Assignment 8 - Ch 9
Jaren Montes
Problem 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. For instance, you can do this as follows:
> x1 <- runif (500) - 0.5
> x2 <- runif (500) - 0.5
> y <- 1 * (x1^2 - x2^2 > 0)
set.seed(421)
x1=runif(500)-.5
x2=runif(500)-.5
y=1*(x1^2-x2^2>0)(b) Plot the observations, colored according to their class labels. Your plot should display X1 on the x axis, and X2 on the yaxis.
par(bg="black", col.lab="white", col.axis="white")
plot(x1[y == 0], x2[y == 0], col = "red", xlab = "X1", ylab = "X2", pch = 5)
points(x1[y == 1], x2[y == 1], col = "blue", pch = 4)
box(col="white")
(c) Fit a logistic regression model to the data, using X1 and X2 as predictors.
lm.data=data.frame(x1 = x1, x2 = x2, y = as.factor(y))
lm.fit=glm(y~., data = lm.data, family="binomial")
summary(lm.fit)##
## Call:
## glm(formula = y ~ ., family = "binomial", data = lm.data)
##
## 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(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.
lm.prob = predict(lm.fit, newdata=lm.data, type = "response")
lm.pred = ifelse(lm.prob > 0.52, 1, 0)
data.pos = lm.data[lm.pred == 1, ]
data.neg = lm.data[lm.pred == 0, ]
par(bg="black", col.lab="white", col.axis="white")
plot(data.pos$x1, data.pos$x2, col = "blue", xlab = "X1", ylab = "X2", pch = 5)
points(data.neg$x1, data.neg$x2, col = "red", pch = 4)
box(col="white")
(e) 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).
lm.fit2=glm(y~poly(x1, 2)+poly(x2, 2)+I(x1 * x2),data=lm.data, family=binomial)## Warning: glm.fit: algorithm did not converge## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredsummary(lm.fit2)##
## Call:
## glm(formula = y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), family = binomial,
## data = lm.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(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.
lm.prob2=predict(lm.fit2,lm.data,type="response")
lm.pred2=ifelse(lm.prob2>.5,1,0)
data.pos2=lm.data[lm.pred2==1, ]
data.neg2=lm.data[lm.pred2==0, ]
par(bg="black", col.lab="white", col.axis="white")
plot(data.pos2$x1, data.pos2$x2,col="green", xlab="x1", ylab="x2", pch=8)
points(data.neg2$x1, data.neg2$x2, col="white", pch=7)
box(col="white")
(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.
library(e1071)
svm.lin=svm(y~.,data=lm.data,kernel="linear",cost=.01)
plot(svm.lin,lm.data)
(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.
train=sample(200,150)
svm.nonlin=svm(y~., data=lm.data[train,], kernel=("radial"),gama=1,cost=1e5)
plot(svm.nonlin,lm.data[train,])
(i) Comment on your results.
When it comes to nonlinear boundaries, SVM linear kernels can work alright with small cost, while nonlinear kernels make a better fit, however irregular they may be.
Problem 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)
library(e1071)
attach(Auto)
summary(Auto)## mpg cylinders displacement horsepower weight
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0 Min. :1613
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0 1st Qu.:2225
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5 Median :2804
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5 Mean :2978
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0 3rd Qu.:3615
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0 Max. :5140
##
## acceleration year origin name
## Min. : 8.00 Min. :70.00 Min. :1.000 amc matador : 5
## 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000 ford pinto : 5
## Median :15.50 Median :76.00 Median :1.000 toyota corolla : 5
## Mean :15.54 Mean :75.98 Mean :1.577 amc gremlin : 4
## 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000 amc hornet : 4
## Max. :24.80 Max. :82.00 Max. :3.000 chevrolet chevette: 4
## (Other) :365(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.
med.mpg=median(Auto$mpg)
bin.var=ifelse(Auto$mpg>med.mpg,1,0)
Auto$mpglevel=as.factor(bin.var)(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. Note you will need to fit the classifier without the gas mileage variable to produce sensible results.
set.seed(333)
tune.oof=tune(svm,mpglevel~.,data = Auto, kernel="linear", ranges = list(cost=c(.01, .1, 1, 5, 10, 100)))
summary(tune.oof)##
## 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 0.07397436 0.03896185
## 2 1e-01 0.04089744 0.04041773
## 3 1e+00 0.01025641 0.01324097
## 4 5e+00 0.01532051 0.01788871
## 5 1e+01 0.02301282 0.02549182
## 6 1e+02 0.03326923 0.02974993(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.
set.seed(221)
tune.ooooff=tune(svm,mpglevel~., data = Auto, kernel="polynomial", ranges = list(cost=c(.1,1,5,10), degree=c(2,3,4)))
summary(tune.ooooff)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost degree
## 10 2
##
## - best performance: 0.5330769
##
## - Detailed performance results:
## cost degree error dispersion
## 1 0.1 2 0.5638462 0.03963569
## 2 1.0 2 0.5638462 0.03963569
## 3 5.0 2 0.5638462 0.03963569
## 4 10.0 2 0.5330769 0.07780372
## 5 0.1 3 0.5638462 0.03963569
## 6 1.0 3 0.5638462 0.03963569
## 7 5.0 3 0.5638462 0.03963569
## 8 10.0 3 0.5638462 0.03963569
## 9 0.1 4 0.5638462 0.03963569
## 10 1.0 4 0.5638462 0.03963569
## 11 5.0 4 0.5638462 0.03963569
## 12 10.0 4 0.5638462 0.03963569set.seed(48)
tune.oooofff=tune(svm,mpglevel~., data = Auto, kernel="radial", ranges = list(cost=c(.1,1,5,10), gama=c(.01,1,5,10,100)))
summary(tune.oooofff)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost gama
## 10 0.01
##
## - best performance: 0.05878205
##
## - Detailed performance results:
## cost gama error dispersion
## 1 0.1 1e-02 0.09955128 0.04443422
## 2 1.0 1e-02 0.07660256 0.04018600
## 3 5.0 1e-02 0.07147436 0.03381912
## 4 10.0 1e-02 0.05878205 0.02985043
## 5 0.1 1e+00 0.09955128 0.04443422
## 6 1.0 1e+00 0.07660256 0.04018600
## 7 5.0 1e+00 0.07147436 0.03381912
## 8 10.0 1e+00 0.05878205 0.02985043
## 9 0.1 5e+00 0.09955128 0.04443422
## 10 1.0 5e+00 0.07660256 0.04018600
## 11 5.0 5e+00 0.07147436 0.03381912
## 12 10.0 5e+00 0.05878205 0.02985043
## 13 0.1 1e+01 0.09955128 0.04443422
## 14 1.0 1e+01 0.07660256 0.04018600
## 15 5.0 1e+01 0.07147436 0.03381912
## 16 10.0 1e+01 0.05878205 0.02985043
## 17 0.1 1e+02 0.09955128 0.04443422
## 18 1.0 1e+02 0.07660256 0.04018600
## 19 5.0 1e+02 0.07147436 0.03381912
## 20 10.0 1e+02 0.05878205 0.02985043(d) 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.lin = svm(mpglevel ~ ., data = Auto, kernel = "linear", cost = 1)
svm.poly = svm(mpglevel ~ ., data = Auto, kernel = "polynomial", cost = 10,
degree = 2)
svm.rad = svm(mpglevel ~ ., data = Auto, kernel = "radial", cost = 10, gamma = 0.01)
plotpairs = function(fit) {
for (name in names(Auto)[!(names(Auto) %in% c("mpg", "mpglevel", "name"))]) {
plot(fit, Auto, as.formula(paste("mpg~", name, sep = "")))
}
}
plotpairs(svm.lin)
plotpairs(svm.poly)
plotpairs(svm.rad)
Problem 8
This problem involves the OJ data set which is part of the ISLR2 Package.
library(ISLR)
library(e1071)(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(96)
train = sample(dim(OJ)[1], 800)
OJ.train = OJ[train, ]
OJ.test = OJ[-train, ](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.
OJ.svm.lin=svm(Purchase~., data = OJ.train, kernel="linear", cost=.01)
summary(OJ.svm.lin)##
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "linear", cost = 0.01)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: linear
## cost: 0.01
##
## Number of Support Vectors: 433
##
## ( 217 216 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MM(c) What are the training and test error rates?
train.pred=predict(OJ.svm.lin, OJ.train)
table(OJ.train$Purchase, train.pred)## train.pred
## CH MM
## CH 436 52
## MM 78 234test.pred=predict(OJ.svm.lin, OJ.test)
table(OJ.test$Purchase, test.pred)## test.pred
## CH MM
## CH 144 21
## MM 29 76- Training Error = 83.75%
- Test Error = 81.48%
(d) Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
set.seed(1554)
tune.ooooofffff=tune(svm,Purchase~., data = OJ.train, kernel = "linear", ranges = list(cost=10^seq(-2,1, by = .25)))
summary(tune.ooooofffff)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 0.5623413
##
## - best performance: 0.1575
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01000000 0.17125 0.03682259
## 2 0.01778279 0.16875 0.03596391
## 3 0.03162278 0.16250 0.03435921
## 4 0.05623413 0.16375 0.03508422
## 5 0.10000000 0.16375 0.03304563
## 6 0.17782794 0.16250 0.03535534
## 7 0.31622777 0.15875 0.03387579
## 8 0.56234133 0.15750 0.03641962
## 9 1.00000000 0.16500 0.03670453
## 10 1.77827941 0.16625 0.04210189
## 11 3.16227766 0.16625 0.04126894
## 12 5.62341325 0.17000 0.03184162
## 13 10.00000000 0.16750 0.03545341(e) Compute the training and test error rates using this new value for cost.
svm.linear = svm(Purchase ~ ., kernel = "linear", data = OJ.train, cost = tune.ooooofffff$best.parameters$cost)
train.pred = predict(svm.linear, OJ.train)
table(OJ.train$Purchase, train.pred)## train.pred
## CH MM
## CH 433 55
## MM 70 242test.pred = predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)## test.pred
## CH MM
## CH 141 24
## MM 28 77- Training Error = 84.38%
- Test Error = 80.74%
(f) Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
set.seed(410)
svm.radial=svm(Purchase~., data = OJ.train, kernel="radial")
summary(svm.radial)##
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "radial")
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: radial
## cost: 1
##
## Number of Support Vectors: 370
##
## ( 188 182 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMtrain.pred=predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)## train.pred
## CH MM
## CH 448 40
## MM 77 235test.pred=predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)## test.pred
## CH MM
## CH 146 19
## MM 33 72- Train Error = 85.38%
- Test Error = 80.74%
set.seed(755)
tune.ooooooffffff=tune(svm,Purchase~., data = OJ.train, kernel = "radial", ranges = list(cost=10^seq(-2,1, by = .25)))
summary(tune.ooooooffffff)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 0.5623413
##
## - best performance: 0.16875
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01000000 0.39000 0.04851976
## 2 0.01778279 0.39000 0.04851976
## 3 0.03162278 0.35125 0.06883202
## 4 0.05623413 0.20250 0.03670453
## 5 0.10000000 0.18125 0.03596391
## 6 0.17782794 0.17500 0.03864008
## 7 0.31622777 0.17125 0.04126894
## 8 0.56234133 0.16875 0.03076005
## 9 1.00000000 0.17125 0.02949223
## 10 1.77827941 0.17375 0.03408018
## 11 3.16227766 0.17000 0.03291403
## 12 5.62341325 0.17250 0.03670453
## 13 10.00000000 0.17625 0.03884174svm.radial=svm(Purchase~., data = OJ.train, kernel = "radial", cost=tune.ooooooffffff$best.parameter$cost)
train.pred=predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)## train.pred
## CH MM
## CH 445 43
## MM 79 233test.pred=predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)## test.pred
## CH MM
## CH 145 20
## MM 30 75- Train Error = 84.75%
- Test Error = 81.48%
(g) Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree = 2.
set.seed(21)
tune.ooooooofffffff=svm(Purchase~., data = OJ.train, kernel = "polynomial", cost=tune.ooooooffffff$best.parameter$cost, degree=2)
summary(tune.ooooooofffffff)##
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "polynomial",
## cost = tune.ooooooffffff$best.parameter$cost, degree = 2)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: polynomial
## cost: 0.5623413
## degree: 2
## coef.0: 0
##
## Number of Support Vectors: 489
##
## ( 248 241 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMsvm.poly=svm(Purchase~., data = OJ.train, kernel = "polynomial", cost=tune.ooooooffffff$best.parameter$cost, degree=2)
train.pred=predict(svm.poly, OJ.train)
table(OJ.train$Purchase, train.pred)## train.pred
## CH MM
## CH 463 25
## MM 125 187test.pred=predict(svm.poly, OJ.test)
table(OJ.test$Purchase, test.pred)## test.pred
## CH MM
## CH 152 13
## MM 44 61- Train Error = 81.25%
- Test Error = 78.89%
(h) Overall, which approach seems to give the best results on this Data?
The polynomial model using the new value for cost ended up with the lowest error rates for the data set.