts.table)

## [1] 0.1735331

## [1] 0.1375465

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

cost <- seq(0.01,10,100)
tune.oj.svm.fit <- tune(svm, Purchase~., data = oj,
                        kernel = 'linear',
                        ranges = list(cost = cost))
tune.oj.svm.fit$best.model

## 
## Call:
## best.tune(method = svm, train.x = Purchase ~ ., data = oj, ranges = list(cost = cost), 
##     kernel = "linear")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
##       gamma:  0.05555556 
## 
## Number of Support Vectors:  560

The optimal cost is 0.01, so no need to calculate

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

cost <- seq(0.01,10,100)
tune.oj.rd.svm.fit <- tune(svm, Purchase~., data = Train.o,
                        kernel = 'radial',
                        ranges = list(cost = cost))

tr.rd.pred <- predict(tune.oj.rd.svm.fit$best.model, Train.o)
ts.rd.pred <- predict(tune.oj.rd.svm.fit$best.model, Test.o)
table(pred = tr.rd.pred, truth = Train.o$Purchase)

##     truth
## pred  CH  MM
##   CH 489 312
##   MM   0   0

table(pred = ts.rd.pred, truth = Test.o$Purchase)

##     truth
## pred  CH  MM
##   CH 164 105
##   MM   0   0

radial is not fit.

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

tune.oj.polu.svm.fit <- tune(svm, Purchase~., data = Train.o,
                        kernel = 'polynomial',
                        ranges = list(cost = cost),
                        degree = 2)
tr.rd.pred <- predict(tune.oj.polu.svm.fit$best.model, Train.o)
ts.rd.pred <- predict(tune.oj.polu.svm.fit$best.model, Test.o)
table(pred = tr.rd.pred, truth = Train.o$Purchase)

##     truth
## pred  CH  MM
##   CH 489 312
##   MM   0   0

table(pred = ts.rd.pred, truth = Test.o$Purchase)

##     truth
## pred  CH  MM
##   CH 164 105
##   MM   0   0

Linear approach is rock!