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)

set.seed(1)
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 yaxis.

#Construct the data frame with deviance and size 
df.svm.x12 <- data.frame(x1 = x1,
                          x2 = x2,
                          y = y)
##Use the ggplot
ggplot(df.svm.x12, aes(x = x1, y = x2, color = y)) +
 geom_point(size = 2) +
  theme(panel.background = element_rect(fill = 'white', colour = 'red'), legend.position = "none") +
  labs(title = "Non-linear Decision Boundary",
       x = "x1",
       y = "x2")

(c) Fit a logistic regression model to the data, using X1 and X2 as predictors.

glm.svm.x12 <- glm(y ~ x1 + x2, family = "binomial")
summary(glm.svm.x12)
## 
## Call:
## glm(formula = y ~ x1 + x2, family = "binomial")
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.179  -1.139  -1.112   1.206   1.257  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.087260   0.089579  -0.974    0.330
## x1           0.196199   0.316864   0.619    0.536
## x2          -0.002854   0.305712  -0.009    0.993
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 692.18  on 499  degrees of freedom
## Residual deviance: 691.79  on 497  degrees of freedom
## AIC: 697.79
## 
## Number of Fisher Scoring iterations: 3

From the above summary, we can identify that none of the predictors are significant against y.

(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.

svm.probs1 <- predict(glm.svm.x12, df.svm.x12, type = "response")
svm.preds1 <- rep(0, 500)
svm.preds1[svm.probs1 > 0.47] <- 1
plot(df.svm.x12[svm.preds1 == 1, ]$x1, df.svm.x12[svm.preds1 == 1, ]$x2, col = "red", pch = "+", xlab = "X1", ylab = "X2")
points(df.svm.x12[svm.preds1 == 0, ]$x1, df.svm.x12[svm.preds1 == 0, ]$x2, col = "blue", pch = 4)

(e) 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).

Using the few transformations on the x1 and x2 and use them as predictors in the logistic regression model.

log.x12.fit <- glm(y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), family = "binomial")
summary(log.x12.fit)
## 
## Call:
## glm(formula = y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), family = "binomial")
## 
## Deviance Residuals: 
##        Min          1Q      Median          3Q         Max  
## -8.240e-04  -2.000e-08  -2.000e-08   2.000e-08   1.163e-03  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)
## (Intercept)    -102.2     4302.0  -0.024    0.981
## poly(x1, 2)1   2715.3   141109.5   0.019    0.985
## poly(x1, 2)2  27218.5   842987.2   0.032    0.974
## poly(x2, 2)1   -279.7    97160.4  -0.003    0.998
## poly(x2, 2)2 -28693.0   875451.3  -0.033    0.974
## I(x1 * x2)     -206.4    41802.8  -0.005    0.996
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 6.9218e+02  on 499  degrees of freedom
## Residual deviance: 3.5810e-06  on 494  degrees of freedom
## AIC: 12
## 
## Number of Fisher Scoring iterations: 25

Again none of the predictors are statistically significant with y.

(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.

svm.probs2 <- predict(log.x12.fit, df.svm.x12, type = "response")
svm.preds2 <- rep(0, 500)
svm.preds2[svm.probs2 > 0.47] <- 1
plot(df.svm.x12[svm.preds2 == 1, ]$x1, df.svm.x12[svm.preds2 == 1, ]$x2, col = "red", pch = "+", xlab = "X1", ylab = "X2")
points(df.svm.x12[svm.preds2 == 0, ]$x1, df.svm.x12[svm.preds2 == 0, ]$x2, col = "blue", pch = 4)

The above plot shows that the decision boundary is non-linear.

(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.

df.svm.x12$y <- as.factor(df.svm.x12$y)
svm.x12.fit1 <- svm(y ~ x1 + x2, df.svm.x12, kernel = "linear", cost = 0.01)
svm.x12.preds1 <- predict(svm.x12.fit1, df.svm.x12)
plot(df.svm.x12[svm.x12.preds1 == 0, ]$x1, df.svm.x12[svm.x12.preds1 == 0, ]$x2, col = "red", pch = "+", xlab = "X1", ylab = "X2")
points(df.svm.x12[svm.x12.preds1 == 1, ]$x1, df.svm.x12[svm.x12.preds1 == 1, ]$x2, col = "blue", pch = 4)

The above plot shows that even with the low cost, the linear kernel fails to find non-linear decision boundary and it classifies all points to a single class.

(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.

svm.x12.fit2 <- svm(as.factor(y) ~ x1 + x2, df.svm.x12, gamma = 1)
svm.x12.preds2 <- predict(svm.x12.fit2, df.svm.x12)
df.svm.x12.pos <- df.svm.x12[svm.x12.preds2 == 1, ]
df.svm.x12.neg <- df.svm.x12[svm.x12.preds2 == 0, ]
plot(df.svm.x12.pos$x1, df.svm.x12.pos$x2, col = "red", xlab = "X1", ylab = "X2", pch = "+")
points(df.svm.x12.neg$x1, df.svm.x12.neg$x2, col = "blue", pch = 4)

The above plot shows that the non-linear decision boundary closely resembles the true decision boundary.

(i) Comment on your results.

From all the above regression models, SVM with non-linear kernel and logistic regression with interaction terms are very powerful in finding the non-linear boundary, and the SVM with linear kernel are NOT good in finding non-linear decision boundaries. In order to make the logistic regression SVM with right interaction terms to find the non-linear decision boundaries, we need to fine tune gamma.

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.

myAuto <- Auto
myAuto <- mutate(myAuto, mpg.median.level = as.factor(ifelse(Auto$mpg > median(Auto$mpg), 1, 0)))

(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(1)
auto.cost <- c(0.01, 0.1, 1, 10, 100, 1000)
tune.auto.linear <- tune(svm, mpg.median.level ~ ., data = myAuto, kernel = "linear", ranges = list(cost = auto.cost))
summary(tune.auto.linear)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.01025641 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-02 0.07653846 0.03617137
## 2 1e-01 0.04596154 0.03378238
## 3 1e+00 0.01025641 0.01792836
## 4 1e+01 0.02051282 0.02648194
## 5 1e+02 0.03076923 0.03151981
## 6 1e+03 0.03076923 0.03151981

The above result shows the cross-validation errors of different cost, i.e., for the costs 0.01, 0.1, 1, 10, 100, 1000 their corresponding cross-validation errors are 0.07653846, 0.04596154, 0.01025641, 0.02051282, 0.03076923, 0.03076923 respectively. The lowest cross-validation error is 0.01025641, which is for the cost = 1 (perform the best) for the linear kernel.

(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.

Running the SVM with radial kernel with different gamma and cost.

set.seed(1)
auto.gamma <- c(0.01, 0.1, 1, 5, 10, 100)
tune.auto.radial <- tune(svm, mpg.median.level ~ ., data = myAuto, kernel = "radial", ranges = list(cost = auto.cost, gamma = auto.gamma))
summary(tune.auto.radial)
## 
## 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 0.55115385 0.04366593
## 2  1e-01 1e-02 0.08929487 0.04382379
## 3  1e+00 1e-02 0.07403846 0.03522110
## 4  1e+01 1e-02 0.02557692 0.02093679
## 5  1e+02 1e-02 0.01282051 0.01813094
## 6  1e+03 1e-02 0.02820513 0.02549818
## 7  1e-02 1e-01 0.21711538 0.09865227
## 8  1e-01 1e-01 0.07903846 0.03874545
## 9  1e+00 1e-01 0.05371795 0.03525162
## 10 1e+01 1e-01 0.03076923 0.03375798
## 11 1e+02 1e-01 0.03583333 0.02759051
## 12 1e+03 1e-01 0.03583333 0.02759051
## 13 1e-02 1e+00 0.55115385 0.04366593
## 14 1e-01 1e+00 0.55115385 0.04366593
## 15 1e+00 1e+00 0.06384615 0.04375618
## 16 1e+01 1e+00 0.05884615 0.04020934
## 17 1e+02 1e+00 0.05884615 0.04020934
## 18 1e+03 1e+00 0.05884615 0.04020934
## 19 1e-02 5e+00 0.55115385 0.04366593
## 20 1e-01 5e+00 0.55115385 0.04366593
## 21 1e+00 5e+00 0.49493590 0.04724924
## 22 1e+01 5e+00 0.48217949 0.05470903
## 23 1e+02 5e+00 0.48217949 0.05470903
## 24 1e+03 5e+00 0.48217949 0.05470903
## 25 1e-02 1e+01 0.55115385 0.04366593
## 26 1e-01 1e+01 0.55115385 0.04366593
## 27 1e+00 1e+01 0.51794872 0.05063697
## 28 1e+01 1e+01 0.51794872 0.04917316
## 29 1e+02 1e+01 0.51794872 0.04917316
## 30 1e+03 1e+01 0.51794872 0.04917316
## 31 1e-02 1e+02 0.55115385 0.04366593
## 32 1e-01 1e+02 0.55115385 0.04366593
## 33 1e+00 1e+02 0.55115385 0.04366593
## 34 1e+01 1e+02 0.55115385 0.04366593
## 35 1e+02 1e+02 0.55115385 0.04366593
## 36 1e+03 1e+02 0.55115385 0.04366593

The lowest cross-validation error (0.01282051) for the radial kernel is for cost = 100 and gamma = 0.01.

set.seed(1)
auto.degree <- c(2, 3, 4)
tune.auto.poly <- tune(svm, mpg.median.level ~ ., data = myAuto, kernel = "polynomial", ranges = list(cost = auto.cost, degree = auto.degree))
summary(tune.auto.poly)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##  1000      2
## 
## - best performance: 0.2454487 
## 
## - Detailed performance results:
##     cost degree     error dispersion
## 1  1e-02      2 0.5511538 0.04366593
## 2  1e-01      2 0.5511538 0.04366593
## 3  1e+00      2 0.5511538 0.04366593
## 4  1e+01      2 0.5130128 0.08963366
## 5  1e+02      2 0.3013462 0.09961961
## 6  1e+03      2 0.2454487 0.11551451
## 7  1e-02      3 0.5511538 0.04366593
## 8  1e-01      3 0.5511538 0.04366593
## 9  1e+00      3 0.5511538 0.04366593
## 10 1e+01      3 0.5511538 0.04366593
## 11 1e+02      3 0.3446154 0.09821588
## 12 1e+03      3 0.2528846 0.09383590
## 13 1e-02      4 0.5511538 0.04366593
## 14 1e-01      4 0.5511538 0.04366593
## 15 1e+00      4 0.5511538 0.04366593
## 16 1e+01      4 0.5511538 0.04366593
## 17 1e+02      4 0.5511538 0.04366593
## 18 1e+03      4 0.5435897 0.05056569

The lowest cross-validation error (0.2454487) for the polynomial kernel is for cost = 1000 and degree = 2.

(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.

Fit the models with the kernels (linear, radial and polynomial) with those cost, gamma and degree for which we got the lowest cross-validation errors in (b) and (c).

svm.auto.best.linear <- svm(mpg.median.level ~ ., data = myAuto, kernel = "linear", cost = 1)
svm.auto.best.radial <- svm(mpg.median.level ~ ., data = myAuto, kernel = "radial", cost = 100, gamma = 0.01)
svm.auto.best.poly <- svm(mpg.median.level ~ ., data = myAuto, kernel = "polynomial", cost = 1000, degree = 2)

Writing a function to plot the pair of predictors with the mpg.median.level so that we can apply for all the three models (models with 3 different kernels).

plotpairedpredictors = function(model.fit) {
    for (name in names(myAuto)[!(names(Auto) %in% c("mpg", "mpg.median.level", "name"))]) {
        plot(model.fit, myAuto, as.formula(paste("mpg~", name, sep = "")))
    }
}

Plot the paired predictors for linear kernel model.

plotpairedpredictors(svm.auto.best.linear)

Plot the paired predictors for `radial`` kernel model.

plotpairedpredictors(svm.auto.best.radial)

Plot the paired predictors for polynomial kernel model.

plotpairedpredictors(svm.auto.best.poly)

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(1)
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 the other variables as predictors. Use the summary() function to produce summary statistics, and describe the results obtained.

svm.oj.linear <- svm(Purchase ~ ., data = oj.train, kernel = "linear", cost = 0.01)
summary(svm.oj.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

The support vector classifier svm using the training data and with cost 0.01 created 435 support vectors with 2 classes, that’s CH level with 219 support vectors and MM level with 216 support vectors.

(c) What are the training and test error rates?

train.oj.pred <- predict(svm.oj.linear, oj.train)
table(oj.train$Purchase, train.oj.pred)
##     train.oj.pred
##       CH  MM
##   CH 420  65
##   MM  75 240

The training error rate is (75+65)/(420+240+75+65) = 0.175, 17.5%.

test.oj.pred <- predict(svm.oj.linear, oj.test)
table(oj.test$Purchase, test.oj.pred)
##     test.oj.pred
##       CH  MM
##   CH 153  15
##   MM  33  69

The test error rate is (33+15)/(153+69+33+15) = 0.178, 17.8%.

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

set.seed(1)
oj.cost <- c(0.01, 0.05, 0.1, 0.5, 1, 5, 10)
tune.oj.linear <- tune(svm, Purchase ~ ., data = oj.train, kernel = "linear", ranges = list(cost = oj.cost))
summary(tune.oj.linear)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##   0.5
## 
## - best performance: 0.16875 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.17625 0.02853482
## 2  0.05 0.17625 0.02853482
## 3  0.10 0.17250 0.03162278
## 4  0.50 0.16875 0.02651650
## 5  1.00 0.17500 0.02946278
## 6  5.00 0.17250 0.03162278
## 7 10.00 0.17375 0.03197764

From the above tune() function summary, we see that the optimal cost is 0.5 as it has the lowest error rate 0.16875.

(e) Compute the training and test error rates using this new value for cost.

svm.oj.cost.linear <- svm(Purchase ~ ., kernel = "linear", data = oj.train, 
                          cost = tune.oj.linear$best.parameter$cost)
train.oj.cost.pred <- predict(svm.oj.cost.linear, oj.train)
table(oj.train$Purchase, train.oj.cost.pred)
##     train.oj.cost.pred
##       CH  MM
##   CH 424  61
##   MM  71 244

The training error rate is (71+61)/(424+244+71+61) = 0.165, 16.5%.

test.oj.cost.pred <- predict(svm.oj.cost.linear, oj.test)
table(oj.test$Purchase, test.oj.cost.pred)
##     test.oj.cost.pred
##       CH  MM
##   CH 155  13
##   MM  29  73

The test error rate is (29+13)/(155+73+29+13) = 0.156, 15.6%.

(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.oj.radial <- svm(Purchase ~ ., kernel = "radial", data = oj.train)
summary(svm.oj.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

The SVM radial kernel with default gamma model creates 373 support vectors, with 188 support vectors for the level CH and remaining 185 support vectors for the level MM.

train.radial.pred <- predict(svm.oj.radial, oj.train)
table(oj.train$Purchase, train.radial.pred)
##     train.radial.pred
##       CH  MM
##   CH 441  44
##   MM  77 238

The training error rate is (77+44)/(441+238+77+44) = 0.151, 15.1%.

test.radial.pred <- predict(svm.oj.radial, oj.test)
table(oj.test$Purchase, test.radial.pred)
##     test.radial.pred
##       CH  MM
##   CH 151  17
##   MM  33  69

The test error rate is (33+17)/(151+69+33+17) = 0.185, 18.5%.

The training error rate of the svm with radial and with default gamma is good improvement from the linear kernel, but the test error rate is bigger than the linear.

Lets try to find the optimal cost using the cross-validation for the radial kernel.

set.seed(1)
tune.radial.out <- tune(svm, Purchase ~ ., data = oj.train, kernel = "radial", 
                          ranges = list(cost = oj.cost))
summary(tune.radial.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##   0.5
## 
## - best performance: 0.1675 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.39375 0.04007372
## 2  0.05 0.20250 0.03374743
## 3  0.10 0.18625 0.02853482
## 4  0.50 0.16750 0.02443813
## 5  1.00 0.17125 0.02128673
## 6  5.00 0.18000 0.02220485
## 7 10.00 0.18625 0.02853482

The optimal cost is 0.5 from the above tune() function for the radial kernel.

set.seed(1)
svm.cost.radial <- svm(Purchase ~ ., kernel = "radial", data = oj.train, 
                        cost = tune.radial.out$best.parameter$cost)
summary(svm.cost.radial)
## 
## Call:
## svm(formula = Purchase ~ ., data = oj.train, kernel = "radial", cost = tune.radial.out$best.parameter$cost)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  0.5 
## 
## Number of Support Vectors:  407
## 
##  ( 205 202 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.radial.cost.pred <- predict(svm.cost.radial, oj.train)
table(oj.train$Purchase, train.radial.cost.pred)
##     train.radial.cost.pred
##       CH  MM
##   CH 438  47
##   MM  71 244

The training error rate is (71+47)/(438+244+71+47) = 0.148, 14.8%.

test.radial.cost.pred <- predict(svm.cost.radial, oj.test)
table(oj.test$Purchase, test.radial.cost.pred)
##     test.radial.cost.pred
##       CH  MM
##   CH 150  18
##   MM  30  72

The test error rate is (30+18)/(150+72+30+18) = 0.178, 17.8%.

Both training and test error rates reduced from the previous models with the radial kernel with tuning (that’s with optimal cost of 0.5).

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

set.seed(1)
svm.oj.poly <- svm(Purchase ~ ., kernel = "polynomial", data = oj.train, degree = 2)
summary(svm.oj.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:  447
## 
##  ( 225 222 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

The SVM polynomial kernel with degree=2 model creates 447 support vectors, with 225 support vectors for the level CH and remaining 222 support vectors for the level MM.

train.poly.pred <- predict(svm.oj.poly, oj.train)
table(oj.train$Purchase, train.poly.pred)
##     train.poly.pred
##       CH  MM
##   CH 449  36
##   MM 110 205

The training error rate is (110+36)/(449+205+110+36) = 0.183, 18.3%.

test.poly.pred <- predict(svm.oj.poly, oj.test)
table(oj.test$Purchase, test.poly.pred)
##     test.poly.pred
##       CH  MM
##   CH 153  15
##   MM  45  57

The test error rate is (45+15)/(153+57+45+15) = 0.222, 22.2%.

The training and test error rates of the svm with polynomial and with degree = 2 is increased considerably compare to the previous models.

Lets try to find the optimal cost using the cross-validation for the polynomial kernel.

set.seed(1)
tune.poly.out <- tune(svm, Purchase ~ ., data = oj.train, kernel = "polynomial", 
                          degree = 2, ranges = list(cost = oj.cost))
summary(tune.poly.out)
## 
## 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  0.01 0.39125 0.04210189
## 2  0.05 0.34875 0.04348132
## 3  0.10 0.32125 0.05001736
## 4  0.50 0.20625 0.04050463
## 5  1.00 0.20250 0.04116363
## 6  5.00 0.18250 0.03496029
## 7 10.00 0.18125 0.02779513

The optimal cost is 10 from the above tune() function for the polynomial kernel.

set.seed(1)
svm.cost.poly <- svm(Purchase ~ ., kernel = "polynomial", data = oj.train, 
                        degree = 2,
                        cost = tune.poly.out$best.parameter$cost)
summary(svm.cost.poly)
## 
## Call:
## svm(formula = Purchase ~ ., data = oj.train, kernel = "polynomial", 
##     degree = 2, cost = tune.poly.out$best.parameter$cost)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  10 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  340
## 
##  ( 171 169 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.poly.cost.pred <- predict(svm.cost.poly, oj.train)
table(oj.train$Purchase, train.poly.cost.pred)
##     train.poly.cost.pred
##       CH  MM
##   CH 447  38
##   MM  82 233

The training error rate is (82+38)/(447+233+82+38) = 0.15, 15.0%.

test.poly.cost.pred <- predict(svm.cost.poly, oj.test)
table(oj.test$Purchase, test.poly.cost.pred)
##     test.poly.cost.pred
##       CH  MM
##   CH 154  14
##   MM  37  65

The test error rate is (37+14)/(154+65+37+14) = 0.189, 18.9%.

Both training and test error rates reduced a lot from the previous model with the polynomial kernel with tuning (that’s with optimal cost of 10).

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

The lowest training rate is 14.8% from the tuned radial model with cost 0.5 and the lowest test rate is 15.6% from the tuned ‘linear’ model with cost 0.5.

The tuned radial basis kernel with cost=0.5 has the lowest training error rate of 14.8% and a decent test error rate of 17.8% as compare to all other kernels. So, I would recommend this one seems to give the best results on this data.