1. 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*(x12-x22 > 0)

ANSWER 5a: Code shown below:

require(ISLR)
## Loading required package: ISLR
require(dplyr)
## Loading required package: dplyr
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
require(caret)
## Loading required package: caret
## Loading required package: lattice
## Loading required package: ggplot2
require(ggplot2)
require(tidyr)
## Loading required package: tidyr
set.seed(421)
x1 = runif(500) - 0.5
x2 = runif(500) - 0.5
y = 1 * (x1^2 - x2^2 > 0)
  1. Plot the observations, colored according to their class labels. Your plot should display X1 on the x-axis, and X2 on the y-axis.

ANSWER 5b: Plot shown below:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)

plot(x1[y == 0], x2[y == 0], col = "orange", xlab = "X1", ylab = "X2", pch = "+")
points(x1[y == 1], x2[y == 1], col = "blue", pch = 4)

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

ANSWER 5c: Logistic Regression Model shown below:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)

lm_fit = glm(y ~ x1 + x2, family = binomial)
summary(lm_fit)
## 
## Call:
## glm(formula = y ~ x1 + x2, family = binomial)
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.278  -1.227   1.089   1.135   1.175  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)
## (Intercept)  0.11999    0.08971   1.338    0.181
## x1          -0.16881    0.30854  -0.547    0.584
## x2          -0.08198    0.31476  -0.260    0.795
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 691.35  on 499  degrees of freedom
## Residual deviance: 690.99  on 497  degrees of freedom
## AIC: 696.99
## 
## Number of Fisher Scoring iterations: 3

require(tidyr)

  1. 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.

ANSWER 5d: Observations plotted shown:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)

data = data.frame(x1 = x1, x2 = x2, y = y)
lm_prob = predict(lm_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 = "orange", pch = 4)

  1. Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X2 1 , X1 × X2, log(X2), and so forth).

ANSWER 5e: Logistic Regression Model is fitted below:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)

lm_fit = glm(y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), data = data, family = binomial)
summary(lm_fit)
## 
## Call:
## glm(formula = y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), family = binomial, 
##     data = data)
## 
## Deviance Residuals: 
##       Min         1Q     Median         3Q        Max  
## -0.003575   0.000000   0.000000   0.000000   0.003720  
## 
## Coefficients:
##                Estimate Std. Error z value Pr(>|z|)
## (Intercept)      236.09   34920.61   0.007    0.995
## poly(x1, 2)1    3608.97  246381.97   0.015    0.988
## poly(x1, 2)2   88150.22 1333540.93   0.066    0.947
## poly(x2, 2)1    3256.75  177352.91   0.018    0.985
## poly(x2, 2)2  -87128.37 1164195.57  -0.075    0.940
## I(x1 * x2)       -33.23  446735.64   0.000    1.000
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 6.9135e+02  on 499  degrees of freedom
## Residual deviance: 3.3069e-05  on 494  degrees of freedom
## AIC: 12
## 
## Number of Fisher Scoring iterations: 25
  1. 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.

ANSWER 5f: Model applied to training data as shown:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)

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 = "blue", xlab = "X1", ylab = "X2", pch = "+")
points(data_neg$x1, data_neg$x2, col = "orange", pch = 4)

  1. 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.

ANSWER 5g: The support vector classifier is fitted with X1 and X2 as predictors shown below:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)
require(e1071)
## Loading required package: e1071
svm_fit = svm(as.factor(y) ~ x1 + x2, data, kernel = "linear", cost = 0.1)
svm_pred = predict(svm_fit, data)
data_pos = data[svm_pred == 1, ]
data_neg = data[svm_pred == 0, ]
plot(data_pos$x1, data_pos$x2, col = "blue", xlab = "X1", ylab = "X2", pch = "+")
points(data_neg$x1, data_neg$x2, col = "orange", pch = 4)

  1. 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.

ANSWER 5h: SVM is fitted using a non-linear kernel to the data as shown.

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)

svm_fit = svm(as.factor(y) ~ x1 + x2, data, gamma = 1)
svm_pred = predict(svm_fit, data)
data_pos = data[svm_pred == 1, ]
data_neg = data[svm_pred == 0, ]
plot(data_pos$x1, data_pos$x2, col = "blue", xlab = "X1", ylab = "X2", pch = "+")
points(data_neg$x1, data_neg$x2, col = "orange", pch = 4)

  1. Comment on your results.

ANSWER 5i: As previously shown, SVM with non-linear kernel and Logistic Regression with interaction terms are very useful for finding non-linear decision boundaries.

  1. 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.

ANSWER 7a: Binary variable shown below:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)

Auto %>% as_tibble() %>% mutate(above_median = as.factor( ifelse(mpg >= median(mpg), 1, 0) ) ) -> auto
  1. 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.

ANSWER 7b: As shown below the lowest error is when cost = 1 .

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)

set.seed(1)
    auto %>%
    tune(svm, above_median ~ ., data = ., kernel = 'linear', ranges = list(cost = c(0.01, 0.1, 1, 10, 100))) -> auto_svc

summary(auto_svc)
## 
## 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
  1. 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.

ANSWER 7c:

As shown below, for radial kernel we have the lowest error when gamma = 0.01 and cost = 10. However, with the polynomial kernel the lowest error is shown with degree = 2 and cost = 10.

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)

set.seed(1)
auto %>% tune(svm, above_median ~ ., data = ., kernel = 'radial', ranges = list(gamma = c(0.01, 0.1, 1, 10, 100), cost = c(.01, .1, 1, 10))) -> auto_svm_radial
summary(auto_svm_radial)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  gamma cost
##   0.01   10
## 
## - best performance: 0.02557692 
## 
## - Detailed performance results:
##    gamma  cost      error dispersion
## 1  1e-02  0.01 0.55115385 0.04366593
## 2  1e-01  0.01 0.21711538 0.09865227
## 3  1e+00  0.01 0.55115385 0.04366593
## 4  1e+01  0.01 0.55115385 0.04366593
## 5  1e+02  0.01 0.55115385 0.04366593
## 6  1e-02  0.10 0.08929487 0.04382379
## 7  1e-01  0.10 0.07903846 0.03874545
## 8  1e+00  0.10 0.55115385 0.04366593
## 9  1e+01  0.10 0.55115385 0.04366593
## 10 1e+02  0.10 0.55115385 0.04366593
## 11 1e-02  1.00 0.07403846 0.03522110
## 12 1e-01  1.00 0.05371795 0.03525162
## 13 1e+00  1.00 0.06384615 0.04375618
## 14 1e+01  1.00 0.51794872 0.05063697
## 15 1e+02  1.00 0.55115385 0.04366593
## 16 1e-02 10.00 0.02557692 0.02093679
## 17 1e-01 10.00 0.03076923 0.03375798
## 18 1e+00 10.00 0.05884615 0.04020934
## 19 1e+01 10.00 0.51794872 0.04917316
## 20 1e+02 10.00 0.55115385 0.04366593
auto %>% tune(svm, above_median ~ ., data = ., kernel = 'polynomial',ranges = list(degree = seq(2, 5), cost = c(.01, .1, 1, 10))) -> auto_svm_poly
summary(auto_svm_poly)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  degree cost
##       2   10
## 
## - best performance: 0.5841667 
## 
## - Detailed performance results:
##    degree  cost     error dispersion
## 1       2  0.01 0.6019231 0.06346118
## 2       3  0.01 0.6019231 0.06346118
## 3       4  0.01 0.6019231 0.06346118
## 4       5  0.01 0.6019231 0.06346118
## 5       2  0.10 0.6019231 0.06346118
## 6       3  0.10 0.6019231 0.06346118
## 7       4  0.10 0.6019231 0.06346118
## 8       5  0.10 0.6019231 0.06346118
## 9       2  1.00 0.6019231 0.06346118
## 10      3  1.00 0.6019231 0.06346118
## 11      4  1.00 0.6019231 0.06346118
## 12      5  1.00 0.6019231 0.06346118
## 13      2 10.00 0.5841667 0.07806609
## 14      3 10.00 0.6019231 0.06346118
## 15      4 10.00 0.6019231 0.06346118
## 16      5 10.00 0.6019231 0.06346118
  1. 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.

ANSWER 7d: Plots shown below:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)
require(stringr)
## Loading required package: stringr
svm_linr <- svm(above_median ~ ., data = auto, kernel = 'linear', cost = 1)
svm_poly <- svm(above_median ~ ., data = auto, kernel = 'polynomial', degree = 2, cost = 10)
svm_radl <- svm(above_median ~ ., data = auto, kernel = 'radial', gamma = 0.01, cost = 10)

plot_pairs <- function(fit, data, dependent, independents) {
    for (independent in independents) {
        formula = as.formula( str_c( dependent, '~', independent) )
        plot(fit, data, formula)
    }
}

plot_pairs(svm_linr, auto, 'mpg', c('acceleration', 'displacement', 'horsepower'))

  1. 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.

ANSWER 8a: Traning set created oj_train_samples shown below:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)
require(modelr)
## Loading required package: modelr
set.seed(1)
oj_train_samples <-OJ %>% resample_partition(c(train = .8, test = .2))
summary(oj_train_samples)
##       Length Class    Mode
## train 2      resample list
## test  2      resample list
  1. 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.

ANSWER 8b: As shown below there are 471 Support Vectors, 226 are connected to class CH and 224 are connected to class MM.

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)

oj_linr_svc <- svm(Purchase ~ ., data = oj_train_samples$train, kernel = 'linear', cost = 0.01)
summary(oj_linr_svc)
## 
## Call:
## svm(formula = Purchase ~ ., data = oj_train_samples$train, kernel = "linear", 
##     cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  450
## 
##  ( 226 224 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
  1. What are the training and test error rates?

ANSWER 8c: Train(0.157) and Test(0.205) Error Rates are shown below:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)

oj_train_samples$train %>% as_tibble() %>% mutate(Purchase_prime = predict(oj_linr_svc, newdata = .)) %>% summarize('Train Error Rate' = mean(Purchase != Purchase_prime))
## # A tibble: 1 x 1
##   `Train Error Rate`
##                <dbl>
## 1              0.157
oj_train_samples$test %>%as_tibble() %>% mutate(Purchase_prime = predict(oj_linr_svc, newdata = .)) %>% summarize('Test Error Rate' = mean(Purchase != Purchase_prime))
## # A tibble: 1 x 1
##   `Test Error Rate`
##               <dbl>
## 1             0.205
  1. Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.

ANSWER 8d: The best parameters are shown below at cost = 0.015625

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)

set.seed(1)
tune(svm, Purchase ~ ., data = as_tibble( oj_train_samples$train ), kernel = 'linear', ranges = list(cost = 2^seq(-8,4))) -> oj_svc_tune

summary(oj_svc_tune)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  0.015625
## 
## - best performance: 0.1603557 
## 
## - Detailed performance results:
##           cost     error dispersion
## 1   0.00390625 0.1638577 0.03795702
## 2   0.00781250 0.1626813 0.03729402
## 3   0.01562500 0.1603557 0.03840728
## 4   0.03125000 0.1650205 0.03522022
## 5   0.06250000 0.1638988 0.03695773
## 6   0.12500000 0.1615458 0.03716560
## 7   0.25000000 0.1615595 0.03601742
## 8   0.50000000 0.1603967 0.03777655
## 9   1.00000000 0.1650479 0.03445874
## 10  2.00000000 0.1662244 0.03721419
## 11  4.00000000 0.1627086 0.03446296
## 12  8.00000000 0.1627223 0.03666065
## 13 16.00000000 0.1638851 0.03643699
  1. Compute the training and test error rates using this new value for cost.

ANSWER 8e: Train and Test Error Rates are shown below:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)

oj_linr_svc <- svm( Purchase ~ ., data = oj_train_samples$train,kernel = 'linear',cost = oj_svc_tune$best.parameters$cost)

oj_train_samples$train %>% as_tibble() %>% mutate(Purchase_prime = predict(oj_linr_svc)) %>% summarize('Train Error Rate' = mean(Purchase != Purchase_prime))
## # A tibble: 1 x 1
##   `Train Error Rate`
##                <dbl>
## 1              0.152
oj_train_samples$test %>% as_tibble() %>% mutate(Purchase_prime = predict(oj_linr_svc, newdata = .)) %>% summarize('Test Error Rate' = mean(Purchase != Purchase_prime))
## # A tibble: 1 x 1
##   `Test Error Rate`
##               <dbl>
## 1               0.2
  1. Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.

ANSWER 8f: Parts (b) through (e) using SVM with Radial Kernel are shown below:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)

oj_radl_svc <- svm( Purchase ~ ., data = oj_train_samples$train, kernel = 'radial')
summary(oj_radl_svc)
## 
## Call:
## svm(formula = Purchase ~ ., data = oj_train_samples$train, kernel = "radial")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  390
## 
##  ( 201 189 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
oj_train_samples$train %>% as_tibble() %>% mutate(Purchase_prime = predict(oj_radl_svc, newdata = .)) %>% summarize('Train Error Rate' = mean(Purchase != Purchase_prime))
## # A tibble: 1 x 1
##   `Train Error Rate`
##                <dbl>
## 1              0.145
oj_train_samples$test %>% as_tibble() %>% mutate(Purchase_prime = predict(oj_radl_svc, newdata = .)) %>% summarize('Test Error Rate' = mean(Purchase != Purchase_prime))
## # A tibble: 1 x 1
##   `Test Error Rate`
##               <dbl>
## 1             0.186
set.seed(1)
tune(svm,Purchase ~ .,data = as_tibble( oj_train_samples$train ),kernel = 'radial',ranges = list(cost = 2^seq(-8,4))) -> oj_radl_tune
summary(oj_radl_tune)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.1673461 
## 
## - Detailed performance results:
##           cost     error dispersion
## 1   0.00390625 0.3764843 0.06228157
## 2   0.00781250 0.3764843 0.06228157
## 3   0.01562500 0.3764843 0.06228157
## 4   0.03125000 0.3366211 0.07569987
## 5   0.06250000 0.1918605 0.04001821
## 6   0.12500000 0.1755267 0.03617482
## 7   0.25000000 0.1755404 0.03986210
## 8   0.50000000 0.1720246 0.04204614
## 9   1.00000000 0.1673461 0.03585901
## 10  2.00000000 0.1720520 0.04109274
## 11  4.00000000 0.1731874 0.04013555
## 12  8.00000000 0.1825581 0.03589686
## 13 16.00000000 0.1837346 0.03185391
oj_radl_svc <- svm(Purchase ~ .,data = oj_train_samples$train,kernel = 'linear',cost = oj_radl_tune$best.parameters$cost)

oj_train_samples$train %>% as_tibble() %>% mutate(Purchase_prime = predict(oj_radl_svc)) %>% summarize('Train Error Rate' = mean(Purchase != Purchase_prime))
## # A tibble: 1 x 1
##   `Train Error Rate`
##                <dbl>
## 1              0.152
oj_train_samples$test %>% as_tibble() %>% mutate(Purchase_prime = predict(oj_radl_svc, newdata = .)) %>% summarize('Test Error Rate' = mean(Purchase != Purchase_prime))
## # A tibble: 1 x 1
##   `Test Error Rate`
##               <dbl>
## 1               0.2
  1. Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree = 2.

ANSWER 8g: Parts (b) through (e) using SVM with polynomial Kernel( degree = 2) are shown below:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tidyr)
oj_poly_svc <- svm(Purchase ~ ., data = oj_train_samples$train, kernel = 'polynomial',degree = 2)
summary(oj_poly_svc)
## 
## Call:
## svm(formula = Purchase ~ ., data = oj_train_samples$train, kernel = "polynomial", 
##     degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  1 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  467
## 
##  ( 238 229 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
oj_train_samples$train %>% as_tibble() %>% mutate(Purchase_prime = predict(oj_poly_svc, newdata = .)) %>% summarize('Train Error Rate' = mean(Purchase != Purchase_prime))
## # A tibble: 1 x 1
##   `Train Error Rate`
##                <dbl>
## 1              0.174
oj_train_samples$test %>% as_tibble() %>% mutate(Purchase_prime = predict(oj_poly_svc, newdata = .)) %>% summarize('Test Error Rate' = mean(Purchase != Purchase_prime))
## # A tibble: 1 x 1
##   `Test Error Rate`
##               <dbl>
## 1             0.214
set.seed(1)
tune(svm,Purchase ~ .,data = as_tibble( oj_train_samples$train ),kernel = 'polynomial',ranges = list(cost = 2^seq(-8,4)),degree = 2) -> oj_poly_tune
summary(oj_poly_tune)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     8
## 
## - best performance: 0.1708208 
## 
## - Detailed performance results:
##           cost     error dispersion
## 1   0.00390625 0.3764843 0.06228157
## 2   0.00781250 0.3776471 0.06301831
## 3   0.01562500 0.3495075 0.07242409
## 4   0.03125000 0.3389740 0.06929994
## 5   0.06250000 0.3179617 0.06929113
## 6   0.12500000 0.3016005 0.07149584
## 7   0.25000000 0.2268673 0.03547776
## 8   0.50000000 0.2012312 0.04937234
## 9   1.00000000 0.1918194 0.04481021
## 10  2.00000000 0.1848290 0.04773197
## 11  4.00000000 0.1801505 0.04354956
## 12  8.00000000 0.1708208 0.04408733
## 13 16.00000000 0.1732011 0.04348912
oj_poly_svc <- svm(Purchase ~ .,data = oj_train_samples$train,kernel = 'polynomial',cost = oj_poly_tune$best.parameters$cost)

oj_train_samples$train %>% as_tibble() %>%mutate(Purchase_prime = predict(oj_poly_svc)) %>%summarize('Train Error Rate' = mean(Purchase != Purchase_prime))
## # A tibble: 1 x 1
##   `Train Error Rate`
##                <dbl>
## 1              0.137
oj_train_samples$test %>% as_tibble() %>% mutate(Purchase_prime = predict(oj_poly_svc, newdata = .)) %>% summarize('Test Error Rate' = mean(Purchase != Purchase_prime))
## # A tibble: 1 x 1
##   `Test Error Rate`
##               <dbl>
## 1             0.219
  1. Overall, which approach seems to give the best results on this data?

ANSWER 8h: Overall we can tell that the Radial Kernel approach gives the best results since our train and test error rates were the lowest compared to Linear and Polynomial kernels.