library(ISLR2)
library(MASS)
library(class)
library(tidyverse)
library(caret)

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 y-axis.

plot(x1,x2,col=ifelse(y,'red','blue'))

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

glm.fit=glm(y~ x1 + x2 ,family='binomial')
summary(glm.fit)
## 
## 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

Using GLM, X1 and X2 are both insignificant

(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)
glm.pred=predict(glm.fit,data.frame(x1,x2))
plot(x1,x2,col=ifelse(glm.pred>0,'red','blue'),pch=ifelse(as.integer(glm.pred>0) == y,1,4))

In this plot, the predicted 1 is in red, with cirles being correctly identifed.

(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).

lm.fit = glm(y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), data = data, family = binomial)

(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.

data = data.frame(x1 = x1, x2 = x2, y = y)
lm.pred=predict(lm.fit,data.frame(x1,x2))
plot(x1,x2,col=ifelse(lm.pred>0,'red','blue'),pch=ifelse(as.integer(lm.pred>0) == y,1,4))

The non-linear boundary shows up, which is similar to the true decision boundary.

(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.

For g and h, this could have alternatively been accomplished in caret telling it which kernel to use; I chose to use the svm package here in order to validate with solutions online.

library(e1071)
svm.fit=svm(y~.,data=data.frame(x1,x2,y=as.factor(y)),kernel='linear')
svm.pred=predict(svm.fit,data.frame(x1,x2),type='response')
plot(x1,x2,col=ifelse(svm.pred!=0,'red','blue'),pch=ifelse(svm.pred == y,1,4))

Here, everything is blue, indicating they are all predicted as the same 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.fit2=svm(y~.,data=data.frame(x1,x2,y=as.factor(y)),kernel='polynomial',degree=2)
svm.pred2=predict(svm.fit2,data.frame(x1,x2),type='response')
plot(x1,x2,col=ifelse(svm.pred2!=0,'red','blue'),pch=ifelse(svm.pred2 == y,1,4))

Here we changed the the kernel to polynomial; using the same identification, the plot is more similar to true, with only some misclassified.

(i) Comment on your results.

The closest fit appears to be the polynomial logistic regression model. Similar, and almost as good, the SVM with a polynomial kernal was close in second. The fact that the linear SVM was not good is not surprising, as SVMs with non-linear kernals are good at finding non-linear boundaries, and the linear SVM is would not succeed as we did not have a linear boundary. It is possible that the non-linear logistic regression performed so well because of how well specified the boundary was.

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.

attach(Auto)

(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(ISLR)

gas.med = median(Auto$mpg)
new.var = ifelse(Auto$mpg > gas.med, 1, 0)
Auto$mpglevel = as.factor(new.var)
table(Auto$mpglevel)
## 
##   0   1 
## 196 196

The new variable, mpglevel is split as expected with 0s and 1s

(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.

Note about problem 7. I struggled for days trying to overcome an error about “logical subscript too long.” Even when copying code word for word from working sources I get the same error.

summary(Auto)
##       mpg          cylinders      displacement     horsepower        weight    
##  Min.   : 9.00   Min.   :3.000   Min.   : 68.0   Min.   : 46.0   Min.   :1613  
##  1st Qu.:17.00   1st Qu.:4.000   1st Qu.:105.0   1st Qu.: 75.0   1st Qu.:2225  
##  Median :22.75   Median :4.000   Median :151.0   Median : 93.5   Median :2804  
##  Mean   :23.45   Mean   :5.472   Mean   :194.4   Mean   :104.5   Mean   :2978  
##  3rd Qu.:29.00   3rd Qu.:8.000   3rd Qu.:275.8   3rd Qu.:126.0   3rd Qu.:3615  
##  Max.   :46.60   Max.   :8.000   Max.   :455.0   Max.   :230.0   Max.   :5140  
##                                                                                
##   acceleration        year           origin                      name    
##  Min.   : 8.00   Min.   :70.00   Min.   :1.000   amc matador       :  5  
##  1st Qu.:13.78   1st Qu.:73.00   1st Qu.:1.000   ford pinto        :  5  
##  Median :15.50   Median :76.00   Median :1.000   toyota corolla    :  5  
##  Mean   :15.54   Mean   :75.98   Mean   :1.577   amc gremlin       :  4  
##  3rd Qu.:17.02   3rd Qu.:79.00   3rd Qu.:2.000   amc hornet        :  4  
##  Max.   :24.80   Max.   :82.00   Max.   :3.000   chevrolet chevette:  4  
##                                                  (Other)           :365  
##  mpglevel
##  0:196   
##  1:196   
##          
##          
##          
##          
## 
str(Auto)
## 'data.frame':    392 obs. of  10 variables:
##  $ mpg         : num  18 15 18 16 17 15 14 14 14 15 ...
##  $ cylinders   : num  8 8 8 8 8 8 8 8 8 8 ...
##  $ displacement: num  307 350 318 304 302 429 454 440 455 390 ...
##  $ horsepower  : num  130 165 150 150 140 198 220 215 225 190 ...
##  $ weight      : num  3504 3693 3436 3433 3449 ...
##  $ acceleration: num  12 11.5 11 12 10.5 10 9 8.5 10 8.5 ...
##  $ year        : num  70 70 70 70 70 70 70 70 70 70 ...
##  $ origin      : num  1 1 1 1 1 1 1 1 1 1 ...
##  $ name        : Factor w/ 304 levels "amc ambassador brougham",..: 49 36 231 14 161 141 54 223 241 2 ...
##  $ mpglevel    : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...

After a lot of investigation and trial and error, I have found that the origin variable is causing errors in my code. The work around I have developed is to factor origin. Going back through my notes, I do see “char - make factor or dummy .” In order to not corrupt my table further, I’ll make a copy and use that from here on out. While I’m modifying the table, mpg01 I recall being a variable I created in another exercise, so I’ll remove that too.

Auto.new=Auto
#Auto.new = subset(Auto.new, select=-c(mpg01)) 
Auto.new$name=as.factor(Auto.new$name)
str(Auto.new)
## 'data.frame':    392 obs. of  10 variables:
##  $ mpg         : num  18 15 18 16 17 15 14 14 14 15 ...
##  $ cylinders   : num  8 8 8 8 8 8 8 8 8 8 ...
##  $ displacement: num  307 350 318 304 302 429 454 440 455 390 ...
##  $ horsepower  : num  130 165 150 150 140 198 220 215 225 190 ...
##  $ weight      : num  3504 3693 3436 3433 3449 ...
##  $ acceleration: num  12 11.5 11 12 10.5 10 9 8.5 10 8.5 ...
##  $ year        : num  70 70 70 70 70 70 70 70 70 70 ...
##  $ origin      : num  1 1 1 1 1 1 1 1 1 1 ...
##  $ name        : Factor w/ 304 levels "amc ambassador brougham",..: 49 36 231 14 161 141 54 223 241 2 ...
##  $ mpglevel    : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...

Now back to the assignment

set.seed(1)
tune.lin <- tune(svm, mpglevel ~ ., data = Auto.new, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100, 1000)))
summary(tune.lin)
## 
## 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 5e+00 0.02051282 0.02648194
## 5 1e+01 0.02051282 0.02648194
## 6 1e+02 0.03076923 0.03151981
## 7 1e+03 0.03076923 0.03151981

Of the costs we looked at, 1 had the lowest error rate.

(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(1)
tune.pol <- tune(svm, mpglevel ~ ., data = Auto.new, kernel = "polynomial", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100), degree = c(2, 3, 4)))
summary(tune.pol)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##   100      2
## 
## - best performance: 0.3013462 
## 
## - 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  5e+00      2 0.5511538 0.04366593
## 5  1e+01      2 0.5130128 0.08963366
## 6  1e+02      2 0.3013462 0.09961961
## 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 5e+00      3 0.5511538 0.04366593
## 11 1e+01      3 0.5511538 0.04366593
## 12 1e+02      3 0.3446154 0.09821588
## 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 5e+00      4 0.5511538 0.04366593
## 17 1e+01      4 0.5511538 0.04366593
## 18 1e+02      4 0.5511538 0.04366593

We see a substantially lower error rate at degree of 2 and cost of 100 than any of the other options.

set.seed(1)
tune.rad <- tune(svm, mpglevel ~ ., data = Auto.new, 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.rad)
## 
## 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  5e+00 1e-02 0.04852564 0.03303346
## 5  1e+01 1e-02 0.02557692 0.02093679
## 6  1e+02 1e-02 0.01282051 0.01813094
## 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 5e+00 1e-01 0.02820513 0.03299190
## 11 1e+01 1e-01 0.03076923 0.03375798
## 12 1e+02 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 5e+00 1e+00 0.05884615 0.04020934
## 17 1e+01 1e+00 0.05884615 0.04020934
## 18 1e+02 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 5e+00 5e+00 0.48217949 0.05470903
## 23 1e+01 5e+00 0.48217949 0.05470903
## 24 1e+02 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 5e+00 1e+01 0.51794872 0.04917316
## 29 1e+01 1e+01 0.51794872 0.04917316
## 30 1e+02 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 5e+00 1e+02 0.55115385 0.04366593
## 35 1e+01 1e+02 0.55115385 0.04366593
## 36 1e+02 1e+02 0.55115385 0.04366593

When we use a radial kernal, we get our lowest error rate at gamma of .01 and cost, once again, at 100. This is also the lowest error rate from anything we ran.

(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.

So first, lets make a fit of our best set ups.

svm_radial = svm (mpglevel ~ ., data = Auto.new, kernel = "radial", scale = T, gamma = 0.01, cost = 100)
svm_poly = svm(mpglevel ~ ., data = Auto.new, kernel = "polynomial", cost = 100, degree = 2)
svm_linear = svm(mpglevel ~ ., data = Auto.new, kernel = "linear", cost = 1)

And now for some plots (Note, I saw several analysts using the same for loop to iterate through all the variables. This seems like a great approach; I had to modify it a little because in our code we had to turn origin into a factor, which prevents this from plotting correctly:

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

plotpairs(svm_poly)

plotpairs(svm_radial)

8

detach(Auto)
attach(OJ)

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.

set.seed(1)
train_index <- sample(1:nrow(OJ), 800)

train = OJ[train_index, ]
test = OJ[-train_index, ]

nrow(train)/nrow(OJ)
## [1] 0.7476636
nrow(test)/nrow(OJ)
## [1] 0.2523364

(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_lin_Auto = svm(Purchase ~ ., data = train, kernel = "linear", scale = T, cost = 0.01)

summary(svm_lin_Auto)
## 
## Call:
## svm(formula = Purchase ~ ., data = train, kernel = "linear", cost = 0.01, 
##     scale = T)
## 
## 
## 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?

data.frame(train_error = mean(predict(svm_lin_Auto, train) != train$Purchase), 
           test_error = mean(predict(svm_lin_Auto, test) != test$Purchase))
##   train_error test_error
## 1       0.175  0.1777778

The test error is only slightly higher than the training error.

(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 = 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
##    10
## 
## - best performance: 0.17125 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.17375 0.03884174
## 2   0.01778279 0.17500 0.03996526
## 3   0.03162278 0.17750 0.03717451
## 4   0.05623413 0.18000 0.03073181
## 5   0.10000000 0.17875 0.03064696
## 6   0.17782794 0.17875 0.03537988
## 7   0.31622777 0.17875 0.03438447
## 8   0.56234133 0.17625 0.03197764
## 9   1.00000000 0.17500 0.03061862
## 10  1.77827941 0.17375 0.02972676
## 11  3.16227766 0.17250 0.03270236
## 12  5.62341325 0.17250 0.03322900
## 13 10.00000000 0.17125 0.03488573

Our best performance is .170 with is occurring at a cost of 1.78 and 3.16. We will use 3.16

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

svm.linear = svm(Purchase ~ ., kernel = "linear", data = train, cost = tune.out$best.parameters$cost)
train.pred = predict(svm.linear, train)
table(train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 423  62
##   MM  69 246
(62+69)/ (62+69+423+246)
## [1] 0.16375
test.pred = predict(svm.linear, test)
table(test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 156  12
##   MM  28  74
(29+12) / (156+73+29+12)
## [1] 0.1518519

Our confusion matrix shows a better performance in test than train. Lets use our original code to verify, as that seems suspicious.

data.frame(train_error = mean(predict(svm.linear, train) != train$Purchase), 
           test_error = mean(predict(svm.linear, test) != test$Purchase))
##   train_error test_error
## 1     0.16375  0.1481481

Both calculation methods show an improvement in the train and test error rates.

(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 = train, kernel = "radial")
summary(svm.radial)
## 
## Call:
## svm(formula = Purchase ~ ., data = 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
data.frame(train_error = mean(predict(svm.radial, train) != train$Purchase), 
           test_error = mean(predict(svm.radial, test) != test$Purchase))
##   train_error test_error
## 1     0.15125  0.1851852

Using radial, our train error improved and our test error got a little worse. All error values have been similar. We had 373 support vectors with 188 and 185 for CH and MM respectively. However, we still need to find optimal gamma:

set.seed(1)
tune.out = tune(svm, Purchase ~ ., data = 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 = train, kernel = "radial", cost = tune.out$best.parameters$cost)
train.pred = predict(svm.radial, train)
table(train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 437  48
##   MM  71 244

Confusion matrix above.

data.frame(train_error = mean(predict(svm.radial, train) != train$Purchase), 
           test_error = mean(predict(svm.radial, test) != test$Purchase))
##   train_error test_error
## 1     0.14875  0.1777778

We were able to reduce our train error rate to its lowest level yet at .149. The test error, while improved, still lags behind the optimal in linear.

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

set.seed(1)
svm.poly = svm(Purchase ~ ., data = train, kernel = "poly", degree = 2)
summary(svm.poly)
## 
## Call:
## svm(formula = Purchase ~ ., data = 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

With our ^2 Poly support vectors increased to 447, with a 225 222 now for CH and MM.

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

Confusion Matrix above.

data.frame(train_error = mean(predict(svm.poly, train) != train$Purchase), 
           test_error = mean(predict(svm.poly, test) != test$Purchase))
##   train_error test_error
## 1      0.1825  0.2222222

Now lets tune it.

set.seed(1)
tune.out = tune(svm, Purchase ~ ., data = 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: 0.1775 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39125 0.04210189
## 2   0.01778279 0.37125 0.03537988
## 3   0.03162278 0.36500 0.03476109
## 4   0.05623413 0.33750 0.04714045
## 5   0.10000000 0.32125 0.05001736
## 6   0.17782794 0.24500 0.04758034
## 7   0.31622777 0.19875 0.03972562
## 8   0.56234133 0.20500 0.03961621
## 9   1.00000000 0.20250 0.04116363
## 10  1.77827941 0.18500 0.04199868
## 11  3.16227766 0.17750 0.03670453
## 12  5.62341325 0.18375 0.03064696
## 13 10.00000000 0.18125 0.02779513
svm.poly = svm(Purchase ~ ., data = train, kernel = "poly", degree = 2, cost = tune.out$best.parameters$cost)
train.pred = predict(svm.poly, train)
table(train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 451  34
##   MM  90 225
data.frame(train_error = mean(predict(svm.poly, train) != train$Purchase), 
           test_error = mean(predict(svm.poly, test) != test$Purchase))
##   train_error test_error
## 1       0.155  0.2037037

The train error was right in the middle of our methods, but the test error was last.

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

Our best results came from the linear kernel both in train, and more importantly, in test.

detach(OJ)