Selena Romero - Assignment 8 - R Code
Selena Romero
7/28/2020
Chapter 09 (page 368): 5, 7, 8
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="#FF6666", xlab="X1", ylab="X2", pch=18)
points(x1[y==1], x2[y==1], col="#66CCFF", 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="#FF6666", xlab="X1", ylab="X2", pch=18)
points(data.negative$x1, data.negative$x2, col="#66CCFF", pch=16)
- Now fit a logistic regression model to the data using non-linear functions of \(X_1\) and \(X_2\) as predictors (e.g. \(X^2_{1}\), \(X^1\) × \(X^2\), \(log(X_2)\), 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 \(x_1\) and log of \(x_2\) are significant variables to estimating y.
glm.fit2 = glm(y~poly(x1,2) + log(x2) + I(x1*x2), data=dat, family = "binomial")## Warning in log(x2): NaNs produced## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredsummary(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")## Warning in log(x2): NaNs producedglm.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="#FF6666", xlab="X1", ylab="X2", pch=18)
points(data.negative2$x1, data.negative2$x2, col="#66CCFF", pch=16)
- Fit a support vector classifier to the data with \(X_1\) and \(X_2\) 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="#FF6666", xlab="X1", ylab="X2", pch=18)
points(svm.negative$x1, svm.negative$x2, col="#66CCFF", 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.
The Non-Linear Kernal SVM does a much better at prediction for this random dataset.
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="#FF6666", xlab="X1", ylab="X2", pch=18)
points(svm.negative2$x1, svm.negative2$x2, col="#66CCFF", pch=16)
- Comment on your results.
The most poweerful model utilized was the Non-Linear SVM. With the Non-Linear SVM, there was less confusion between the classes. The Linear and Non-Linear Logistic Regressions very poorly predicted the dataset.
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)
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) :365- 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.
gas.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.
For this problem, the C Value of 100 is selected by cross-validation. With C = 100, the error rate is minimized at 0.01262821.
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$best.parameters## cost
## 7 100Per cross-validation, R has selected a Radial SVM model with C = 100 and 63 Support Vectors. Of the 63 Support Vectors, 30 of them are classified as “low MPG” or 0 while 33 Support Vectors are classified as “high MPG” or 1.
best.svmLinear = tune.out$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- 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.
For the Radial Kernal, the model selected has a Cost of 100 and Gamma of 0.01. There are 57 Support Vectors in which 27 are classified as “Low MPG” or 0 and the rest of the 30 observations are classified as “High MPG” or 1. The error for this model is 0.01025641.
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 1For the Polynomial Kernal, the model chosen has a Cost of 100 and Polynomial Degree of 2. This model utilizes 63 Support Vectors where 30 of the them are classified as “Low MPG” and the remaining 33 Support Vectors are classified as “High MPG”. The error for this model is 0.01282051.
set.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).
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, data)
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 Support Vector Classifier Plots
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)
Radial Support Vector Classifiers
plotpairs(svm.rad)
Polynomial Support Vector Classifier
plotpairs(svm.poly)
Problem 8
This problem involves the OJ data set which is part of the ISLR package.
library(ISLR)
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.
library(caret)## Loading required package: lattice## Loading required package: ggplot2set.seed(1234)
oj.intrain <- createDataPartition(OJ$Purchase, p = 0.746, list = FALSE)
oj.train <- OJ[oj.intrain,]
oj.test <- OJ[-oj.intrain,]
dim(oj.train)## [1] 800 18- Fit a support vector classifier to the training data using cost=0.01, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics, and describe the results obtained.
The Support Vector Classifier with Cost = 0.01 results in 439 Support Vectors with 221 of the Support Vectors being classified as CH and 218 of the Support Vectors being classified as MM.
library(e1071)
oj.svm <- svm(Purchase~., data = oj.train, kernel = "linear", cost = 0.01)
summary(oj.svm)##
## Call:
## svm(formula = Purchase ~ ., data = oj.train, kernel = "linear", cost = 0.01)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: linear
## cost: 0.01
##
## Number of Support Vectors: 439
##
## ( 221 218 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MM- What are the training and test error rates?
The Train Error Rate is 17.63%.
oj.train.pred <- predict(oj.svm, oj.train)
table(oj.train$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$Purchase, oj.test.pred)## oj.test.pred
## CH MM
## CH 147 18
## MM 21 84(21+18)/270## [1] 0.1444444- Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
The optimal Cost value is 0.1 with an error of 0.17625.
set.seed(1234)
oj.tune.out = tune(svm, Purchase ~., data = oj.train, kernel = "linear", ranges = list(cost=c(0.001, 0.01, 0.1, 1, 5, 10)))
summary(oj.tune.out)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 0.1
##
## - best performance: 0.17625
##
## - Detailed performance results:
## cost error dispersion
## 1 1e-03 0.33750 0.07264832
## 2 1e-02 0.18000 0.04937104
## 3 1e-01 0.17625 0.05816941
## 4 1e+00 0.18000 0.05986095
## 5 5e+00 0.17625 0.05478810
## 6 1e+01 0.17750 0.05583955oj.tune.out$best.parameters## cost
## 3 0.1- Compute the training and test error rates using this new value for cost.
With the Cost Value = 0.1, the Training Error Rate is 16.88%.
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$Purchase, oj.new.train.pred)## oj.new.train.pred
## CH MM
## CH 428 60
## MM 75 237(75 + 60)/800## [1] 0.16875The Test Error Rate with Cost = 0.1 is 14.07%.
oj.new.test.pred = predict(oj.new.svm, oj.test)
table(oj.test$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.
With a Radial SVM where Cost = 0.01, 625 Support Vectors are used. Of the 625 Support Vectors, 313 of them are classified as CH and the rest of the 312 are classified as MM.On the training dataset, this Radial SVM results in a training error rate of 39%.
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$Purchase, oj.rad.train.pred)## oj.rad.train.pred
## CH MM
## CH 488 0
## MM 312 0312/800## [1] 0.39The Radial SVM with the Cost = 0.01, the Test Error Rate is 38.89%.
oj.rad.test.pred = predict(oj.svm.rad, oj.test)
table(oj.test$Purchase, oj.rad.test.pred)## oj.rad.test.pred
## CH MM
## CH 165 0
## MM 105 0105/270## [1] 0.3888889After tuning the Radial SVM, the best Cost Value is 1 where the error = 0.17750.
set.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 0.03747684The Radial SVM with Cost = 1 results in 375 Support Vectors being used with 189 of the Support Vectors being classified as CH and 186 being classified as MM. On the Training Dataset, the Error Rate is 15.25%.
oj.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$Purchase, oj.train.rad.pred2)## oj.train.rad.pred2
## CH MM
## CH 445 43
## MM 79 233(79 + 43)/800## [1] 0.1525On the Test Data, the Error Rate is 15.19% for the Radial SVM with Cost = 1.
oj.test.rad.pred2 = predict(oj.svm.rad2, oj.test)
table(oj.test$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.
The Polynomial SVM with Cost = 0.01 and Degree = 2 results in 625 Support Vectors with 313 classified as CH and 312 as MM. The Training Error is 39%.
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$Purchase, oj.poly.train.pred)## oj.poly.train.pred
## CH MM
## CH 488 0
## MM 312 0312/800## [1] 0.39The Polynomial SVM with the Cost = 0.01 and Degree = 2, the Test Error Rate is 38.89%.
oj.poly.test.pred = predict(oj.svm.poly, oj.test)
table(oj.test$Purchase, oj.poly.test.pred)## oj.poly.test.pred
## CH MM
## CH 165 0
## MM 105 0105/270## [1] 0.3888889After tuning the Polynomial SVM, the best Cost value = 10 where the error = 0.18625.
set.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$best.parameters## cost
## 6 10The Polynomial SVM with Cost = 10 results in 344 Support Vectors being used with 175 of the Support Vectors being classified as CH and 169 being classified as MM. On the Training Dataset, the Error Rate is 15%.
oj.svm.poly2 = svm(Purchase~., data = oj.train, kernel = "poly", cost = oj.poly.tune$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] 0.15On the Test Data, the Error Rate is 17.78% for the Polynomial SVM with Cost = 10 and Degree = 2.
oj.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.