Assignment 8
Tim Harrison
2023-04-19
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.
- 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:
set.seed(12)
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 yaxis.
plot(x1[y==0], x2[y==0], col="red", xlab="X1", ylab="X2", pch=18)
points(x1[y==1], x2[y==1], col="blue", pch=16)
- Fit a logistic regression model to the data, using X1 and X2 as predictors.
dat = data.frame(x1 = x1, x2 = 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
## -1.350 -1.165 1.050 1.151 1.291
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.04927 0.08978 0.549 0.583
## x1 -0.23002 0.31534 -0.729 0.466
## x2 0.51072 0.31560 1.618 0.106
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 692.86 on 499 degrees of freedom
## Residual deviance: 689.58 on 497 degrees of freedom
## AIC: 695.58
##
## 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.prob = predict(glm.fit, newdata = dat, type = "response")
glm.pred = ifelse(glm.prob >= 0.5, 1, 0)
data.positive = dat[glm.pred == 1,]
data.negative = dat[glm.pred == 0,]
plot(data.positive$x1, data.positive$x2, col="red", xlab="X1", ylab="X2", pch=18)
points(data.negative$x1, data.negative$x2, col="blue", pch=16)
- Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X21, X1 × X2, log(X2) , and so forth).
The logistic model that I ran for this problem is - $ y = x^2_{1} +log(x_2)+(x1*x2) $ With the model the 2nd Degree Polynomial of x1 and log of x2 are significant variables to estimating y.
glm.fit2 = glm(y~poly(x1,2) + log(x2) + I(x1*x2), data=dat, family = "binomial")
summary(glm.fit2)##
## Call:
## glm(formula = y ~ poly(x1, 2) + log(x2) + I(x1 * x2), family = "binomial",
## data = dat)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -3.02566 -0.12847 0.00022 0.14337 1.60700
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -8.817 1.504 -5.864 4.53e-09 ***
## poly(x1, 2)1 -10.434 20.032 -0.521 0.602
## poly(x1, 2)2 90.385 14.996 6.027 1.67e-09 ***
## log(x2) -6.237 1.025 -6.085 1.16e-09 ***
## I(x1 * x2) 7.625 9.572 0.797 0.426
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 341.847 on 247 degrees of freedom
## Residual deviance: 90.027 on 243 degrees of freedom
## (252 observations deleted due to missingness)
## AIC: 100.03
##
## Number of Fisher Scoring iterations: 8- 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.
The decision boundary between this plot and the one above are completely different. The plot presented below cannot be split with a linear decision boundary.
glm.probs2 = predict(glm.fit2, newdata = dat, type = "response")
glm.pred2 = ifelse(glm.probs2 >= 0.5, 1, 0)
data.positive2 = dat[glm.pred2 == 1,]
data.negative2 = dat[glm.pred2 == 0,]
plot(data.positive2$x1, data.positive2$x2, col="red", xlab="X1", ylab="X2", pch=18)
points(data.negative2$x1, data.negative2$x2, col="blue", pch=16)
- 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.
The Support Vector Classifier almost perfectly predictions the classes of the observations, but has blurriness towards the middle of the plot..
library(e1071)
svm.fit = svm(as.factor(y)~ x1 + x2, data = dat, kernal = "linear", cost = 0.1)
svm.pred = predict(svm.fit, dat)
svm.positive = dat[svm.pred == 1,]
svm.negative = dat[svm.pred == 0,]
plot(svm.positive$x1, svm.positive$x2, col="red", xlab="X1", ylab="X2", pch=18)
points(svm.negative$x1, svm.negative$x2, col="blue", pch=16)
- 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)
svm.fit2=svm(as.factor(y)~x1+x2, dat, kernel="radial", gamma=1, cost=1)
svm.pred2=predict(svm.fit2, dat)
svm.positive2= dat[svm.pred2==1,]
svm.negative2= dat[svm.pred2==0,]
plot(svm.positive2$x1, svm.positive2$x2, col="red", xlab="X1", ylab="X2", pch=18)
points(svm.negative2$x1, svm.negative2$x2, col="blue", pch=16)
- Comment on your results.
Non-Linear SVM was the best.
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.
- 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
data(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) :365gas.median = median(Auto$mpg)
gas.class = ifelse(Auto$mpg > gas.median, 1, 0)
Auto$mpglevel = as.factor(gas.class)
str(Auto$mpglevel)## 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 value 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.
set.seed(123)
set.seed(10)
tune.out=tune(svm, mpglevel~., data=Auto, kernal="linear", ranges=list(cost=c(0.001, 0.01, 0.1, 1,5,10,100)))
summary(tune.out)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 100
##
## - best performance: 0.01262821
##
## - Detailed performance results:
## cost error dispersion
## 1 1e-03 0.55365385 0.03306661
## 2 1e-02 0.55365385 0.03306661
## 3 1e-01 0.09942308 0.04714670
## 4 1e+00 0.07897436 0.03260883
## 5 5e+00 0.06878205 0.03175943
## 6 1e+01 0.05352564 0.03055668
## 7 1e+02 0.01262821 0.02437031tune.out$best.parameters## cost
## 7 100best.svmLinear = tune.out$best.model
summary(best.svmLinear)##
## Call:
## best.tune(METHOD = svm, train.x = mpglevel ~ ., data = Auto, ranges = list(cost = c(0.001,
## 0.01, 0.1, 1, 5, 10, 100)), kernal = "linear")
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: radial
## cost: 100
##
## Number of Support Vectors: 63
##
## ( 30 33 )
##
##
## Number of Classes: 2
##
## Levels:
## 0 1(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(123)
tune.out.rad = tune(svm, mpglevel~., data=Auto, kernal="radial", ranges=list(cost=c(0.001, 0.01, 0.1, 1, 5, 10 ,100), gamma=c(0.001, 0.01, 0.1, 1, 5, 10, 100)))
summary(tune.out.rad)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost gamma
## 100 0.01
##
## - best performance: 0.01025641
##
## - Detailed performance results:
## cost gamma error dispersion
## 1 1e-03 1e-03 0.58173077 0.04740051
## 2 1e-02 1e-03 0.58173077 0.04740051
## 3 1e-01 1e-03 0.56891026 0.06627739
## 4 1e+00 1e-03 0.09173077 0.03990003
## 5 5e+00 1e-03 0.07634615 0.03928191
## 6 1e+01 1e-03 0.07121795 0.04410874
## 7 1e+02 1e-03 0.02288462 0.01427008
## 8 1e-03 1e-02 0.58173077 0.04740051
## 9 1e-02 1e-02 0.58173077 0.04740051
## 10 1e-01 1e-02 0.08916667 0.04345384
## 11 1e+00 1e-02 0.07378205 0.04185248
## 12 5e+00 1e-02 0.04589744 0.03136327
## 13 1e+01 1e-02 0.02032051 0.02305327
## 14 1e+02 1e-02 0.01025641 0.01792836
## 15 1e-03 1e-01 0.58173077 0.04740051
## 16 1e-02 1e-01 0.21391026 0.09431095
## 17 1e-01 1e-01 0.07634615 0.03928191
## 18 1e+00 1e-01 0.05852564 0.03960325
## 19 5e+00 1e-01 0.03057692 0.02611396
## 20 1e+01 1e-01 0.03314103 0.02942215
## 21 1e+02 1e-01 0.03326923 0.02434857
## 22 1e-03 1e+00 0.58173077 0.04740051
## 23 1e-02 1e+00 0.58173077 0.04740051
## 24 1e-01 1e+00 0.58173077 0.04740051
## 25 1e+00 1e+00 0.05865385 0.04942437
## 26 5e+00 1e+00 0.05608974 0.04595880
## 27 1e+01 1e+00 0.05608974 0.04595880
## 28 1e+02 1e+00 0.05608974 0.04595880
## 29 1e-03 5e+00 0.58173077 0.04740051
## 30 1e-02 5e+00 0.58173077 0.04740051
## 31 1e-01 5e+00 0.58173077 0.04740051
## 32 1e+00 5e+00 0.51544872 0.06790600
## 33 5e+00 5e+00 0.51544872 0.06790600
## 34 1e+01 5e+00 0.51544872 0.06790600
## 35 1e+02 5e+00 0.51544872 0.06790600
## 36 1e-03 1e+01 0.58173077 0.04740051
## 37 1e-02 1e+01 0.58173077 0.04740051
## 38 1e-01 1e+01 0.58173077 0.04740051
## 39 1e+00 1e+01 0.54602564 0.06355090
## 40 5e+00 1e+01 0.54102564 0.06959451
## 41 1e+01 1e+01 0.54102564 0.06959451
## 42 1e+02 1e+01 0.54102564 0.06959451
## 43 1e-03 1e+02 0.58173077 0.04740051
## 44 1e-02 1e+02 0.58173077 0.04740051
## 45 1e-01 1e+02 0.58173077 0.04740051
## 46 1e+00 1e+02 0.58173077 0.04740051
## 47 5e+00 1e+02 0.58173077 0.04740051
## 48 1e+01 1e+02 0.58173077 0.04740051
## 49 1e+02 1e+02 0.58173077 0.04740051tune.out.rad$best.performance## [1] 0.01025641best.rad.model = tune.out.rad$best.model
summary(best.rad.model)##
## Call:
## best.tune(METHOD = svm, train.x = mpglevel ~ ., data = Auto, ranges = list(cost = c(0.001,
## 0.01, 0.1, 1, 5, 10, 100), gamma = c(0.001, 0.01, 0.1, 1, 5,
## 10, 100)), kernal = "radial")
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: radial
## cost: 100
##
## Number of Support Vectors: 57
##
## ( 27 30 )
##
##
## Number of Classes: 2
##
## Levels:
## 0 1set.seed(123)
tune.out.poly = tune(svm, mpglevel~., data=Auto, kernal="polynomial", ranges=list(cost=c(0.001, 0.01, 0.1, 1, 5, 10 ,100), degree=c(2,3,4,5)))
summary(tune.out.poly)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost degree
## 100 2
##
## - best performance: 0.01282051
##
## - Detailed performance results:
## cost degree error dispersion
## 1 1e-03 2 0.58173077 0.04740051
## 2 1e-02 2 0.58173077 0.04740051
## 3 1e-01 2 0.10692308 0.05900981
## 4 1e+00 2 0.07891026 0.03828837
## 5 5e+00 2 0.06608974 0.04785032
## 6 1e+01 2 0.05602564 0.03551922
## 7 1e+02 2 0.01282051 0.01813094
## 8 1e-03 3 0.58173077 0.04740051
## 9 1e-02 3 0.58173077 0.04740051
## 10 1e-01 3 0.10692308 0.05900981
## 11 1e+00 3 0.07891026 0.03828837
## 12 5e+00 3 0.06608974 0.04785032
## 13 1e+01 3 0.05602564 0.03551922
## 14 1e+02 3 0.01282051 0.01813094
## 15 1e-03 4 0.58173077 0.04740051
## 16 1e-02 4 0.58173077 0.04740051
## 17 1e-01 4 0.10692308 0.05900981
## 18 1e+00 4 0.07891026 0.03828837
## 19 5e+00 4 0.06608974 0.04785032
## 20 1e+01 4 0.05602564 0.03551922
## 21 1e+02 4 0.01282051 0.01813094
## 22 1e-03 5 0.58173077 0.04740051
## 23 1e-02 5 0.58173077 0.04740051
## 24 1e-01 5 0.10692308 0.05900981
## 25 1e+00 5 0.07891026 0.03828837
## 26 5e+00 5 0.06608974 0.04785032
## 27 1e+01 5 0.05602564 0.03551922
## 28 1e+02 5 0.01282051 0.01813094tune.out.poly$best.performance## [1] 0.01282051best.poly.model = tune.out.poly$best.model
summary(best.poly.model)##
## Call:
## best.tune(METHOD = svm, train.x = mpglevel ~ ., data = Auto, ranges = list(cost = c(0.001,
## 0.01, 0.1, 1, 5, 10, 100), degree = c(2, 3, 4, 5)), kernal = "polynomial")
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: radial
## cost: 100
##
## Number of Support Vectors: 63
##
## ( 30 33 )
##
##
## Number of Classes: 2
##
## Levels:
## 0 1- Make some plots to back up your assertions in (b) and (c).
svm.linear=svm(mpglevel~., data=Auto, kernal="linear", cost=100)
svm.rad=svm(mpglevel~., data=Auto, kernal="radial", cost=100, gamma=0.01)
svm.poly=svm(mpglevel~., data=Auto, kernal="polynomial", cost=100, degree=2)
plotpairs = function(autofit) {
for (name in names(Auto)[!(names(Auto) %in% c("mpg", "mpglevel", "name"))]) {
plot(autofit, Auto, as.formula(paste("mpg~", name, sep = "")))
}
}
plotpairs(svm.linear)
plotpairs(svm.rad)
plotpairs(svm.poly)
Problem 8
This problem involves the OJ data set which is part of the ISLR package.
data(OJ)
summary(OJ)## Purchase WeekofPurchase StoreID PriceCH PriceMM
## CH:653 Min. :227.0 Min. :1.00 Min. :1.690 Min. :1.690
## MM:417 1st Qu.:240.0 1st Qu.:2.00 1st Qu.:1.790 1st Qu.:1.990
## Median :257.0 Median :3.00 Median :1.860 Median :2.090
## Mean :254.4 Mean :3.96 Mean :1.867 Mean :2.085
## 3rd Qu.:268.0 3rd Qu.:7.00 3rd Qu.:1.990 3rd Qu.:2.180
## Max. :278.0 Max. :7.00 Max. :2.090 Max. :2.290
## DiscCH DiscMM SpecialCH SpecialMM
## Min. :0.00000 Min. :0.0000 Min. :0.0000 Min. :0.0000
## 1st Qu.:0.00000 1st Qu.:0.0000 1st Qu.:0.0000 1st Qu.:0.0000
## Median :0.00000 Median :0.0000 Median :0.0000 Median :0.0000
## Mean :0.05186 Mean :0.1234 Mean :0.1477 Mean :0.1617
## 3rd Qu.:0.00000 3rd Qu.:0.2300 3rd Qu.:0.0000 3rd Qu.:0.0000
## Max. :0.50000 Max. :0.8000 Max. :1.0000 Max. :1.0000
## LoyalCH SalePriceMM SalePriceCH PriceDiff Store7
## Min. :0.000011 Min. :1.190 Min. :1.390 Min. :-0.6700 No :714
## 1st Qu.:0.325257 1st Qu.:1.690 1st Qu.:1.750 1st Qu.: 0.0000 Yes:356
## Median :0.600000 Median :2.090 Median :1.860 Median : 0.2300
## Mean :0.565782 Mean :1.962 Mean :1.816 Mean : 0.1465
## 3rd Qu.:0.850873 3rd Qu.:2.130 3rd Qu.:1.890 3rd Qu.: 0.3200
## Max. :0.999947 Max. :2.290 Max. :2.090 Max. : 0.6400
## PctDiscMM PctDiscCH ListPriceDiff STORE
## Min. :0.0000 Min. :0.00000 Min. :0.000 Min. :0.000
## 1st Qu.:0.0000 1st Qu.:0.00000 1st Qu.:0.140 1st Qu.:0.000
## Median :0.0000 Median :0.00000 Median :0.240 Median :2.000
## Mean :0.0593 Mean :0.02731 Mean :0.218 Mean :1.631
## 3rd Qu.:0.1127 3rd Qu.:0.00000 3rd Qu.:0.300 3rd Qu.:3.000
## Max. :0.4020 Max. :0.25269 Max. :0.440 Max. :4.000str(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 ...- Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1234)
oj.intrain <- createDataPartition(OJ$Purchase, p = 0.746, list = FALSE)
oj.train <- OJ[oj.intrain,]
oj.test <- OJ[-oj.intrain,]
dim(oj.train)## [1] 800 18- Fit a support vector classifier to the training data using cost=0.01, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics, and describe the results obtained.
oj.svm <- svm(Purchase~., data = oj.train, kernel = "linear", cost = 0.01)
summary(oj.svm)##
## Call:
## svm(formula = Purchase ~ ., data = oj.train, kernel = "linear", cost = 0.01)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: linear
## cost: 0.01
##
## Number of Support Vectors: 439
##
## ( 221 218 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MM- What are the training and test error rates?
The Train Error Rate is 17.63%.
oj.train.pred <- predict(oj.svm, oj.train)
table(oj.train$Purchase, oj.train.pred)## oj.train.pred
## CH MM
## CH 430 58
## MM 83 229(83+58)/800## [1] 0.17625The Test Error Rate is 14.44%.
oj.test.pred <- predict(oj.svm, oj.test)
table(oj.test$Purchase, oj.test.pred)## oj.test.pred
## CH MM
## CH 147 18
## MM 21 84(21+18)/270## [1] 0.1444444- Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
set.seed(1234)
oj.tune.out = tune(svm, Purchase ~., data = oj.train, kernel = "linear", ranges = list(cost=c(0.001, 0.01, 0.1, 1, 5, 10)))
summary(oj.tune.out)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 0.1
##
## - best performance: 0.17625
##
## - Detailed performance results:
## cost error dispersion
## 1 1e-03 0.33750 0.07264832
## 2 1e-02 0.18000 0.04937104
## 3 1e-01 0.17625 0.05816941
## 4 1e+00 0.18000 0.05986095
## 5 5e+00 0.17625 0.05478810
## 6 1e+01 0.17750 0.05583955oj.tune.out$best.parameters## cost
## 3 0.1- Compute the training and test error rates using this new value for cost.
oj.new.svm = svm(Purchase ~., kernel = "linear", data = oj.train, cost = oj.tune.out$best.parameters$cost)
oj.new.train.pred = predict(oj.new.svm, oj.train)
table(oj.train$Purchase, oj.new.train.pred)## oj.new.train.pred
## CH MM
## CH 428 60
## MM 75 237(75 + 60)/800## [1] 0.16875oj.new.test.pred = predict(oj.new.svm, oj.test)
table(oj.test$Purchase, oj.new.test.pred)## oj.new.test.pred
## CH MM
## CH 147 18
## MM 20 85(20 + 18)/270## [1] 0.1407407- Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
oj.svm.rad = svm(Purchase~., data = oj.train, kernel = "radial", cost = 0.01)
summary(oj.svm.rad)##
## Call:
## svm(formula = Purchase ~ ., data = oj.train, kernel = "radial", cost = 0.01)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: radial
## cost: 0.01
##
## Number of Support Vectors: 625
##
## ( 313 312 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMoj.rad.train.pred = predict(oj.svm.rad, oj.train)
table(oj.train$Purchase, oj.rad.train.pred)## oj.rad.train.pred
## CH MM
## CH 488 0
## MM 312 0312/800## [1] 0.39oj.rad.test.pred = predict(oj.svm.rad, oj.test)
table(oj.test$Purchase, oj.rad.test.pred)## oj.rad.test.pred
## CH MM
## CH 165 0
## MM 105 0105/270## [1] 0.3888889set.seed(1234)
oj.rad.tune = tune(svm, Purchase~., data = oj.train, kernal = "radial", ranges = list(cost=c(0.001, 0.01, 0.1, 1, 5, 10)))
summary(oj.rad.tune)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 1
##
## - best performance: 0.1775
##
## - Detailed performance results:
## cost error dispersion
## 1 1e-03 0.39000 0.04706674
## 2 1e-02 0.39000 0.04706674
## 3 1e-01 0.18750 0.03632416
## 4 1e+00 0.17750 0.04440971
## 5 5e+00 0.18250 0.03446012
## 6 1e+01 0.18875 0.03747684oj.svm.rad2 = svm(Purchase~., data = oj.train, kernel = "radial", cost = oj.rad.tune$best.parameters$cost)
summary(oj.svm.rad2)##
## Call:
## svm(formula = Purchase ~ ., data = oj.train, kernel = "radial", cost = oj.rad.tune$best.parameters$cost)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: radial
## cost: 1
##
## Number of Support Vectors: 375
##
## ( 189 186 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMoj.train.rad.pred2 = predict(oj.svm.rad2, oj.train)
table(oj.train$Purchase, oj.train.rad.pred2)## oj.train.rad.pred2
## CH MM
## CH 445 43
## MM 79 233(79 + 43)/800## [1] 0.1525oj.test.rad.pred2 = predict(oj.svm.rad2, oj.test)
table(oj.test$Purchase, oj.test.rad.pred2)## oj.test.rad.pred2
## CH MM
## CH 150 15
## MM 26 79(26+15)/270## [1] 0.1518519- Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree=2.
oj.svm.poly = svm(Purchase~., data = oj.train, kernel = "poly", cost = 0.01, degree = 2)
summary(oj.svm.poly)##
## Call:
## svm(formula = Purchase ~ ., data = oj.train, kernel = "poly", cost = 0.01,
## degree = 2)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: polynomial
## cost: 0.01
## degree: 2
## coef.0: 0
##
## Number of Support Vectors: 629
##
## ( 317 312 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMoj.poly.train.pred = predict(oj.svm.poly, oj.train)
table(oj.train$Purchase, oj.poly.train.pred)## oj.poly.train.pred
## CH MM
## CH 488 0
## MM 312 0312/800## [1] 0.39oj.poly.test.pred = predict(oj.svm.poly, oj.test)
table(oj.test$Purchase, oj.poly.test.pred)## oj.poly.test.pred
## CH MM
## CH 165 0
## MM 105 0105/270## [1] 0.3888889set.seed(1234)
oj.poly.tune = tune(svm, Purchase~., data = oj.train, kernel = "poly", ranges = list(cost=c(0.001, 0.01, 0.1, 1, 5, 10)), degree = 2)
summary(oj.poly.tune)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 10
##
## - best performance: 0.18625
##
## - Detailed performance results:
## cost error dispersion
## 1 1e-03 0.39000 0.04706674
## 2 1e-02 0.39000 0.04706674
## 3 1e-01 0.30875 0.05804991
## 4 1e+00 0.20500 0.05109903
## 5 5e+00 0.19125 0.03998698
## 6 1e+01 0.18625 0.04767147oj.poly.tune$best.parameters## cost
## 6 10oj.svm.poly2 = svm(Purchase~., data = oj.train, kernel = "poly", cost = oj.poly.tune$best.parameters$cost, degree = 2)
summary(oj.svm.poly2)##
## Call:
## svm(formula = Purchase ~ ., data = oj.train, kernel = "poly", cost = oj.poly.tune$best.parameters$cost,
## degree = 2)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: polynomial
## cost: 10
## degree: 2
## coef.0: 0
##
## Number of Support Vectors: 344
##
## ( 175 169 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMoj.train.poly.pred2 = predict(oj.svm.poly2, oj.train)
table(oj.train$Purchase, oj.train.poly.pred2)## oj.train.poly.pred2
## CH MM
## CH 447 41
## MM 79 233(79+41)/800## [1] 0.15oj.test.poly.pred2 = predict(oj.svm.poly2, oj.test)
table(oj.test$Purchase, oj.test.poly.pred2)## oj.test.poly.pred2
## CH MM
## CH 152 13
## MM 35 70(35+13)/270## [1] 0.1777778- Overall, which approach seems to give the best results on this data?
The Tuned Radial SVM with Cost = 1 results in the lowest Test Error Rate for the OJ data set.