##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 > x1=runif (500) -0.5 > y=1*(x12-x22 > 0)
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 yaxis.
plot(x1, x2, col=(4-y))

  1. Fit a logistic regression model to the data, using X1 and X2 as predictors.
model.log <- glm(y~ x1 + x2, family = "binomial")
summary(model.log)
## 
## Call:
## glm(formula = y ~ x1 + x2, family = "binomial")
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.4377  -1.1516  -0.9283   1.1725   1.4544  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)  
## (Intercept) -0.02589    0.09032  -0.287   0.7744  
## x1           0.70982    0.32517   2.183   0.0290 *
## x2           0.69319    0.32154   2.156   0.0311 *
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 693.08  on 499  degrees of freedom
## Residual deviance: 684.35  on 497  degrees of freedom
## AIC: 690.35
## 
## Number of Fisher Scoring iterations: 4
  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.
data5 <- data.frame(x1,x2, y=as.factor(y))
probs <- predict(model.log, data5, type = 'response')
preds <- ifelse(probs >= 0.5, 1,0)
plot(x1, x2, col=(4-preds))

  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).
x3 <- as.numeric(I(x1^2))
x4 <- as.numeric(I(x2^2))
x5 <- as.numeric(I(x1 * x2))
#x6 <- as.numeric(I(log(x2)))
data5e <-data.frame(x1,x2,x3,x4,x5, y)
model.log2 <- glm(y~ x1 + x2 + I(x1^2) + I(x2^2) + I(x1 * x2), family = "binomial")
## Warning: glm.fit: algorithm did not converge
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
#model.log2 <- glm(y~., data = data5e, family = "binomial")
  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.
y_class <- predict(model.log2, data5e, type = 'response')
preds2 <- ifelse(y_class >= 0.5, 1,0)
plot(x1, x2, col=(4-preds2))
grid()

  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.
library(e1071)
model.svm <- svm(y~., data=data5, kernel = "linear", cost = 10, scale = FALSE)
plot(model.svm,data5)

  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.
library(e1071)
model.svmr <- svm(y~., data=data5, kernel = "radial", gamma = 1, cost = .01)
plot(model.svmr,data5)

  1. Comment on your results. Good problem to review linear vs non-linear data and which models to use. If the data is linear, which in this example we can see that the data is non linear (based on the plot of the data), a simple linear regression model is easy, efficient and reduces computing cost. For non-linear data, the logistec regression model or the svm model (radial or poly kernals) will work. This was a good exercise to see how the data “was” linear or non-linear and based on that, see how the different models logit, svm linear or svm radial would predict and plot those predictions.

##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.

library(ISLR)
## Warning: package 'ISLR' was built under R version 4.1.2
str(Auto)
## 'data.frame':    392 obs. of  9 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 ...
  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.
library(tidyverse)
## -- Attaching packages --------------------------------------- tidyverse 1.3.1 --
## v ggplot2 3.3.5     v purrr   0.3.4
## v tibble  3.1.4     v dplyr   1.0.8
## v tidyr   1.2.0     v stringr 1.4.0
## v readr   2.1.2     v forcats 0.5.1
## Warning: package 'ggplot2' was built under R version 4.1.3
## Warning: package 'tidyr' was built under R version 4.1.3
## Warning: package 'readr' was built under R version 4.1.3
## Warning: package 'purrr' was built under R version 4.1.3
## Warning: package 'dplyr' was built under R version 4.1.3
## -- Conflicts ------------------------------------------ tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag()    masks stats::lag()
Auto2 = Auto
med_mpg = median(Auto2$mpg)
Auto2 = mutate(Auto2, mpg01 = ifelse(mpg > med_mpg, 1,0))
Auto2$mpg01 = as.factor(Auto2$mpg01)
Auto2$origin = as.factor(Auto2$origin)
Auto2 = subset(Auto2, select = -c(mpg,name))
str(Auto2)
## 'data.frame':    392 obs. of  8 variables:
##  $ 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      : Factor w/ 3 levels "1","2","3": 1 1 1 1 1 1 1 1 1 1 ...
##  $ mpg01       : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
#table(Auto2$mpg01)
  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.
library(caret)
## Warning: package 'caret' was built under R version 4.1.2
## Loading required package: lattice
## 
## Attaching package: 'caret'
## The following object is masked from 'package:purrr':
## 
##     lift
# Set up Repeated k-fold Cross Validation
train_control1 <- trainControl(method="repeatedcv", number=10, repeats=3) ##(taking too long with repeats = 3)

Creating SVM model with c held constant at a value of 1: Accuracy = 0.913977

svm7 <- train(mpg01 ~., data = Auto2, method = "svmLinear", trControl = train_control1, preProcess = c ("center","scale"))
svm7
## Support Vector Machines with Linear Kernel 
## 
## 392 samples
##   7 predictor
##   2 classes: '0', '1' 
## 
## Pre-processing: centered (8), scaled (8) 
## Resampling: Cross-Validated (10 fold, repeated 3 times) 
## Summary of sample sizes: 353, 352, 354, 354, 353, 353, ... 
## Resampling results:
## 
##   Accuracy   Kappa    
##   0.9113417  0.8226492
## 
## Tuning parameter 'C' was held constant at a value of 1

Results saved for svm7 C held constant svmLinear

res1 <- as_tibble(svm7$results[which.min(svm7$results[,2]),])
res1
## # A tibble: 1 x 5
##       C Accuracy Kappa AccuracySD KappaSD
##   <dbl>    <dbl> <dbl>      <dbl>   <dbl>
## 1     1    0.911 0.823     0.0456  0.0912

Creating SVM model with tuneGrid for “C”: Accuracy = 0.9200675, final value of C = 1.895263 SVM7a is slightly more accurate than SVM7 when using cross-validation for grid values of C

svm7a <- train(mpg01 ~., data = Auto2, method = "svmLinear", trControl = train_control1, preProcess = c ("center","scale"), tuneGrid = expand.grid(C = seq(0.01,2,length = 20)))
svm7a
## Support Vector Machines with Linear Kernel 
## 
## 392 samples
##   7 predictor
##   2 classes: '0', '1' 
## 
## Pre-processing: centered (8), scaled (8) 
## Resampling: Cross-Validated (10 fold, repeated 3 times) 
## Summary of sample sizes: 352, 354, 353, 353, 352, 352, ... 
## Resampling results across tuning parameters:
## 
##   C          Accuracy   Kappa    
##   0.0100000  0.9096727  0.8192702
##   0.1147368  0.9054611  0.8108582
##   0.2194737  0.9037506  0.8074549
##   0.3242105  0.9038147  0.8075664
##   0.4289474  0.9021491  0.8042376
##   0.5336842  0.9055038  0.8109369
##   0.6384211  0.9037731  0.8074967
##   0.7431579  0.9063158  0.8125934
##   0.8478947  0.9089249  0.8177970
##   0.9526316  0.9123673  0.8246989
##   1.0573684  0.9131781  0.8263150
##   1.1621053  0.9106129  0.8211991
##   1.2668421  0.9132006  0.8263745
##   1.3715789  0.9132006  0.8263745
##   1.4763158  0.9123673  0.8247079
##   1.5810526  0.9149100  0.8297687
##   1.6857895  0.9148662  0.8296809
##   1.7905263  0.9157434  0.8314353
##   1.8952632  0.9157434  0.8314353
##   2.0000000  0.9157434  0.8314353
## 
## Accuracy was used to select the optimal model using the largest value.
## The final value used for the model was C = 1.790526.
#svm7a$bestTune
res2 <- as_tibble(svm7a$results[which.min(svm7a$results[,2]),])
res2
## # A tibble: 1 x 5
##       C Accuracy Kappa AccuracySD KappaSD
##   <dbl>    <dbl> <dbl>      <dbl>   <dbl>
## 1 0.429    0.902 0.804     0.0362  0.0723
  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.

Radial: I used Caret (based on class R Lab to identify the optimal gamma and C for the radial SVM model) Optimal Gamma: 0.2228925 C: 8

svm7r <- train(mpg01 ~., data = Auto2, method = "svmRadial", trControl = train_control1, preProcess = c ("center","scale"), tuneLength = 10)
svm7r$bestTune
##       sigma  C
## 7 0.1593645 16
res3 <- as_tibble(svm7r$results[which.min(svm7r$results[,2]),])
res3
## # A tibble: 1 x 6
##   sigma     C Accuracy Kappa AccuracySD KappaSD
##   <dbl> <dbl>    <dbl> <dbl>      <dbl>   <dbl>
## 1 0.159  0.25    0.908 0.816     0.0584   0.116

Polynomial

svm7p <- train(mpg01 ~., data = Auto2, method = "svmPoly", trControl = train_control1, preProcess = c ("center","scale"), tuneLength = 4)
svm7p$bestTune
##    degree scale C
## 44      3   0.1 2
res4 <- as_tibble(svm7p$results[which.min(svm7p$results[,2]),])
res4
## # A tibble: 1 x 7
##   degree scale     C Accuracy Kappa AccuracySD KappaSD
##    <int> <dbl> <dbl>    <dbl> <dbl>      <dbl>   <dbl>
## 1      1 0.001  0.25    0.702 0.411      0.124   0.240

Accuracy results per svm model Comments: Poly took way too long to run with tunelength of 10, so I changed it to 4. Accuracy is fairly close, SVM linear however has the highest accuracy.

df<-tibble(Model=c('SVM Linear','SVM Linear w/ choice of cost','SVM Radial','SVM Poly'),Accuracy=c(svm7$results[2][[1]],res2$Accuracy,res3$Accuracy,res4$Accuracy))
df #%>% arrange(Accuracy)
## # A tibble: 4 x 2
##   Model                        Accuracy
##   <chr>                           <dbl>
## 1 SVM Linear                      0.911
## 2 SVM Linear w/ choice of cost    0.902
## 3 SVM Radial                      0.908
## 4 SVM Poly                        0.702
  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.

svm linear with optimal C chosen via Caret

plot(svm7a)

svm radial with optimal parameters chosen via Caret

plot(svm7r)

svm poly with optimal parameters chosen via Caret

plot(svm7p)

  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.
library(ISLR)
str(OJ)
## 'data.frame':    1070 obs. of  18 variables:
##  $ Purchase      : Factor w/ 2 levels "CH","MM": 1 1 1 2 1 1 1 1 1 1 ...
##  $ WeekofPurchase: num  237 239 245 227 228 230 232 234 235 238 ...
##  $ StoreID       : num  1 1 1 1 7 7 7 7 7 7 ...
##  $ PriceCH       : num  1.75 1.75 1.86 1.69 1.69 1.69 1.69 1.75 1.75 1.75 ...
##  $ PriceMM       : num  1.99 1.99 2.09 1.69 1.69 1.99 1.99 1.99 1.99 1.99 ...
##  $ DiscCH        : num  0 0 0.17 0 0 0 0 0 0 0 ...
##  $ DiscMM        : num  0 0.3 0 0 0 0 0.4 0.4 0.4 0.4 ...
##  $ SpecialCH     : num  0 0 0 0 0 0 1 1 0 0 ...
##  $ SpecialMM     : num  0 1 0 0 0 1 1 0 0 0 ...
##  $ LoyalCH       : num  0.5 0.6 0.68 0.4 0.957 ...
##  $ SalePriceMM   : num  1.99 1.69 2.09 1.69 1.69 1.99 1.59 1.59 1.59 1.59 ...
##  $ SalePriceCH   : num  1.75 1.75 1.69 1.69 1.69 1.69 1.69 1.75 1.75 1.75 ...
##  $ PriceDiff     : num  0.24 -0.06 0.4 0 0 0.3 -0.1 -0.16 -0.16 -0.16 ...
##  $ Store7        : Factor w/ 2 levels "No","Yes": 1 1 1 1 2 2 2 2 2 2 ...
##  $ PctDiscMM     : num  0 0.151 0 0 0 ...
##  $ PctDiscCH     : num  0 0 0.0914 0 0 ...
##  $ ListPriceDiff : num  0.24 0.24 0.23 0 0 0.3 0.3 0.24 0.24 0.24 ...
##  $ STORE         : num  1 1 1 1 0 0 0 0 0 0 ...

Training and test set

set.seed(22)
train.oj=sample(1:nrow(OJ), 800)
OJ.train = OJ[train.oj,]
OJ.test = OJ[-train.oj,]
  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.
svm8 = svm(Purchase~., data = OJ.train, kernel = 'linear', cost = 0.01, scale = FALSE)
summary(svm8)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "linear", cost = 0.01, 
##     scale = FALSE)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  621
## 
##  ( 310 311 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
  1. What are the training and test error rates?
svm8trainerror <- mean(predict(svm8, OJ.train) != OJ.train$Purchase)
svm8testerror <- mean(predict(svm8, OJ.test) != OJ.test$Purchase)
error<-tibble(Model=c('Train Error','Test Error'),Error=c(svm8trainerror,svm8testerror))
error
## # A tibble: 2 x 2
##   Model       Error
##   <chr>       <dbl>
## 1 Train Error 0.236
## 2 Test Error  0.219
  1. Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
set.seed(22)
tune.out=tune(svm,Purchase~., data = OJ.train, kernel = 'linear',ranges=list(cost=c(0.001, 0.01, 0.1, 1,10)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##   0.1
## 
## - best performance: 0.18375 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1 1e-03 0.34500 0.05342440
## 2 1e-02 0.18625 0.03304563
## 3 1e-01 0.18375 0.03866254
## 4 1e+00 0.18375 0.03775377
## 5 1e+01 0.18750 0.03952847

best model for svm with range of C

svm.best <- tune.out$best.model
summary(svm.best)
## 
## Call:
## best.tune(method = svm, train.x = Purchase ~ ., data = OJ.train, 
##     ranges = list(cost = c(0.001, 0.01, 0.1, 1, 10)), kernel = "linear")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.1 
## 
## Number of Support Vectors:  365
## 
##  ( 182 183 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
  1. Compute the training and test error rates using this new value for cost.
svm8atrainerror <- mean(predict(svm.best, OJ.train) != OJ.train$Purchase)
svm8atesterror <- mean(predict(svm.best, OJ.test) != OJ.test$Purchase)
error1<-tibble(Model=c('Train Error SVM Best','Test Error SVM Best'),Error=c(svm8atrainerror,svm8atesterror))
error1
## # A tibble: 2 x 2
##   Model                Error
##   <chr>                <dbl>
## 1 Train Error SVM Best 0.176
## 2 Test Error SVM Best  0.141
  1. Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
svm8r=svm(Purchase~., data=OJ.train, kernel ="radial", gamma=1, cost=1)
summary(svm8r)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "radial", gamma = 1, 
##     cost = 1)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  501
## 
##  ( 230 271 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
svm8rtrainerror <- mean(predict(svm8r, OJ.train) != OJ.train$Purchase)
svm8rtesterror <- mean(predict(svm8r, OJ.test) != OJ.test$Purchase)
error.r<-tibble(Model=c('Train Error radial','Test Error radial'),Error=c(svm8rtrainerror,svm8rtesterror))
error.r
## # A tibble: 2 x 2
##   Model              Error
##   <chr>              <dbl>
## 1 Train Error radial 0.119
## 2 Test Error radial  0.196

using caret for the optimal radial model

svm8ro <- train(Purchase ~., data = OJ.train, method = "svmRadial", trControl = train_control1, preProcess = c ("center","scale"), tuneLength = 10)
svm8ro$bestTune
##        sigma   C
## 2 0.05532364 0.5
svm8rotrainerror <- mean(predict(svm8ro, OJ.train) != OJ.train$Purchase)
svm8rotesterror <- mean(predict(svm8ro, OJ.test) != OJ.test$Purchase)
error.ro<-tibble(Model=c('Train Error radial','Test Error radial'),Error=c(svm8rotrainerror,svm8rotesterror))
error.ro
## # A tibble: 2 x 2
##   Model              Error
##   <chr>              <dbl>
## 1 Train Error radial 0.17 
## 2 Test Error radial  0.159
  1. Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree=2.
svm8p=svm(Purchase~., data=OJ.train, kernel ="polynomial", degree=2, cost=1)
summary(svm8r)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "radial", gamma = 1, 
##     cost = 1)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  501
## 
##  ( 230 271 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
svm8ptrainerror <- mean(predict(svm8p, OJ.train) != OJ.train$Purchase)
svm8ptesterror <- mean(predict(svm8p, OJ.test) != OJ.test$Purchase)
error.p<-tibble(Model=c('Train Error poly','Test Error poly'),Error=c(svm8ptrainerror,svm8ptesterror))
error.p
## # A tibble: 2 x 2
##   Model            Error
##   <chr>            <dbl>
## 1 Train Error poly 0.194
## 2 Test Error poly  0.167
svm8po <- train(Purchase ~., data = OJ.train, method = "svmPoly", trControl = train_control1, preProcess = c ("center","scale"), tuneLength = 4)
svm8potrainerror <- mean(predict(svm8po, OJ.train) != OJ.train$Purchase)
svm8potesterror <- mean(predict(svm8po, OJ.test) != OJ.test$Purchase)
error.po<-tibble(Model=c('Train Error polyo','Test Error polyo'),Error=c(svm8potrainerror,svm8potesterror))
error.po
## # A tibble: 2 x 2
##   Model             Error
##   <chr>             <dbl>
## 1 Train Error polyo 0.169
## 2 Test Error polyo  0.144
  1. Overall, which approach seems to give the best results on this data? Train error: Test error: SVM-Linear:0.23 SVM-Linear:0.21 SVM-Linear optimal:0.17 SVM-Linear optimal:0.14
    SVM-Radial:0.12 SVM-Radial:0.19
    SVM-Radial optimal:0.16 SVM-Radial optimal:0.15 SVM-Poly:0.19 SVM-Poly:0.17 SVM-Poly optimal:0.17 SVM-Poly optimal:0.14

Poly seems to be the best approach, though the training error for radial (standard) had the best error results but not on the test dataset. Overall, as the data becomes more non-linear, the non-linear kernals performed better than the linear kernal. Tuning the parameters for radial and poly produced the best performance, in conclusion the non-linear kernal models with optimal parameters performed the best