STA 6543 - HW 7
Farheen Ali
4/30/2021
QUESTION 3: Consider the Gini index, classification error, and entorpy 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.
p = seq(0, 1, 0.01)
gini = p * (1 - p) * 2
ent = -(p * log(p) + (1 - p) * log(1 - p))
class.err = 1 - pmax(p, 1 - p)
matplot(p, cbind(gini, ent, class.err), col = c("red", "green", "blue"))
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.
library(ISLR)## Warning: package 'ISLR' was built under R version 4.0.3library(tree)## Warning: package 'tree' was built under R version 4.0.4attach(Carseats)- Split the data set into a training set and a test set.
set.seed(1)
row.number = sample(1:nrow(Carseats), 0.7*nrow(Carseats))
Carseats_train = Carseats[row.number,]
Carseats_test = Carseats[-row.number,]- Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
carseats.tree1<-tree(Sales~.,data=Carseats_train)
summary(carseats.tree1)##
## Regression tree:
## tree(formula = Sales ~ ., data = Carseats_train)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Age" "Income" "CompPrice"
## [6] "Advertising"
## Number of terminal nodes: 18
## Residual mean deviance: 2.409 = 631.1 / 262
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -4.77800 -0.96100 -0.08865 0.00000 1.01800 4.14100plot(carseats.tree1)
text(carseats.tree1,pretty=0)
carseats.preds1<-predict(carseats.tree1,newdata=Carseats_test)
carseats.mse1<-mean((Carseats_test$Sales-carseats.preds1)^2)
carseats.mse1## [1] 4.208383MSE is at 4.21.
- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv.carseats <- cv.tree(carseats.tree1)
plot(cv.carseats$size, cv.carseats$dev, type = "b")
prune.carseats <- prune.tree(carseats.tree1, best = 10)
plot(prune.carseats)
text(prune.carseats, pretty = 0)
prune.predict2<-predict(prune.carseats,Carseats_test)
mean((prune.predict2-Carseats_test$Sales)^2)## [1] 4.815186pruning the tree did not improve the Test MSE Increased only to 4.82
- 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)## Warning: package 'randomForest' was built under R version 4.0.5## randomForest 4.6-14## Type rfNews() to see new features/changes/bug fixes.bag.cs<-randomForest(Sales~.,Carseats_train,importance=TRUE,mtry=13)## Warning in randomForest.default(m, y, ...): invalid mtry: reset to within valid
## rangebag.carseats.preds3<-predict(bag.cs,Carseats_test)
mean((bag.carseats.preds3-Carseats_test$Sales)^2)## [1] 2.540992Test MSE obtained is 2.61.
importance(bag.cs)## %IncMSE IncNodePurity
## CompPrice 33.7743862 228.91321
## Income 6.0942550 118.68202
## Advertising 18.6617139 153.90381
## Population -0.5498118 65.13690
## Price 66.5408316 687.49056
## ShelveLoc 67.2755302 634.93813
## Age 20.2323914 227.02999
## Education 1.7403062 63.19110
## Urban -0.0555473 11.43354
## US -0.3669702 11.04510Price and ShelveLoc are the most important variables.
- Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
random.cs <- randomForest(Sales ~ ., data = Carseats_train, mtry = 3, ntree = 500, importance = TRUE)
random.cs.preds4 <- predict(random.cs, newdata = Carseats_test)
mean((random.cs.preds4 - Carseats_test$Sales)^2)## [1] 2.862987The test MSE is 2.91.
importance(random.cs)## %IncMSE IncNodePurity
## CompPrice 17.6208035 210.20144
## Income 4.2057215 161.10896
## Advertising 12.9084644 162.02173
## Population -2.5611283 134.21794
## Price 45.4429707 548.63871
## ShelveLoc 44.4055881 509.39047
## Age 18.6482348 270.10708
## Education 1.9219385 97.33615
## Urban -0.7258283 20.87099
## US 5.6233465 29.07053Changing m and the number of variables increased the MSE Price and ShelveLoc are the two most important variables
QUESTION 9: This problem involves the OJ data set which is part of the ISLR package.
attach(OJ)- Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)
row.number1 <- sample(1:nrow(OJ), 800)
OJ_train <- OJ[row.number1, ]
OJ_test <- OJ[-row.number1, ]- 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 ~ ., data = OJ_train)
summary(OJ.tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ_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 / 800Terminal nodes = 9 training error rate = 0.1588. “PriceDiff”, “SpecialCH”, “LoyalCH”, “ListPriceDiff”, and “PctDiscMM” as the variables in the model.
- 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 ) *There are 109 observations in the leaf with the residual variance of 147 Selecting node 11 PriceDiff, the node splits for when PriceDiff>0.05 The overall prediction is CH = 59.63% taking CH value
The overall prediction is CH = 40.37% taking the MM value
- Create a plot of the tree, and interpret the results.
plot(OJ.tree)
text(OJ.tree, pretty = 0)
Most important indicator of Purchase is “LoyalCH” (top 3 nodes contain “LoyalCH”)
- 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?
set.seed(1)
OJ.preds1<-predict(OJ.tree,newdata = OJ_test,type = "class")
caret::confusionMatrix(OJ.preds1, OJ_test$Purchase)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 160 38
## MM 8 64
##
## Accuracy : 0.8296
## 95% CI : (0.7794, 0.8725)
## No Information Rate : 0.6222
## P-Value [Acc > NIR] : 8.077e-14
##
## Kappa : 0.6154
##
## Mcnemar's Test P-Value : 1.904e-05
##
## Sensitivity : 0.9524
## Specificity : 0.6275
## Pos Pred Value : 0.8081
## Neg Pred Value : 0.8889
## Prevalence : 0.6222
## Detection Rate : 0.5926
## Detection Prevalence : 0.7333
## Balanced Accuracy : 0.7899
##
## 'Positive' Class : CH
## OJ.te<-(8+38)/nrow(OJ_test)
OJ.te## [1] 0.1703704Test error rate = 17.04%.
- Apply the cv.tree() function to the training set in order to determine the optimal tree size.
OJ.tree.cv<-cv.tree(OJ.tree,FUN = prune.misclass)
OJ.tree.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
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"- Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(OJ.tree.cv$size, OJ.tree.cv$dev, type = "b", xlab = "Tree size", ylab = "CV Deviance")
points(4,min(OJ.tree.cv$dev),col="red")
- Which tree size corresponds to the lowest cross-validated classification error rate?
tree size = 4.
- 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.
OJ.prune <- prune.misclass(OJ.tree, best = 4)
plot(OJ.prune)
text(OJ.prune, pretty = 0)
- Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(OJ.prune)##
## Classification tree:
## snip.tree(tree = OJ.tree, nodes = c(4L, 10L, 3L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff"
## Number of terminal nodes: 4
## Residual mean deviance: 0.8922 = 710.2 / 796
## Misclassification error rate: 0.1788 = 143 / 800summary(OJ.tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ_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 / 800Misclassification error rate of the pruned tree = 0.1788 This is higher than the original tree at 0.1588.
- Compare the test error rates between the pruned and unpruned trees. Which is higher?
set.seed(1)
OJ.prune.preds6<-predict(OJ.prune,newdata = OJ_test,type = "class")
caret::confusionMatrix(OJ.prune.preds6, OJ_test$Purchase)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 161 41
## MM 7 61
##
## Accuracy : 0.8222
## 95% CI : (0.7713, 0.8659)
## No Information Rate : 0.6222
## P-Value [Acc > NIR] : 6.769e-13
##
## Kappa : 0.5954
##
## Mcnemar's Test P-Value : 1.906e-06
##
## Sensitivity : 0.9583
## Specificity : 0.5980
## Pos Pred Value : 0.7970
## Neg Pred Value : 0.8971
## Prevalence : 0.6222
## Detection Rate : 0.5963
## Detection Prevalence : 0.7481
## Balanced Accuracy : 0.7782
##
## 'Positive' Class : CH
## OJ.prune.te<-(7+41)/nrow(OJ_test)
OJ.prune.te## [1] 0.1777778OJ.te## [1] 0.1703704The test error rate slightly increased with the pruned tree Test error rate = 17.78% Unpruned tree had a test error rate of 17.03%