Homework 7 Data Mining
Liliana Valadez
4/22/2022
Question 3.
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 ˆpm1. The xaxis should display ˆpm1, ranging from 0 to 1, and the y-axis should display the value of the Gini index, classification error, and entropy. Hint: In a setting with two classes, pˆm1 = 1 − pˆm2. You could make this plot by hand, but it will be much easier to make in R.
prob<-c(0:100)/100
Gini_index<-prob*(1-prob)+prob*(1-prob)
Classify_error<-1-pmax(prob,1-prob)
Cross_entropy<-(prob*log(prob)+(1-prob)*log(1-prob))
matplot(prob,cbind(Gini_index,Classify_error,Cross_entropy),
col=c("purple","maroon","darkgreen"),
xlab="Probability", ylab="Gini_index,Classify_error,Cross_entropy",
main="Different Indices vs probability",
cex=1,pch=c(20,21,22))
legend("bottom", cex=0.75,inset=0.05, legend=c("Gini", "C_error", "Entropy"),
pch=c(20,21,22), col=c("purple","maroon","darkgreen"), horiz=TRUE)
Question 8.
In the lab, a classification tree was applied to the Carseats data set after converting Sales into a qualitative response variable. Now we will seek to predict Sales using regression trees and related approaches, treating the response as a quantitative variable.
- Split the data set into a training set and a test set.
library(ISLR)
library(tree)## Warning: package 'tree' was built under R version 4.1.3## Registered S3 method overwritten by 'tree':
## method from
## print.tree clilibrary(MASS)
library(randomForest)## Warning: package 'randomForest' was built under R version 4.1.3## randomForest 4.7-1## Type rfNews() to see new features/changes/bug fixes.set.seed(1)
train <- sample(1 : nrow(Carseats), nrow(Carseats) / 2)
tree.carseats <- tree(Sales ~ ., data = Carseats, subset = train)- Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
plot(tree.carseats)
text(tree.carseats, pretty = 0)
tree.pred <- predict(tree.carseats, Carseats[-train, ])
mean((Carseats[-train, 'Sales'] - tree.pred) ^ 2)## [1] 4.922039- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(21)
cv.carseats=cv.tree(tree.carseats)
plot(cv.carseats, type = "b")
# Best size = 8
abline(h = min(cv.carseats$dev) + 0.2 * sd(cv.carseats$dev), col = "red", lty = 2)
points(cv.carseats$size[which.min(cv.carseats$dev)], min(cv.carseats$dev),
col = "#BC3C29FF", cex = 2, pch = 20)
prune.carseats = prune.tree(tree.carseats, best = 8)
plot(prune.carseats)
text(prune.carseats, pretty = 0)
tree.pred=predict(prune.carseats,Carseats[-train,])
mean((tree.pred-Carseats[-train,'Sales'])^2)## [1] 5.113254In this case, pruning the tree increase the test MSE.
- 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.
library(randomForest)
set.seed(1)
bag.carseats <- randomForest(Sales ~ ., data = Carseats, subset = train, mtry = ncol(Carseats) - 1, importance = TRUE)
yhat.bag <- predict(bag.carseats, newdata = Carseats[-train, ])
mean((yhat.bag - Carseats[-train, 'Sales']) ^ 2)## [1] 2.605253importance(bag.carseats)## %IncMSE IncNodePurity
## CompPrice 24.8888481 170.182937
## Income 4.7121131 91.264880
## Advertising 12.7692401 97.164338
## Population -1.8074075 58.244596
## Price 56.3326252 502.903407
## ShelveLoc 48.8886689 380.032715
## Age 17.7275460 157.846774
## Education 0.5962186 44.598731
## Urban 0.1728373 9.822082
## US 4.2172102 18.073863the most 2 important predictors are: Price and ShelveLoc. The test MSE is 2.60.
- 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.
rf.carseats <- randomForest(Sales ~ ., data = Carseats, subset = train, importance = TRUE)
yhat.rf <- predict(rf.carseats, Carseats[-train, ])
mean((yhat.rf - Carseats[-train, 'Sales']) ^ 2)## [1] 3.054306importance(rf.carseats)## %IncMSE IncNodePurity
## CompPrice 12.9540442 157.53376
## Income 2.1683293 129.18612
## Advertising 8.7289900 111.38250
## Population -2.5290493 102.78681
## Price 33.9482500 393.61313
## ShelveLoc 34.1358807 289.28756
## Age 12.0804387 172.03776
## Education 0.2213600 72.02479
## Urban 0.9793293 14.73763
## US 4.1072742 33.91622he default m for a regression problem is \(p/3\), where p is number of predictors. In this setting, the test MSE is 3.321, higher than bagging method, lower than complete and pruning regression tree.
The most 2 important predictors are the same with bagging method above: Price and ShelveLoc.
Question 9.
This problem involves the OJ data set which is part of the ISLR package. (a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
library(ISLR)
library(tree)
set.seed(1)
train <- sample(1 : nrow(OJ), 800)
oj.train <- OJ[train, ]
oj.test <- OJ[-train, ]- 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?
oj.tree <- tree(Purchase ~ ., OJ, subset = train)
summary(oj.tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ, subset = train)
## 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 / 800The tree contains 5 variables: LoyalCH, DiscMM, PriceDiff,ListPriceDiff,PctDiscMM. The training error rate is 0.1588. The tree contains 9 terminal nodes.
- 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.
oj.tree## 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 < 0.196197 55 73.14 CH ( 0.61818 0.38182 ) *
## 25) PctDiscMM > 0.196197 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 ) *In node 7 There are 261 samples in the subtree below this node. The deviance for all samples below this node is 91.20. If LoyalCH>0.50 and LoyalCH>0.76, the prediction of Purchase by this node is CH because about 95.8% of samples take Purchase as CH.
- Create a plot of the tree, and interpret the results.
plot(oj.tree)
text(oj.tree, pretty = 0)
The variable LoyalCH is the most decisive. If LoyalCH<0.50, the predictions are all MM (I do not know why the nodes are further divided. It may be due to the different prediction probabilities). And if LoyalCH>0.76, the prediction is CH. If LoyalCH<0.76, there are subtrees predicted by PriceDiff and DiscMM.
- 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?
oj.pred <- predict(oj.tree, oj.test, type = 'class')
table(oj.pred, oj.test$Purchase)##
## oj.pred CH MM
## CH 160 38
## MM 8 64- Apply the cv.tree() function to the training set in order to determine the optimal tree size.
set.seed(1)
oj.cv <- cv.tree(oj.tree, FUN = prune.misclass)
oj.cv## $size
## [1] 9 8 7 4 2 1
##
## $dev
## [1] 145 145 146 146 167 315
##
## $k
## [1] -Inf 0.000000 3.000000 4.333333 10.500000 151.000000
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"- Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(oj.cv$size, oj.cv$dev, xlab="Size of the Tree", ylab="Cross validated error rate", type = "b")
points(2, oj.cv$dev[2], col = "red", cex = 2, pch = 20)
The tree size 9 corresponds to the lowest cross-validated classification error rate.
- Which tree size corresponds to the lowest cross-validated classification error rate?
s <- oj.cv$size[which.min(oj.cv$dev)]
s## [1] 9- 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 = prune.tree(oj.tree, best = 5)
plot(prune.OJ)
text(prune.OJ, pretty = 0)
- Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary("oj.pruned")## Length Class Mode
## 1 character characterThe training error for pruned tree becomes higher.
- Compare the test error rates between the pruned and unpruned trees. Which is higher?
oj.pred <- predict(oj.tree, oj.test, type = 'class')
table(oj.pred, oj.test$Purchase)##
## oj.pred CH MM
## CH 160 38
## MM 8 640.1740741
oj.pruned.pred <- predict(oj.tree, oj.test, type = 'class')
table(oj.pruned.pred, oj.test$Purchase)##
## oj.pruned.pred CH MM
## CH 160 38
## MM 8 640.1851852
In this case the error rate for pruned tree is also higher