Assignment 8
Arsalan Bhojani
8/13/2021
Exercise 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.
- 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.
set.seed(1)
x1 = runif(500) - 0.5
x2 = runif(500) - 0.5
y = 1 * (x1^2 - x2^2 > 0)- 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','navy'),xlab='X1',ylab='X2')
- Fit a logistic regression model to the data, using X1 and X2 as predictors.
dat = data.frame(x1, x2, y = as.factor(y))
glm.fit = glm(y~., data = dat, family = "binomial")
summary(glm.fit)##
## Call:
## glm(formula = y ~ ., family = "binomial", data = dat)
##
## 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- 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.
glm.preds = predict(glm.fit, newdata = dat, type ="response")
plot(x1,x2,col=ifelse(glm.preds>=0.5,'navy','red'),xlab='X1',ylab='X2')
- Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X12, X1 ×X2, log(X2), and so forth).
glm.fit2 = glm(y~I(x1*x2) + poly(x2,2) + poly(x1,2), data=dat, family = "binomial")## Warning: glm.fit: algorithm did not converge## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred- 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.
glm.preds2 = predict(glm.fit2, newdata = dat, type ="response")
plot(x1,x2,col=ifelse(glm.preds2>=0.5,'navy','red'),xlab='X1',ylab='X2')
- 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(y~.,data=dat,kernal='linear',cost=0.01)
svm.preds=predict(svm.fit,newdata=dat,type='response')
plot(x1,x2,col=ifelse(svm.preds!=0,'navy','red'),xlab='X1',ylab='X2')
- 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=dat,kernel='radial',gamma=1)
svm.preds2=predict(svm.fit2,newdata=dat,type='response')
plot(x1,x2,col=ifelse(svm.preds2!=0,'navy','red'),xlab='X1',ylab='X2')
- Comment on your results.
- From the results we can determine that SVM with non-linear kernel and logistic regression with non linear functions of X1 and X2 are better and powerful. SVM with linear kernel and logistic regression without any interaction term are not good.
Exercise 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.
- 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.
df = Auto
df$y = as.factor(ifelse(df$mpg > median(df$mpg), 1, 0))
str(df)## '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 ...
## $ y : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 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.
set.seed(42)
tune.out = tune(svm, y ~ . -mpg -name, data=df, kernel="linear", ranges=list(cost=c(0.1, 1, 10)))
summary(tune.out)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 10
##
## - best performance: 0.08653846
##
## - Detailed performance results:
## cost error dispersion
## 1 0.1 0.09673077 0.05699840
## 2 1.0 0.09423077 0.04632467
## 3 10.0 0.08653846 0.03776796- Now repeat (b), this time using SVMs with radial and polynomial basis kernels, with different values of gamma and degree and cost. train(target~ months_since_donate + num_child + last_gift, data=train, method=‘knn’, trControl = train_control, tuneLength=30)
set.seed(42)
tune.out = tune(svm, y ~ . -mpg -name, data=df, kernel="radial", ranges=list(cost=c(0.1, 1, 10), gamma=c(0.5, 1, 2)))
summary(tune.out)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost gamma
## 1 0.5
##
## - best performance: 0.07628205
##
## - Detailed performance results:
## cost gamma error dispersion
## 1 0.1 0.5 0.08660256 0.04961909
## 2 1.0 0.5 0.07628205 0.04267196
## 3 10.0 0.5 0.08634615 0.04391746
## 4 0.1 1.0 0.09673077 0.05699840
## 5 1.0 1.0 0.07878205 0.04472958
## 6 10.0 1.0 0.09403846 0.04383004
## 7 0.1 2.0 0.15269231 0.10000813
## 8 1.0 2.0 0.08378205 0.04837755
## 9 10.0 2.0 0.10173077 0.05012408- Make some plots to back up your assertions in (b) and (c).
df = df[, -c(1,9)]
set.seed(42)
svm_fit = svm(y ~ ., data=df, kernel='linear', cost=10)
plot(svm_fit, df, displacement~cylinders)
set.seed(42)
svm_fit = svm(y ~ ., data=df, kernel='radial', cost=1, gamma=0.5)
plot(svm_fit, df, weight~acceleration)
set.seed(42)
svm_fit = svm(y ~ ., data=df, kernel='polynomial', cost=10, degree=3)
plot(svm_fit, df, year~horsepower)
Exercise 8: This problem involves the OJ data set which is part of the ISLR package.
- Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
df = OJ
df$Purchase = as.factor(df$Purchase)
set.seed(42)
index = sample(nrow(df), 800)
train = df[index, ]
test = df[-index, ]- 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_fit = svm(Purchase~., data=train, kernel='linear', cost=0.01)
summary(svm_fit)##
## Call:
## svm(formula = Purchase ~ ., data = train, kernel = "linear", cost = 0.01)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: linear
## cost: 0.01
##
## Number of Support Vectors: 432
##
## ( 215 217 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MM- What are the training and test error rates?
svm_preds = predict(svm_fit, train)
confusionMatrix(data=svm_preds, reference=train$Purchase)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 432 77
## MM 60 231
##
## Accuracy : 0.8288
## 95% CI : (0.8008, 0.8542)
## No Information Rate : 0.615
## P-Value [Acc > NIR] : <2e-16
##
## Kappa : 0.6346
##
## Mcnemar's Test P-Value : 0.1716
##
## Sensitivity : 0.8780
## Specificity : 0.7500
## Pos Pred Value : 0.8487
## Neg Pred Value : 0.7938
## Prevalence : 0.6150
## Detection Rate : 0.5400
## Detection Prevalence : 0.6362
## Balanced Accuracy : 0.8140
##
## 'Positive' Class : CH
## svm_preds = predict(svm_fit, test)
confusionMatrix(data=svm_preds, reference=test$Purchase)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 142 25
## MM 19 84
##
## Accuracy : 0.837
## 95% CI : (0.7875, 0.879)
## No Information Rate : 0.5963
## P-Value [Acc > NIR] : <2e-16
##
## Kappa : 0.6585
##
## Mcnemar's Test P-Value : 0.451
##
## Sensitivity : 0.8820
## Specificity : 0.7706
## Pos Pred Value : 0.8503
## Neg Pred Value : 0.8155
## Prevalence : 0.5963
## Detection Rate : 0.5259
## Detection Prevalence : 0.6185
## Balanced Accuracy : 0.8263
##
## 'Positive' Class : CH
## - Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
set.seed(42)
tune.out = tune(svm, Purchase~., data=train, kernel="linear", ranges=list(cost=c(0.1, 1, 10)))
summary(tune.out)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 1
##
## - best performance: 0.175
##
## - Detailed performance results:
## cost error dispersion
## 1 0.1 0.17625 0.03356689
## 2 1.0 0.17500 0.02886751
## 3 10.0 0.18625 0.02729087- Compute the training and test error rates using this new value for cost.
svm_fit = svm(Purchase~., data=train, kernel='linear', cost=1)
#train
svm_preds = predict(svm_fit, train)
confusionMatrix(data=svm_preds, reference=train$Purchase)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 434 76
## MM 58 232
##
## Accuracy : 0.8325
## 95% CI : (0.8048, 0.8577)
## No Information Rate : 0.615
## P-Value [Acc > NIR] : <2e-16
##
## Kappa : 0.6424
##
## Mcnemar's Test P-Value : 0.1419
##
## Sensitivity : 0.8821
## Specificity : 0.7532
## Pos Pred Value : 0.8510
## Neg Pred Value : 0.8000
## Prevalence : 0.6150
## Detection Rate : 0.5425
## Detection Prevalence : 0.6375
## Balanced Accuracy : 0.8177
##
## 'Positive' Class : CH
## # test
svm_preds = predict(svm_fit, test)
confusionMatrix(data=svm_preds, reference=test$Purchase)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 140 23
## MM 21 86
##
## Accuracy : 0.837
## 95% CI : (0.7875, 0.879)
## No Information Rate : 0.5963
## P-Value [Acc > NIR] : <2e-16
##
## Kappa : 0.6605
##
## Mcnemar's Test P-Value : 0.8802
##
## Sensitivity : 0.8696
## Specificity : 0.7890
## Pos Pred Value : 0.8589
## Neg Pred Value : 0.8037
## Prevalence : 0.5963
## Detection Rate : 0.5185
## Detection Prevalence : 0.6037
## Balanced Accuracy : 0.8293
##
## 'Positive' Class : CH
## - Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
svm_fit = svm(Purchase~., data=train, kernel='radial')
summary(svm_fit)##
## Call:
## svm(formula = Purchase ~ ., data = train, kernel = "radial")
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: radial
## cost: 1
##
## Number of Support Vectors: 375
##
## ( 183 192 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MM# train
svm_preds = predict(svm_fit, train)
confusionMatrix(data=svm_preds, reference=train$Purchase)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 453 81
## MM 39 227
##
## Accuracy : 0.85
## 95% CI : (0.8233, 0.874)
## No Information Rate : 0.615
## P-Value [Acc > NIR] : < 2.2e-16
##
## Kappa : 0.675
##
## Mcnemar's Test P-Value : 0.000182
##
## Sensitivity : 0.9207
## Specificity : 0.7370
## Pos Pred Value : 0.8483
## Neg Pred Value : 0.8534
## Prevalence : 0.6150
## Detection Rate : 0.5663
## Detection Prevalence : 0.6675
## Balanced Accuracy : 0.8289
##
## 'Positive' Class : CH
## # test
svm_preds = predict(svm_fit, test)
confusionMatrix(data=svm_preds, reference=test$Purchase)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 146 28
## MM 15 81
##
## Accuracy : 0.8407
## 95% CI : (0.7915, 0.8823)
## No Information Rate : 0.5963
## P-Value [Acc > NIR] : < 2e-16
##
## Kappa : 0.6627
##
## Mcnemar's Test P-Value : 0.06725
##
## Sensitivity : 0.9068
## Specificity : 0.7431
## Pos Pred Value : 0.8391
## Neg Pred Value : 0.8438
## Prevalence : 0.5963
## Detection Rate : 0.5407
## Detection Prevalence : 0.6444
## Balanced Accuracy : 0.8250
##
## 'Positive' Class : CH
## set.seed(42)
tune.out = tune(svm, Purchase~., data=train, kernel="radial", ranges=list(gamma=c(0.5, 1, 2)))
summary(tune.out)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## gamma
## 0.5
##
## - best performance: 0.19375
##
## - Detailed performance results:
## gamma error dispersion
## 1 0.5 0.19375 0.04299952
## 2 1.0 0.20250 0.04158325
## 3 2.0 0.21750 0.04257347svm_fit = svm(Purchase~., data=train, kernel='radial', gamma=0.5)
# train
svm_preds = predict(svm_fit, train)
confusionMatrix(data=svm_preds, reference=train$Purchase)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 457 74
## MM 35 234
##
## Accuracy : 0.8638
## 95% CI : (0.838, 0.8868)
## No Information Rate : 0.615
## P-Value [Acc > NIR] : < 2.2e-16
##
## Kappa : 0.7053
##
## Mcnemar's Test P-Value : 0.0002729
##
## Sensitivity : 0.9289
## Specificity : 0.7597
## Pos Pred Value : 0.8606
## Neg Pred Value : 0.8699
## Prevalence : 0.6150
## Detection Rate : 0.5713
## Detection Prevalence : 0.6637
## Balanced Accuracy : 0.8443
##
## 'Positive' Class : CH
## # test
svm_preds = predict(svm_fit, test)
confusionMatrix(data=svm_preds, reference=test$Purchase)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 144 32
## MM 17 77
##
## Accuracy : 0.8185
## 95% CI : (0.7673, 0.8626)
## No Information Rate : 0.5963
## P-Value [Acc > NIR] : 3.798e-15
##
## Kappa : 0.6145
##
## Mcnemar's Test P-Value : 0.0455
##
## Sensitivity : 0.8944
## Specificity : 0.7064
## Pos Pred Value : 0.8182
## Neg Pred Value : 0.8191
## Prevalence : 0.5963
## Detection Rate : 0.5333
## Detection Prevalence : 0.6519
## Balanced Accuracy : 0.8004
##
## 'Positive' Class : CH
## - Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree=2.
svm_fit = svm(Purchase~., data=train, kernel='polynomial', degree=2)
summary(svm_fit)##
## Call:
## svm(formula = Purchase ~ ., data = train, kernel = "polynomial",
## degree = 2)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: polynomial
## cost: 1
## degree: 2
## coef.0: 0
##
## Number of Support Vectors: 443
##
## ( 217 226 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MM# train
svm_preds = predict(svm_fit, train)
confusionMatrix(data=svm_preds, reference=train$Purchase)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 461 112
## MM 31 196
##
## Accuracy : 0.8212
## 95% CI : (0.7929, 0.8472)
## No Information Rate : 0.615
## P-Value [Acc > NIR] : < 2.2e-16
##
## Kappa : 0.603
##
## Mcnemar's Test P-Value : 2.233e-11
##
## Sensitivity : 0.9370
## Specificity : 0.6364
## Pos Pred Value : 0.8045
## Neg Pred Value : 0.8634
## Prevalence : 0.6150
## Detection Rate : 0.5763
## Detection Prevalence : 0.7163
## Balanced Accuracy : 0.7867
##
## 'Positive' Class : CH
## # test
svm_preds = predict(svm_fit, test)
confusionMatrix(data=svm_preds, reference=test$Purchase)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 149 41
## MM 12 68
##
## Accuracy : 0.8037
## 95% CI : (0.7512, 0.8494)
## No Information Rate : 0.5963
## P-Value [Acc > NIR] : 2.768e-13
##
## Kappa : 0.574
##
## Mcnemar's Test P-Value : 0.00012
##
## Sensitivity : 0.9255
## Specificity : 0.6239
## Pos Pred Value : 0.7842
## Neg Pred Value : 0.8500
## Prevalence : 0.5963
## Detection Rate : 0.5519
## Detection Prevalence : 0.7037
## Balanced Accuracy : 0.7747
##
## 'Positive' Class : CH
## set.seed(42)
tune.out = tune(svm, Purchase~., data=train, kernel="polynomial", ranges=list(degree=c(1, 2, 3)))
summary(tune.out)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## degree
## 1
##
## - best performance: 0.18125
##
## - Detailed performance results:
## degree error dispersion
## 1 1 0.18125 0.03498512
## 2 2 0.19250 0.04216370
## 3 3 0.19625 0.03586723svm_fit = svm(Purchase~., data=train, kernel='polynomial', degree=1)
# train
svm_preds = predict(svm_fit, train)
confusionMatrix(data=svm_preds, reference=train$Purchase)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 432 74
## MM 60 234
##
## Accuracy : 0.8325
## 95% CI : (0.8048, 0.8577)
## No Information Rate : 0.615
## P-Value [Acc > NIR] : <2e-16
##
## Kappa : 0.6433
##
## Mcnemar's Test P-Value : 0.2614
##
## Sensitivity : 0.8780
## Specificity : 0.7597
## Pos Pred Value : 0.8538
## Neg Pred Value : 0.7959
## Prevalence : 0.6150
## Detection Rate : 0.5400
## Detection Prevalence : 0.6325
## Balanced Accuracy : 0.8189
##
## 'Positive' Class : CH
## # test
svm_preds = predict(svm_fit, test)
confusionMatrix(data=svm_preds, reference=test$Purchase)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 141 23
## MM 20 86
##
## Accuracy : 0.8407
## 95% CI : (0.7915, 0.8823)
## No Information Rate : 0.5963
## P-Value [Acc > NIR] : <2e-16
##
## Kappa : 0.6677
##
## Mcnemar's Test P-Value : 0.7604
##
## Sensitivity : 0.8758
## Specificity : 0.7890
## Pos Pred Value : 0.8598
## Neg Pred Value : 0.8113
## Prevalence : 0.5963
## Detection Rate : 0.5222
## Detection Prevalence : 0.6074
## Balanced Accuracy : 0.8324
##
## 'Positive' Class : CH
## - Overall, which approach seems to give the best results on this data?
- Overall, the radial kernal approach gives the better results.