Homework 7
Thomas Farrell
Results
Q3
a.Consider the Gini index, classification error, and entropy in a simple classification setting with two classes. Create a single plot that displays each of these quantities as a function of pˆm1. The x- axis should display pˆm1, ranging from 0 to 1, and the y-axis should display the value of the Gini index, classification error, and entropy.
p = seq(0, 1, 0.001)
gini.index = 2 * p * (1 - p)
class.error = 1 - pmax(p, 1 - p)
cross.entropy = - (p * log(p) + (1 - p) * log(1 - p))
matplot(p, cbind(gini.index, class.error, cross.entropy), ylab = "gini.index, class.error, cross.entropy", col = c("red", "green", "blue"))
Q8
a.Split the data set into a training set and a test set.
library(ISLR)
library(randomForest)
library(caret)
b1 = Carseatsset.seed(1)
tr_ind = sample(nrow(b1),.8*nrow(b1), replace = F)
b1train = b1[tr_ind,]
b1test = b1[-tr_ind,]b.Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
tree.seats = tree::tree(Sales ~ ., data = b1train)## Registered S3 method overwritten by 'tree':
## method from
## print.tree clisummary(tree.seats)##
## Regression tree:
## tree::tree(formula = Sales ~ ., data = b1train)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Age" "Income" "CompPrice"
## [6] "Advertising"
## Number of terminal nodes: 16
## Residual mean deviance: 2.572 = 781.9 / 304
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -4.45400 -1.07000 -0.05544 0.00000 1.14500 4.69600plot(tree.seats)
text(tree.seats, pretty = 0)
treeseat.pred = predict(tree.seats, newdata = b1test)
mean((treeseat.pred - b1test$Sales)^2)## [1] 4.936081c.Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(1)
cv.seats = tree::cv.tree(tree.seats)
plot(cv.seats$size, cv.seats$dev, type = "b")
prune.car = tree::prune.tree(tree.seats, best = 10)
plot(prune.car)
text(prune.car,pretty=0)
treeseat.pred = predict(prune.car, newdata = b1test)
mean((treeseat.pred - b1test$Sales)^2)## [1] 5.088731In this situation pruning did not improve the MSE of my data set as it went up from 4.93 to 5.08.
d.Use the bagging approach in order to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important.
set.seed(1)
bag.seats = randomForest(Sales~., data = b1train, mtry = 10, ntree = 551, importance = TRUE)
bagseat.pred = predict(bag.seats, newdata = b1test)
mean((bagseat.pred - b1test$Sales)^2)## [1] 2.94674importance(bag.seats)## %IncMSE IncNodePurity
## CompPrice 40.223153 260.94508
## Income 12.662829 138.23658
## Advertising 23.611579 190.71765
## Population -2.386952 73.81902
## Price 80.660141 744.61378
## ShelveLoc 78.381654 692.52750
## Age 26.637401 232.40567
## Education 2.557387 62.04542
## Urban -2.563052 10.15531
## US 3.486152 12.13078e.Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
set.seed(1)
rando.seats = randomForest(Sales~., data = b1train, mtry = 10, importance = TRUE)
randseat.pred = predict(rando.seats, newdata = b1test)
mean((randseat.pred - b1test$Sales)^2)## [1] 2.945423importance(rando.seats)## %IncMSE IncNodePurity
## CompPrice 38.143176 259.77221
## Income 11.839999 138.20846
## Advertising 22.964249 191.25839
## Population -2.309923 74.13451
## Price 76.903940 744.20064
## ShelveLoc 73.841154 692.64875
## Age 25.449768 231.66005
## Education 2.547928 61.58542
## Urban -2.600879 10.15212
## US 3.572899 12.28877Q9
a.Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)
train = sample(1:nrow(OJ), 800)
OJtrain = OJ[train, ]
OJtest = OJ[-train, ]b.Fit a tree to the training data, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics about the tree, and describe the results obtained. What is the training error rate? How many terminal nodes does the tree have?
The tree had 5 variables in it with 9 different nodes. The training error rate of my tree was .1588.
tree.OJ = tree::tree(Purchase ~ ., data = OJtrain)
summary(tree.OJ)##
## Classification tree:
## tree::tree(formula = Purchase ~ ., data = OJtrain)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800plot(tree.OJ)
text(tree.OJ, pretty = 0)
c.Type in the name of the tree object in order to get a detailed text output. Pick one of the terminal nodes, and interpret the information displayed.
tree.OJ## node), split, n, deviance, yval, (yprob)
## * denotes terminal node
##
## 1) root 800 1073.00 CH ( 0.60625 0.39375 )
## 2) LoyalCH < 0.5036 365 441.60 MM ( 0.29315 0.70685 )
## 4) LoyalCH < 0.280875 177 140.50 MM ( 0.13559 0.86441 )
## 8) LoyalCH < 0.0356415 59 10.14 MM ( 0.01695 0.98305 ) *
## 9) LoyalCH > 0.0356415 118 116.40 MM ( 0.19492 0.80508 ) *
## 5) LoyalCH > 0.280875 188 258.00 MM ( 0.44149 0.55851 )
## 10) PriceDiff < 0.05 79 84.79 MM ( 0.22785 0.77215 )
## 20) SpecialCH < 0.5 64 51.98 MM ( 0.14062 0.85938 ) *
## 21) SpecialCH > 0.5 15 20.19 CH ( 0.60000 0.40000 ) *
## 11) PriceDiff > 0.05 109 147.00 CH ( 0.59633 0.40367 ) *
## 3) LoyalCH > 0.5036 435 337.90 CH ( 0.86897 0.13103 )
## 6) LoyalCH < 0.764572 174 201.00 CH ( 0.73563 0.26437 )
## 12) ListPriceDiff < 0.235 72 99.81 MM ( 0.50000 0.50000 )
## 24) PctDiscMM < 0.196196 55 73.14 CH ( 0.61818 0.38182 ) *
## 25) PctDiscMM > 0.196196 17 12.32 MM ( 0.11765 0.88235 ) *
## 13) ListPriceDiff > 0.235 102 65.43 CH ( 0.90196 0.09804 ) *
## 7) LoyalCH > 0.764572 261 91.20 CH ( 0.95785 0.04215 ) *Line 10 has a total of 79 observations. Over 77.21% of the observations take the value of MM and 22.78% take the value of CH.
d.Create a plot of the tree, and interpret the results.
The most important variables shown are LoyalCH, SpecialCH, PriceDiff, PctDiscMM, and ListPriceDiff.
plot(tree.OJ)
text(tree.OJ)
e. Predict the response on the test data, and produce a confusion matrix comparing the test labels to the predicted test labels. What is the test error rate?
treeOJ.pred = predict(tree.OJ, newdata = OJtest, type = "class")
table(treeOJ.pred, OJtest$Purchase)##
## treeOJ.pred CH MM
## CH 160 38
## MM 8 64(38 + 8) / 270## [1] 0.1703704The test error rate .17037.
f. Apply the cv.tree() function to the training set in order to determine the optimal tree size.
OJcv = tree::cv.tree(tree.OJ, FUN = prune.misclass)
OJcv## $size
## [1] 9 8 7 4 2 1
##
## $dev
## [1] 150 150 149 158 172 315
##
## $k
## [1] -Inf 0.000000 3.000000 4.333333 10.500000 151.000000
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"g.Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(OJcv$size, OJcv$dev, type = "b", xlab = "Tree Size", ylab = "cv classification error rate")
h.Which tree size corresponds to the lowest cross-validated classification error rate?
The tree size with 7 nodes corresponds to the lowest error rate.
i.Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
prune.OJ=tree::prune.tree(tree.OJ,best=7)j.Compare the training error rates between the pruned and un-pruned trees. Which is higher?
The pruned tree has a higher error rate of .1625 than the unpruned tree which had a rate of .1588.
summary(tree.OJ)##
## Classification tree:
## tree::tree(formula = Purchase ~ ., data = OJtrain)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800summary(prune.OJ)##
## Classification tree:
## snip.tree(tree = tree.OJ, nodes = c(10L, 4L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff" "PctDiscMM"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7748 = 614.4 / 793
## Misclassification error rate: 0.1625 = 130 / 800k.Compare the test error rates between the pruned and unpruned trees. Which is higher?
treeOJ.pred = predict(tree.OJ, newdata = OJtest, type = "class")
table(treeOJ.pred, OJtest$Purchase)##
## treeOJ.pred CH MM
## CH 160 38
## MM 8 64unprunedOJvalerr = (38 + 8) / 270
unprunedOJvalerr## [1] 0.1703704pruneOJ.pred = predict(prune.OJ, newdata = OJtest, type = "class")
table(pruneOJ.pred, OJtest$Purchase)##
## pruneOJ.pred CH MM
## CH 160 36
## MM 8 66prunedOJvalerr = (36 + 8) / 270
prunedOJvalerr## [1] 0.162963