Assignment #8
Briley Winge
Applied
Question 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.
x1 <- runif(500) - 0.5
x2 <- runif(500) - 0.5
y <- 1 * (x1^2 - x2^2 > 0)
data <- data.frame(x1 = x1, x2 = x2, y = as.factor(y))- Plot the observations, colored according to their class labels. Your plot should display X1 on the x-axis, and X2 on the y-axis.
library(ggplot2)data_plot <- ggplot(data,
aes(
x = x1,
y = x2,
color = y
)) +
geom_point()
data_plot
- Fit a logistic regression model to the data, using X1 and X2 as predictors.
set.seed(1)
data_glm <- glm(y ~., data, family = 'binomial')
summary(data_glm)##
## Call:
## glm(formula = y ~ ., family = "binomial", data = data)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.03364 0.08977 0.375 0.708
## x1 0.09568 0.31439 0.304 0.761
## x2 0.47105 0.31456 1.498 0.134
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 692.95 on 499 degrees of freedom
## Residual deviance: 690.59 on 497 degrees of freedom
## AIC: 696.59
##
## 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.
set.seed(1)
glm_preds <- predict(data_glm, type = "response")
glm_probs <- ifelse(glm_preds > 0.5, 1, 0)
data$glm_preds <- as.factor(glm_probs)
data_coef <- coef(data_glm)
ggplot(
data,
aes(
x = x1,
y = x2,
color = glm_probs
)) +
geom_point() +
geom_abline(intercept = -data_coef[1] / data_coef[3], slope = -data_coef[2] / data_coef[3])
- Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X^2_1 , X_1 × X_2, log(X_2),and so forth).
set.seed(1)
square_glm <- glm(y ~ poly(x1, 2) + x1*x2, data, family = "binomial")
summary(square_glm)##
## Call:
## glm(formula = y ~ poly(x1, 2) + x1 * x2, family = "binomial",
## data = data)
##
## Coefficients: (1 not defined because of singularities)
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.2293 0.1179 1.945 0.0518 .
## poly(x1, 2)1 2.0903 2.8383 0.736 0.4614
## poly(x1, 2)2 37.4311 3.4377 10.888 <2e-16 ***
## x1 NA NA NA NA
## x2 0.1757 0.3896 0.451 0.6520
## x1:x2 -0.2041 1.5816 -0.129 0.8973
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 692.95 on 499 degrees of freedom
## Residual deviance: 492.23 on 495 degrees of freedom
## AIC: 502.23
##
## Number of Fisher Scoring iterations: 5- Apply this model to the training data in order to obtain a predicted class label for each training observation. Plot the observations, colored according to the predicted class labels. The decision boundary should be obviously non-linear. If it is not, then repeat (a)-(e) until you come up with an example in which the predicted class labels are obviously non-linear.
set.seed(1)
square_glm_preds <- predict(square_glm, type = "response")
square_glm_probs <- ifelse(square_glm_preds > 0.5, 1, 0)
data$square_glm_preds <- as.factor(square_glm_probs)
square_coef <- coef(square_glm)
ggplot(
data,
aes(
x = x1,
y = x2,
color = square_glm_probs
)) +
geom_point()
- 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)##
## Attaching package: 'e1071'## The following object is masked from 'package:ggplot2':
##
## elementset.seed(1)
svm_data <- svm(y ~ x1 + x2,
data = data,
kernel = "linear")
svm_data##
## Call:
## svm(formula = y ~ x1 + x2, data = data, kernel = "linear")
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: linear
## cost: 1
##
## Number of Support Vectors: 479svm_pred <- predict(svm_data, data)
data$svm_pred <- svm_predggplot(data,
aes(
x = x1,
y = x2,
color = svm_pred
)) +
geom_point()
- Fit a SVM using a non-linear kernel to the data. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.
set.seed(1)
nonlinear_svm_data <- svm(y ~ x1 + x2,
data = data,
kernel = "radial")
nonlinear_svm_data##
## Call:
## svm(formula = y ~ x1 + x2, data = data, kernel = "radial")
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: radial
## cost: 1
##
## Number of Support Vectors: 160nonlinear_svm_pred <- predict(nonlinear_svm_data, data)
data$svm_pred <- nonlinear_svm_predggplot(data,
aes(
x = x1,
y = x2,
color = nonlinear_svm_pred
)) +
geom_point()
- Comment on your results.
When the support machince vector has a kernal of linear it creates a linear boundary like the logistic regression model. However, when manipulating the predictor variables (by squaring the value) or using radial as the kernel for the support machine vector has a boundary line more adapt to the data.
Question 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(ISLR2)attach(Auto)## The following object is masked from package:ggplot2:
##
## mpg- 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.
median_mpg <- median(Auto$mpg)
Auto$target <- ifelse(Auto$mpg > median_mpg, 1, 0)
Auto$target <- as.factor(Auto$target)- 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.
set.seed(1)
updated_auto <- Auto[, -c(1,9)]
auto_svm <- svm(target ~ ., data = updated_auto, kernel = "linear")
summary(auto_svm)##
## Call:
## svm(formula = target ~ ., data = updated_auto, kernel = "linear")
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: linear
## cost: 1
##
## Number of Support Vectors: 88
##
## ( 43 45 )
##
##
## Number of Classes: 2
##
## Levels:
## 0 1set.seed(1)
tune.out <- tune(svm, target ~ ., data = updated_auto, kernel = "linear",
ranges = list(cost = c(0.001, 0.01, 0.1, 1, 5, 10)))
summary(tune.out)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 1
##
## - best performance: 0.08435897
##
## - Detailed performance results:
## cost error dispersion
## 1 1e-03 0.13525641 0.05661708
## 2 1e-02 0.08923077 0.04698309
## 3 1e-01 0.09185897 0.04393409
## 4 1e+00 0.08435897 0.03662670
## 5 5e+00 0.08948718 0.03898410
## 6 1e+01 0.08948718 0.03898410From the tune out summary, it can be seen that when cost is equal to 1, it has the lowest cross-validation error of 0.08435897.
- 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)
radial_auto_svm <- svm(target ~ ., data = updated_auto, kernel = "radial")
summary(radial_auto_svm)##
## Call:
## svm(formula = target ~ ., data = updated_auto, kernel = "radial")
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: radial
## cost: 1
##
## Number of Support Vectors: 102
##
## ( 51 51 )
##
##
## Number of Classes: 2
##
## Levels:
## 0 1set.seed(1)
radial_tune.out <- tune(svm, target ~ ., data = updated_auto, kernel = "radial",
ranges = list(cost = c(0.001, 0.01, 0.1, 1, 5, 10),
gamma = c(0.5, 1, 2, 3, 4)))
summary(radial_tune.out)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost gamma
## 1 1
##
## - best performance: 0.06634615
##
## - Detailed performance results:
## cost gamma error dispersion
## 1 1e-03 0.5 0.55115385 0.04366593
## 2 1e-02 0.5 0.55115385 0.04366593
## 3 1e-01 0.5 0.08666667 0.04687413
## 4 1e+00 0.5 0.06884615 0.02963114
## 5 5e+00 0.5 0.07903846 0.03051601
## 6 1e+01 0.5 0.08923077 0.02732003
## 7 1e-03 1.0 0.55115385 0.04366593
## 8 1e-02 1.0 0.55115385 0.04366593
## 9 1e-01 1.0 0.08673077 0.04535158
## 10 1e+00 1.0 0.06634615 0.03244101
## 11 5e+00 1.0 0.08916667 0.02708952
## 12 1e+01 1.0 0.08923077 0.02732003
## 13 1e-03 2.0 0.55115385 0.04366593
## 14 1e-02 2.0 0.55115385 0.04366593
## 15 1e-01 2.0 0.14282051 0.07578262
## 16 1e+00 2.0 0.08673077 0.04371113
## 17 5e+00 2.0 0.09942308 0.04881948
## 18 1e+01 2.0 0.09429487 0.05387705
## 19 1e-03 3.0 0.55115385 0.04366593
## 20 1e-02 3.0 0.55115385 0.04366593
## 21 1e-01 3.0 0.31878205 0.13973969
## 22 1e+00 3.0 0.08416667 0.04171436
## 23 5e+00 3.0 0.09179487 0.05416218
## 24 1e+01 3.0 0.09179487 0.05416218
## 25 1e-03 4.0 0.55115385 0.04366593
## 26 1e-02 4.0 0.55115385 0.04366593
## 27 1e-01 4.0 0.51788462 0.06842176
## 28 1e+00 4.0 0.08923077 0.03843042
## 29 5e+00 4.0 0.09173077 0.05268076
## 30 1e+01 4.0 0.09179487 0.05139393From the radial tune out summary, it can be seen that when cost is equal to 1 and gamma is equal to 1, it has the lowest cross-validation error of 0.08435897.
set.seed(1)
poly_auto_svm <- svm(target ~ ., data = updated_auto, kernel = "polynomial")
summary(poly_auto_svm)##
## Call:
## svm(formula = target ~ ., data = updated_auto, kernel = "polynomial")
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: polynomial
## cost: 1
## degree: 3
## coef.0: 0
##
## Number of Support Vectors: 151
##
## ( 74 77 )
##
##
## Number of Classes: 2
##
## Levels:
## 0 1set.seed(1)
poly_tune.out <- tune(svm, target ~ ., data = updated_auto, kernel = "polynomial",
ranges = list(cost = c(0.001, 0.01, 0.1, 1, 5, 10),
gamma = c(0.5, 1, 2, 3, 4)))
summary(poly_tune.out)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost gamma
## 0.001 4
##
## - best performance: 0.07666667
##
## - Detailed performance results:
## cost gamma error dispersion
## 1 1e-03 0.5 0.25794872 0.09147506
## 2 1e-02 0.5 0.10217949 0.03617516
## 3 1e-01 0.5 0.08173077 0.03779968
## 4 1e+00 0.5 0.08685897 0.03861230
## 5 5e+00 0.5 0.08673077 0.03641774
## 6 1e+01 0.5 0.09698718 0.05063701
## 7 1e-03 1.0 0.09185897 0.03993389
## 8 1e-02 1.0 0.08679487 0.04533461
## 9 1e-01 1.0 0.08179487 0.03391477
## 10 1e+00 1.0 0.09442308 0.04647330
## 11 5e+00 1.0 0.11467949 0.04965956
## 12 1e+01 1.0 0.12487179 0.04202595
## 13 1e-03 2.0 0.08679487 0.04533461
## 14 1e-02 2.0 0.07923077 0.03084659
## 15 1e-01 2.0 0.08935897 0.04210128
## 16 1e+00 2.0 0.11467949 0.04965956
## 17 5e+00 2.0 0.11743590 0.03875223
## 18 1e+01 2.0 0.12512821 0.04455710
## 19 1e-03 3.0 0.08435897 0.04544023
## 20 1e-02 3.0 0.08429487 0.04194489
## 21 1e-01 3.0 0.10198718 0.05484627
## 22 1e+00 3.0 0.10974359 0.04369624
## 23 5e+00 3.0 0.13006410 0.03044171
## 24 1e+01 3.0 0.12506410 0.03318569
## 25 1e-03 4.0 0.07666667 0.03209996
## 26 1e-02 4.0 0.08673077 0.03641774
## 27 1e-01 4.0 0.11467949 0.04965956
## 28 1e+00 4.0 0.12512821 0.04455710
## 29 5e+00 4.0 0.12756410 0.03410915
## 30 1e+01 4.0 0.12756410 0.03410915From the radial tune out summary, it can be seen that when cost is equal to 0.001 and gamma is equal to 4, it has the lowest cross-validation error of 0.08435897.
- 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.
linear_svm <- svm(target ~ ., data = updated_auto, kernel = "linear", cost = tune.out$best.parameters$cost)
radial_svm <- svm(target ~ ., data = updated_auto, kernel = "radial", cost = radial_tune.out$best.parameters$cost, gamma = radial_tune.out$best.parameters$gamma)
poly_svm <- svm(target ~ ., data = updated_auto, kernel = "polynomial", cost = poly_tune.out$best.parameters$cost, gamma = poly_tune.out$best.parameters$gamma)plot(linear_svm, updated_auto, displacement ~ weight)
plot(radial_svm, updated_auto, displacement ~ weight)
plot(poly_svm, updated_auto, displacement ~ weight)
Question 8
This problem involves the OJ data set which is part of the ISLR2 package.
attach(OJ)- Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)
trainindex <- sample(1:nrow(OJ), 800)
train <- OJ[trainindex, ]
test<- OJ[-trainindex, ]- Fit a support vector classifier to the training data using cost = 0.01, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics, and describe the results obtained.
set.seed(1)
oj_svm <- svm(Purchase ~., data = train, kernel = "linear", cost = 0.01)
summary(oj_svm)##
## Call:
## svm(formula = Purchase ~ ., data = train, kernel = "linear", cost = 0.01)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: linear
## cost: 0.01
##
## Number of Support Vectors: 435
##
## ( 219 216 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMBased on the SVM summary, we can see that there are two classes and the binary response will be either CH or MM. Also 435 support vectors were used to create the model. Of those 435, 219 are in on class and 216 are in the other.
- What are the training and test error rates?
Train
train_pred <- predict(oj_svm, train)
table(train_pred, train$Purchase)##
## train_pred CH MM
## CH 420 75
## MM 65 240((65 + 75) / 800)* 100## [1] 17.5Test
test_pred <- predict(oj_svm, test)
table(test_pred, test$Purchase)##
## test_pred CH MM
## CH 153 33
## MM 15 69((15 + 33) / 270) * 100## [1] 17.77778- Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
oj_tune <- tune(svm, Purchase ~., data = train, kernel = "linear",
ranges = list(cost = c(0.01, 0.1, 1, 5, 10)))
summary(oj_tune)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 0.1
##
## - best performance: 0.1725
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01 0.17625 0.02853482
## 2 0.10 0.17250 0.03162278
## 3 1.00 0.17500 0.02946278
## 4 5.00 0.17250 0.03162278
## 5 10.00 0.17375 0.03197764- Compute the training and test error rates using this new value for cost.
updated_oj_svm <- svm(Purchase ~., data = train, kernel = "linear", cost = oj_tune$best.parameters$cost)Train
updated_train_pred <- predict(updated_oj_svm, train)
table(updated_train_pred, train$Purchase)##
## updated_train_pred CH MM
## CH 422 69
## MM 63 246((62 + 69) / 800) * 100## [1] 16.375Test
updated_train_pred <- predict(updated_oj_svm, test)
table(updated_train_pred, test$Purchase)##
## updated_train_pred CH MM
## CH 155 31
## MM 13 71((12 + 28) / 270) * 100## [1] 14.81481- Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
set.seed(1)
oj_svm_radial <- svm(Purchase ~., data = train, kernel = "radial", cost = 0.01)
summary(oj_svm_radial)##
## Call:
## svm(formula = Purchase ~ ., data = train, kernel = "radial", cost = 0.01)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: radial
## cost: 0.01
##
## Number of Support Vectors: 634
##
## ( 319 315 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMBased on the SVM summary, we can see that there are two classes and the binary response will be either CH or MM. Also 634 support vectors were used to create the model. Of those 435, 319 are in on class and 315 are in the other.
- What are the training and test error rates?
Train
radial_train_pred <- predict(oj_svm_radial, train)
table(radial_train_pred, train$Purchase)##
## radial_train_pred CH MM
## CH 485 315
## MM 0 0((315 + 0) / 800)* 100## [1] 39.375Test
radial_test_pred <- predict(oj_svm_radial, test)
table(radial_test_pred, test$Purchase)##
## radial_test_pred CH MM
## CH 168 102
## MM 0 0((102 + 0) / 270) * 100## [1] 37.77778- Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
radial_oj_tune <- tune(svm, Purchase ~., data = train, kernel = "radial",
ranges = list(cost = c(0.01, 0.1, 1, 5, 10)))
summary(radial_oj_tune)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 1
##
## - best performance: 0.17125
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01 0.39375 0.04007372
## 2 0.10 0.18625 0.02853482
## 3 1.00 0.17125 0.02128673
## 4 5.00 0.18000 0.02220485
## 5 10.00 0.18625 0.02853482- Compute the training and test error rates using this new value for cost.
updated_radial_oj_svm <- svm(Purchase ~., data = train, kernel = "radial", cost = radial_oj_tune$best.parameters$cost)Train
updated_radial_train_pred <- predict(updated_radial_oj_svm, train)
table(updated_radial_train_pred, train$Purchase)##
## updated_radial_train_pred CH MM
## CH 441 77
## MM 44 238((44 + 77) / 800) * 100## [1] 15.125Test
updated_radial_test_pred <- predict(updated_radial_oj_svm, test)
table(updated_radial_test_pred, test$Purchase)##
## updated_radial_test_pred CH MM
## CH 151 33
## MM 17 69((17 + 33) / 270) * 100## [1] 18.51852- Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree = 2.
set.seed(1)
oj_svm_poly <- svm(Purchase ~., data = train, kernel = "polynomial", degree = 2, cost = 0.01)
summary(oj_svm_poly)##
## Call:
## svm(formula = Purchase ~ ., data = train, kernel = "polynomial",
## degree = 2, cost = 0.01)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: polynomial
## cost: 0.01
## degree: 2
## coef.0: 0
##
## Number of Support Vectors: 636
##
## ( 321 315 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMBased on the SVM summary, we can see that there are two classes and the binary response will be either CH or MM. Also 636 support vectors were used to create the model. Of those 435, 321 are in on class and 315 are in the other.
- What are the training and test error rates?
Train
poly_train_pred <- predict(oj_svm_poly, train)
table(poly_train_pred, train$Purchase)##
## poly_train_pred CH MM
## CH 484 297
## MM 1 18((297 + 1) / 800)* 100## [1] 37.25Test
poly_test_pred <- predict(oj_svm_poly, test)
table(poly_test_pred, test$Purchase)##
## poly_test_pred CH MM
## CH 167 98
## MM 1 4((98 + 1) / 270) * 100## [1] 36.66667- Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
poly_oj_tune <- tune(svm, Purchase ~., data = train, kernel = "polynomial", degree = 2,
ranges = list(cost = c(0.01, 0.1, 1, 5, 10)))
summary(poly_oj_tune)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 10
##
## - best performance: 0.18125
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01 0.39125 0.04210189
## 2 0.10 0.32125 0.05001736
## 3 1.00 0.20250 0.04116363
## 4 5.00 0.18250 0.03496029
## 5 10.00 0.18125 0.02779513- Compute the training and test error rates using this new value for cost.
updated_poly_oj_svm <- svm(Purchase ~., data = train, kernel = "polynomial", cost = poly_oj_tune$best.parameters$cost)Train
updated_poly_train_pred <- predict(updated_poly_oj_svm, train)
table(updated_poly_train_pred, train$Purchase)##
## updated_poly_train_pred CH MM
## CH 446 75
## MM 39 240((75 + 39) / 800) * 100## [1] 14.25Test
updated_poly_test_pred <- predict(updated_poly_oj_svm, test)
table(updated_poly_test_pred, test$Purchase)##
## updated_poly_test_pred CH MM
## CH 155 42
## MM 13 60((13 + 42) / 270) * 100## [1] 20.37037- Overall, which approach seems to give the best results on this data?
| Model | Error Rate (%) |
|---|---|
| Linear SVM Train | 16.375 |
| Linear SVM Test | 14.81481 |
| Radial SVM Train | 15.125 |
| Radial SVM Test | 18.51852 |
| Polynomial SVM Train | 14.25 |
| Polynomial SVM Test | 20.37037 |
Of all the models (after using tune), the model that performed the best was the Polynomial SVM.