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.

library(knitr)
knitr::opts_chunk$set(echo = TRUE)
library(tidyverse)

## Registered S3 methods overwritten by 'ggplot2':
##   method         from 
##   [.quosures     rlang
##   c.quosures     rlang
##   print.quosures rlang

## Registered S3 method overwritten by 'rvest':
##   method            from
##   read_xml.response xml2

## -- Attaching packages -------------------------------------------- tidyverse 1.2.1 --

## v ggplot2 3.1.1       v purrr   0.3.2  
## v tibble  2.1.1       v dplyr   0.8.0.1
## v tidyr   0.8.3       v stringr 1.4.0  
## v readr   1.3.1       v forcats 0.4.0

## -- Conflicts ----------------------------------------------- tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag()    masks stats::lag()

library(ggthemes)
library(caret)

## Loading required package: lattice

## 
## Attaching package: 'caret'

## The following object is masked from 'package:purrr':
## 
##     lift

library(e1071)

theme_set(theme_tufte(base_size = 14))

df <- data.frame(replicate(2, rnorm(500)))
df <- as.tibble(df)

## Warning: `as.tibble()` is deprecated, use `as_tibble()` (but mind the new semantics).
## This warning is displayed once per session.

class_func <- function(x, y) {
    x^2 + y < 1
}

df <- df %>%
    rename(Var1 = X1, Var2 = X2) %>%
    mutate(Class = ifelse(class_func(Var1, Var2),
                          'Class A',
                          'Class B'),
           Class = factor(Class))

b) Plot the observations, colored according to their class labels. Your plot should display X1 on the x-axis, and X2 on the yaxis.

inTrain <- sample(nrow(df), nrow(df)*0.6, replace = FALSE)

training <- df[inTrain,]
testing <- df[-inTrain,]

ggplot(df, aes(Var1, Var2, col = Class)) +
    geom_point(size = 2)

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

logreg_fit <- glm(Class ~ ., data = training, family = 'binomial')
summary(logreg_fit)

## 
## Call:
## glm(formula = Class ~ ., family = "binomial", data = training)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.8249  -0.7919  -0.2596   0.6825   2.4252  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  -0.4213     0.1505  -2.800  0.00512 ** 
## Var1         -0.1080     0.1429  -0.756  0.44992    
## Var2          2.0213     0.2400   8.421  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 412.47  on 299  degrees of freedom
## Residual deviance: 279.31  on 297  degrees of freedom
## AIC: 285.31
## 
## Number of Fisher Scoring iterations: 5

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.

pred <- predict(logreg_fit, testing, type = 'response')
pred <- ifelse(pred >= 0.5, 'Class B', 'Class A')
mean(pred == testing$Class)

## [1] 0.73

e) Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors.

logreg_fit <- glm(Class ~ poly(Var1, 2) + Var2, data = training, family = 'binomial')

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

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

summary(logreg_fit)

## 
## Call:
## glm(formula = Class ~ poly(Var1, 2) + Var2, family = "binomial", 
##     data = training)
## 
## Deviance Residuals: 
##        Min          1Q      Median          3Q         Max  
## -8.848e-04  -2.000e-08  -2.000e-08   2.000e-08   9.595e-04  
## 
## Coefficients:
##                  Estimate Std. Error z value Pr(>|z|)
## (Intercept)         64.51    5701.25   0.011    0.991
## poly(Var1, 2)1   -2363.93  127037.62  -0.019    0.985
## poly(Var1, 2)2   25924.46 1078037.74   0.024    0.981
## Var2               912.27   37674.57   0.024    0.981
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 4.1247e+02  on 299  degrees of freedom
## Residual deviance: 1.9387e-06  on 296  degrees of freedom
## AIC: 8
## 
## Number of Fisher Scoring iterations: 25

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

pred <- predict(logreg_fit, testing, type = 'response')
pred <- ifelse(pred >= 0.5, 'Class B', 'Class A')
mean(pred == testing$Class)

## [1] 0.995

data_frame(pred = pred, class = testing$Class) %>%
    ggplot(aes(pred, class)) +
    geom_jitter()

## Warning: `data_frame()` is deprecated, use `tibble()`.
## This warning is displayed once per session.

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.

svm_fit <- svm(Class ~ ., data = training,
               kernel = 'linear',
               scale = FALSE)
plot(svm_fit, testing)

mean(predict(svm_fit, testing) == testing$Class)

## [1] 0.745

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_poly <- svm(Class ~ ., data = training,
                kernel = 'polynomial', degree = 2,
                scale = FALSE)
plot(svm_poly, testing)

mean(predict(svm_poly, testing) == testing$Class)

## [1] 0.79

svm_radial <- svm(Class ~ ., data = training,
                  kernel = 'radial',
                  scale = FALSE)
plot(svm_radial, testing)

mean(predict(svm_radial, testing) == testing$Class)

## [1] 0.96

i) Comment on your results: There are strong similarities between Support Vector Machines and Logistic Regression. The best model seemed to be a regression model with the right covariates.

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.

library(ISLR)
theme_set(theme_tufte(base_size = 14))
set.seed(1)

data(Auto)
Auto <- as.tibble(Auto)

Auto <- Auto %>%
    mutate(highmpg = as.integer(mpg > median(mpg))) %>%
    mutate(highmpg = factor(highmpg),
           cylinders = factor(cylinders))

Auto %>%
    sample_n(5) %>%
    select(mpg, highmpg)

## # A tibble: 5 x 2
##     mpg highmpg
##   <dbl> <fct>  
## 1  44.3 1      
## 2  23   1      
## 3  26   1      
## 4  23.9 1      
## 5  23.2 1

Auto <- Auto %>%
    select(-mpg, -name)

dummy_trans <- dummyVars(highmpg ~ ., data = Auto)
Auto_dummy <- predict(dummy_trans, Auto)

## Warning in model.frame.default(Terms, newdata, na.action = na.action, xlev
## = object$lvls): variable 'highmpg' is not a factor

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.

svm_linear <- train(x = Auto_dummy, y = Auto$highmpg,
                    method = 'svmLinear2',
                    trControl = trainControl(method = 'cv', number = 10, allowParallel = TRUE),
                    preProcess = c('center', 'scale'),
                    tuneGrid = expand.grid(cost = seq(1, 20, by = 1)))

svm_linear$finalModel

## 
## Call:
## svm.default(x = as.matrix(x), y = y, kernel = "linear", cost = param$cost, 
##     probability = classProbs)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  2 
##       gamma:  0.09090909 
## 
## Number of Support Vectors:  75

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.

svm_poly <- train(x = Auto_dummy, y = Auto$highmpg,
                  method = 'svmPoly',
                  trControl = trainControl(method = 'cv', number = 10, allowParallel = TRUE),
                  preProcess = c('center', 'scale'),
                  tuneGrid = expand.grid(degree = seq(1, 8, by = 1),
                                         C = seq(1, 5, by = 1),
                                         scale = TRUE))
svm_poly$finalModel

## Support Vector Machine object of class "ksvm" 
## 
## SV type: C-svc  (classification) 
##  parameter : cost C = 1 
## 
## Polynomial kernel function. 
##  Hyperparameters : degree =  2  scale =  TRUE  offset =  1 
## 
## Number of Support Vectors : 71 
## 
## Objective Function Value : -45.587 
## Training error : 0.045918

svm_radial <- train(x = Auto_dummy, y = Auto$highmpg,
                  method = 'svmRadial',
                  trControl = trainControl(method = 'cv', number = 10, allowParallel = TRUE),
                  preProcess = c('center', 'scale'),
                  tuneGrid = expand.grid(C = seq(0.001, 3, length.out = 10),
                                         sigma = seq(0.2, 2, length.out = 5)))
svm_radial$finalModel

## Support Vector Machine object of class "ksvm" 
## 
## SV type: C-svc  (classification) 
##  parameter : cost C = 1.00066666666667 
## 
## Gaussian Radial Basis kernel function. 
##  Hyperparameter : sigma =  1.55 
## 
## Number of Support Vectors : 230 
## 
## Objective Function Value : -73.7206 
## Training error : 0.02551

d) Make some plots to back up your assertions in (b) and (c).

plot(svm_linear)

plot(svm_poly)

plot(svm_radial)

postResample(predict(svm_linear), Auto$highmpg)

##  Accuracy     Kappa 
## 0.9285714 0.8571429

postResample(predict(svm_poly), Auto$highmpg)

##  Accuracy     Kappa 
## 0.9540816 0.9081633

postResample(predict(svm_radial), Auto$highmpg)

##  Accuracy     Kappa 
## 0.9744898 0.9489796

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)

data('OJ')

inTrain <- sample(nrow(OJ), 800, replace = FALSE)

training <- OJ[inTrain,]
testing <- OJ[-inTrain,]

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_linear <- svm(Purchase ~ ., data = training,
                  kernel = 'linear',
                  cost = 0.01)
summary(svm_linear)

## 
## Call:
## svm(formula = Purchase ~ ., data = training, kernel = "linear", 
##     cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
##       gamma:  0.05555556 
## 
## Number of Support Vectors:  435
## 
##  ( 219 216 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

c) What are the training and test error rates?

postResample(predict(svm_linear, training), training$Purchase)

##  Accuracy     Kappa 
## 0.8250000 0.6313971

postResample(predict(svm_linear, testing), testing$Purchase)

##  Accuracy     Kappa 
## 0.8222222 0.6082699

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

svm_linear_tune <- train(Purchase ~ ., data = training,
                         method = 'svmLinear2',
                         trControl = trainControl(method = 'cv', number = 10),
                         preProcess = c('center', 'scale'),
                         tuneGrid = expand.grid(cost = seq(0.01, 10, length.out = 20)))
svm_linear_tune

## Support Vector Machines with Linear Kernel 
## 
## 800 samples
##  17 predictor
##   2 classes: 'CH', 'MM' 
## 
## Pre-processing: centered (17), scaled (17) 
## Resampling: Cross-Validated (10 fold) 
## Summary of sample sizes: 721, 720, 720, 720, 721, 719, ... 
## Resampling results across tuning parameters:
## 
##   cost        Accuracy   Kappa    
##    0.0100000  0.8199215  0.6202565
##    0.5357895  0.8273760  0.6360834
##    1.0615789  0.8236101  0.6284665
##    1.5873684  0.8261105  0.6333280
##    2.1131579  0.8261105  0.6333280
##    2.6389474  0.8273605  0.6362121
##    3.1647368  0.8261105  0.6338114
##    3.6905263  0.8248605  0.6309732
##    4.2163158  0.8248605  0.6309732
##    4.7421053  0.8261105  0.6338114
##    5.2678947  0.8273605  0.6361662
##    5.7936842  0.8273605  0.6361662
##    6.3194737  0.8260947  0.6331693
##    6.8452632  0.8260947  0.6331693
##    7.3710526  0.8260947  0.6331693
##    7.8968421  0.8273605  0.6361662
##    8.4226316  0.8273605  0.6361662
##    8.9484211  0.8273605  0.6361662
##    9.4742105  0.8248447  0.6308145
##   10.0000000  0.8248447  0.6308145
## 
## Accuracy was used to select the optimal model using the largest value.
## The final value used for the model was cost = 0.5357895.

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

postResample(predict(svm_linear_tune, training), training$Purchase)

##  Accuracy     Kappa 
## 0.8350000 0.6524601

postResample(predict(svm_linear_tune, testing), testing$Purchase)

##  Accuracy     Kappa 
## 0.8444444 0.6585983

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

svm_radial <- svm(Purchase ~ ., data = training,
                  method = 'radial',
                  cost = 0.01)
summary(svm_radial)

## 
## Call:
## svm(formula = Purchase ~ ., data = training, method = "radial", 
##     cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  0.01 
##       gamma:  0.05555556 
## 
## Number of Support Vectors:  634
## 
##  ( 319 315 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

postResample(predict(svm_radial, training), training$Purchase)

## Accuracy    Kappa 
##  0.60625  0.00000

postResample(predict(svm_radial, testing), testing$Purchase)

##  Accuracy     Kappa 
## 0.6222222 0.0000000

svm_radial_tune <- train(Purchase ~ ., data = training,
                         method = 'svmRadial',
                         trControl = trainControl(method = 'cv', number = 10),
                         preProcess = c('center', 'scale'),
                         tuneGrid = expand.grid(C = seq(0.01, 10, length.out = 20),
                                                sigma = 0.05691))
svm_radial_tune

## Support Vector Machines with Radial Basis Function Kernel 
## 
## 800 samples
##  17 predictor
##   2 classes: 'CH', 'MM' 
## 
## Pre-processing: centered (17), scaled (17) 
## Resampling: Cross-Validated (10 fold) 
## Summary of sample sizes: 720, 719, 719, 721, 719, 720, ... 
## Resampling results across tuning parameters:
## 
##   C           Accuracy   Kappa    
##    0.0100000  0.6062600  0.0000000
##    0.5357895  0.8274369  0.6315267
##    1.0615789  0.8249527  0.6267051
##    1.5873684  0.8199986  0.6165675
##    2.1131579  0.8174982  0.6105624
##    2.6389474  0.8149824  0.6041027
##    3.1647368  0.8112166  0.5964807
##    3.6905263  0.8112166  0.5964807
##    4.2163158  0.8124512  0.5993391
##    4.7421053  0.8137170  0.6021336
##    5.2678947  0.8137174  0.6017074
##    5.7936842  0.8137174  0.6017074
##    6.3194737  0.8124828  0.5988491
##    6.8452632  0.8124828  0.5988491
##    7.3710526  0.8137641  0.6020343
##    7.8968421  0.8112324  0.5967764
##    8.4226316  0.8112324  0.5967764
##    8.9484211  0.8099666  0.5939493
##    9.4742105  0.8124982  0.5992398
##   10.0000000  0.8124982  0.5992398
## 
## Tuning parameter 'sigma' was held constant at a value of 0.05691
## Accuracy was used to select the optimal model using the largest value.
## The final values used for the model were sigma = 0.05691 and C = 0.5357895.

postResample(predict(svm_radial_tune, training), training$Purchase)

## Accuracy    Kappa 
## 0.851250 0.684392

postResample(predict(svm_radial_tune, testing), testing$Purchase)

##  Accuracy     Kappa 
## 0.8185185 0.6040582

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

svm_poly <- svm(Purchase ~ ., data = training,
                  method = 'polynomial', degree = 2,
                  cost = 0.01)
summary(svm_poly)

## 
## Call:
## svm(formula = Purchase ~ ., data = training, method = "polynomial", 
##     degree = 2, cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  0.01 
##       gamma:  0.05555556 
## 
## Number of Support Vectors:  634
## 
##  ( 319 315 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

postResample(predict(svm_poly, training), training$Purchase)

## Accuracy    Kappa 
##  0.60625  0.00000

postResample(predict(svm_poly, testing), testing$Purchase)

##  Accuracy     Kappa 
## 0.6222222 0.0000000

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

Overall the models are very similar, but the radial kernel does best by a small margin.

Data Mining HW 8

Daniel Upchurch PTH886

5/3/2019

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.

b) Plot the observations, colored according to their class labels. Your plot should display X1 on the x-axis, and X2 on the yaxis.

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

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.

e) Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors.

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.

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.

i) Comment on your results: There are strong similarities between Support Vector Machines and Logistic Regression. The best model seemed to be a regression model with the right covariates.

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.

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.

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.

d) Make some plots to back up your assertions in (b) and (c).

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.

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.

c) What are the training and test error rates?

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

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

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

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

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

Overall the models are very similar, but the radial kernel does best by a small margin.