Assignment 7
Cray Lester
2023-04-21
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- 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 x-axis should display ˆpm1, ranging from 0 to 1, and the y-axis should display the value of the Gini index, classification error, and entropy.
prop1 <- seq(0,1,.01)
prop2 <- 1 - prop1
classerror <- 1 - pmax(prop1, prop2)
gini <- prop1 * (1-prop1) + prop2 * (1-prop2)
entropy <- -prop1 * log(prop1) - prop2 * log(prop2)
data.frame(prop1, prop2, classerror, gini, entropy) %>%
pivot_longer(cols = c(classerror,gini, entropy), names_to = "metric") %>%
ggplot(aes(prop1, value, col = factor(metric))) +
geom_line() +
scale_color_hue(labels = c("Classification Error", "Entropy", "Gini")) +
labs(col = "Metric", x = "proportion", y = "value")## Warning: Removed 2 rows containing missing values (`geom_line()`).
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.
set.seed(1)
car.index <- createDataPartition(Carseats$Sales, p = .75, list = FALSE, times = 1)
carTrain <- Carseats[car.index,]
carTest <- Carseats[-car.index,]- Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
car.tree <- tree(Sales~., data = carTrain)
plot(car.tree)
text(car.tree, pretty = 0, cex = .5)
summary(car.tree)##
## Regression tree:
## tree(formula = Sales ~ ., data = carTrain)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Income" "Age" "Advertising"
## [6] "CompPrice"
## Number of terminal nodes: 16
## Residual mean deviance: 2.295 = 654 / 285
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -3.5490 -1.0160 -0.1187 0.0000 0.9648 3.5840car.pred.t <- predict(car.tree, carTest)
mean((car.pred.t-carTest$Sales)^2)## [1] 4.935788We get an MSE if 4.935, which is pretty good.
- 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.cars <- cv.tree(car.tree)
plot(cv.cars$size, cv.cars$dev, type = 'b')
It appears that pruning our model down to 12 terminal nodes would lower
the deviance the most.
prune.carseats <- prune.tree(car.tree, best = 12)
plot(prune.carseats)
text(prune.carseats, pretty = 0, cex = .5)
prune.pred <- predict(prune.carseats, carTest)
mean((prune.pred-carTest$Sales)^2)## [1] 4.970822This raised the MSE by .04.
- Use the bagging approach in order to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important.
car.bag <- randomForest(Sales~., data = carTrain, mtry =10, ntree = 500, importance = TRUE)
bag.preds <- predict(car.bag, carTest)
mean((bag.preds - carTest$Sales)^2)## [1] 3.179889importance(car.bag) %>%
as.data.frame() %>%
rownames_to_column('variable') %>%
arrange(desc(IncNodePurity))## variable %IncMSE IncNodePurity
## 1 Price 68.2206989 743.44922
## 2 ShelveLoc 80.1387119 694.76788
## 3 CompPrice 29.7704182 215.53404
## 4 Advertising 24.7563156 185.71361
## 5 Age 20.3956172 181.77479
## 6 Income 8.8936255 104.74986
## 7 Population 0.9713052 70.37262
## 8 Education 2.8936326 61.95442
## 9 US 3.9202545 14.31920
## 10 Urban -0.7213293 10.73275It appears that Price and ShelveLoc are the two 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.
car.rf <- randomForest(Sales~., data = carTrain, mtry = 3, ntree = 500, importance = TRUE)
rf.preds <- predict(car.rf, carTest)
mean((rf.preds - carTest$Sales)^2)## [1] 3.639842importance(car.rf) %>%
as.data.frame() %>%
rownames_to_column('variable') %>%
arrange(desc(IncNodePurity))## variable %IncMSE IncNodePurity
## 1 Price 46.405424 571.77646
## 2 ShelveLoc 48.426476 552.90732
## 3 Age 16.616958 229.12277
## 4 CompPrice 13.685658 200.89817
## 5 Advertising 18.826769 200.14971
## 6 Income 2.149908 157.18510
## 7 Population -1.204152 134.97045
## 8 Education 1.874559 106.63373
## 9 US 5.656612 33.74457
## 10 Urban -2.098901 22.43028Again, Price and ShelveLoc are the two most important variables. Increasing mtry reduces the test error, and we can see an reduction in test MSE as mtry approaches 10.
- Now analyze the data using BART, and report your results.
set.seed(1)
train <- sample(1:nrow(Carseats), nrow(Carseats) /2)
x <- Carseats[, 2:11]
y <- Carseats[, 'Sales']
xtrain <- x[train, ]
ytrain <- y[train]
xtest <- x[-train,]
ytest <-y[-train]
car.bart <- gbart(xtrain, ytrain, x.test = xtest)## *****Calling gbart: type=1
## *****Data:
## data:n,p,np: 200, 14, 200
## y1,yn: 2.781850, 1.091850
## x1,x[n*p]: 107.000000, 1.000000
## xp1,xp[np*p]: 111.000000, 1.000000
## *****Number of Trees: 200
## *****Number of Cut Points: 63 ... 1
## *****burn,nd,thin: 100,1000,1
## *****Prior:beta,alpha,tau,nu,lambda,offset: 2,0.95,0.273474,3,0.23074,7.57815
## *****sigma: 1.088371
## *****w (weights): 1.000000 ... 1.000000
## *****Dirichlet:sparse,theta,omega,a,b,rho,augment: 0,0,1,0.5,1,14,0
## *****printevery: 100
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## MCMC
## done 0 (out of 1100)
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## done 500 (out of 1100)
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## done 700 (out of 1100)
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## time: 3s
## trcnt,tecnt: 1000,1000bart.preds <- car.bart$yhat.test.mean
mean((ytest - bart.preds)^2)## [1] 1.499118Question 9
- This problem involves the OJ data set which is part of the ISLR2 package.
- Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
OJ.index <- createDataPartition(OJ$Purchase, p = .75, list = FALSE, times = 1)
OJtrain <- OJ[OJ.index,]
OJtest <- OJ[-OJ.index,]- 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 = OJtrain)
summary(oj.tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJtrain)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff" "DiscMM"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7389 = 588.2 / 796
## Misclassification error rate: 0.1569 = 126 / 803There are 7 terminal nodes and a missclassification rate of around 15%.
- 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
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## 1) root 803 1074.00 CH ( 0.61021 0.38979 )
## 2) LoyalCH < 0.48285 290 306.10 MM ( 0.22069 0.77931 )
## 4) LoyalCH < 0.280875 167 118.30 MM ( 0.11377 0.88623 ) *
## 5) LoyalCH > 0.280875 123 161.60 MM ( 0.36585 0.63415 )
## 10) PriceDiff < -0.24 13 0.00 MM ( 0.00000 1.00000 ) *
## 11) PriceDiff > -0.24 110 148.80 MM ( 0.40909 0.59091 ) *
## 3) LoyalCH > 0.48285 513 467.10 CH ( 0.83041 0.16959 )
## 6) LoyalCH < 0.764572 237 298.90 CH ( 0.67511 0.32489 )
## 12) ListPriceDiff < 0.235 95 129.30 MM ( 0.42105 0.57895 )
## 24) DiscMM < 0.15 56 75.84 CH ( 0.58929 0.41071 ) *
## 25) DiscMM > 0.15 39 36.71 MM ( 0.17949 0.82051 ) *
## 13) ListPriceDiff > 0.235 142 122.50 CH ( 0.84507 0.15493 ) *
## 7) LoyalCH > 0.764572 276 85.99 CH ( 0.96377 0.03623 ) *I chose 4) to be my terminal node. This is a set of observations where LoyalCH is less than .280875. There are 167 observations with a deviance 118.3. This branch overall predicts for MM, based off of a 11-89 split.
- Create a plot of the tree, and interpret the results.
plot(oj.tree)
text(oj.tree, pretty = 0, cex = .8)
It seems that generally if LoyaltyCH is less than .48, it is MM, but
otherwise it is CH with one node exception.
- 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.preds <- predict(oj.tree, OJtest, type = 'class')
table(OJ.preds, OJtest$Purchase)##
## OJ.preds CH MM
## CH 125 16
## MM 38 88(38+16)/270## [1] 0.2We have a 20% test error rate.
- Apply the cv.tree() function to the training set in order to determine the optimal tree size.
oj.cv <- cv.tree(oj.tree, FUN = prune.misclass)
oj.cv## $size
## [1] 7 5 2 1
##
## $dev
## [1] 151 151 169 313
##
## $k
## [1] -Inf 0.000000 8.333333 162.000000
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## $method
## [1] "misclass"
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## attr(,"class")
## [1] "prune" "tree.sequence"- Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(oj.cv$size, oj.cv$dev, type = 'b')
- Which tree size corresponds to the lowest cross-validated classification error rate?
It looks like a 5 node tree is our best bet.
- Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
oj.prune <- prune.misclass(oj.tree, best = 5)
plot(oj.prune)
text(oj.prune, pretty = 0, cex = .7)
- Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(oj.tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJtrain)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff" "DiscMM"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7389 = 588.2 / 796
## Misclassification error rate: 0.1569 = 126 / 803summary(oj.prune)##
## Classification tree:
## snip.tree(tree = oj.tree, nodes = 2L)
## Variables actually used in tree construction:
## [1] "LoyalCH" "ListPriceDiff" "DiscMM"
## Number of terminal nodes: 5
## Residual mean deviance: 0.7858 = 627.1 / 798
## Misclassification error rate: 0.1569 = 126 / 803It appears that the pruned tree and the original tree have the same misclassification error rate.
- Compare the test error rates between the pruned and unpruned trees. Which is higher?
prune.preds <- predict(oj.prune, OJtest, type = 'class')
table(prune.preds, OJtest$Purchase)##
## prune.preds CH MM
## CH 125 16
## MM 38 88(38+16)/270## [1] 0.2They again have the same test missclassification rate.