Forests
2024-01-02
In the lab, we applied random forests to the Boston data using mtry = 6 and using ntree = 25 and ntree = 500. Create a plot displaying the test error resulting from random forests on this data set for a more comprehensive range of values for mtry and ntree. You can model your plot after Figure 8.10. Describe the results obtained.
set.seed (123)
train_index<-sample (1:nrow(Boston), nrow(Boston)/2)
train<-Boston[train_index,]
test<-Boston[-train_index,]
# building models after 8.10
rf1 <- randomForest(x = train[,-14], y = train$medv, xtest = test[,-14], ytest = test$medv, mtry = ncol(Boston) - 1, ntree = 500)
rf2 <- randomForest(x = train[,-14], y = train$medv, xtest = test[,-14], ytest = test$medv, mtry = (ncol(Boston) - 1) / 2, ntree = 500)
rf3 <- randomForest(x = train[,-14], y = train$medv, xtest = test[,-14], ytest = test$medv, mtry = sqrt(ncol(Boston) - 1), ntree = 500)plot(1:500, rf1$test$mse, col = "orange", type = "l", xlab = "Number of Trees", ylab = "Test MSE", ylim = c(0, 10))
lines(1:500, rf2$test$mse, col = "green", type = "l")
lines(1:500, rf3$test$mse, col = "maroon", type = "l")
legend("topright", c("m = p", "m = p/2", "m = sqrt(p)"), col = c("orange", "green", "maroon"), cex = 1, lty = 1)
- 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(123)
train<-sample(1:nrow(Carseats), nrow(Carseats)/2)
carseats.train<-Carseats[train, ]
y.train<-Carseats[train, "Sales"]
carseats.test<-Carseats[-train, ]
y.test<-Carseats[-train, "Sales"]- Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
tree.car<-tree(Sales ~ ., data = carseats.train)
summary(tree.car)##
## Regression tree:
## tree(formula = Sales ~ ., data = carseats.train)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Income" "Age" "Population"
## [6] "Education" "CompPrice" "Advertising"
## Number of terminal nodes: 18
## Residual mean deviance: 2.132 = 388.1 / 182
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -4.08000 -0.92870 0.06244 0.00000 0.87020 3.71700plot(tree.car)
text(tree.car, pretty = 0, cex = 0.5)
yhat = predict(tree.car, newdata = carseats.test)
mean((yhat - y.test)^2)## [1] 4.395357- 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.car = cv.tree(tree.car)
plot(cv.car$size , cv.car$dev, type = 'b')
prune.car<-prune.tree(tree.car, best = 12)
plot(prune.car)
text(prune.car, pretty = 0, cex = 0.5)
yhat<- predict(prune.car, carseats.test)
mean((yhat - y.test)^2)## [1] 4.662975It looks like the test MSE was slightly higher as a result of the pruning process.
- 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.car<-randomForest(Sales ~ ., data = Carseats,
subset = train , mtry = 10, importance = TRUE)
yhat.bag <- predict(bag.car, newdata = carseats.test)
mean((yhat.bag - y.test)^2)## [1] 2.719623importance(bag.car)## %IncMSE IncNodePurity
## CompPrice 19.45546048 157.493242
## Income 6.54808190 88.084674
## Advertising 9.35661044 76.584090
## Population -0.03332117 52.232916
## Price 46.38347866 383.642775
## ShelveLoc 50.78194122 388.510326
## Age 19.66414864 181.155199
## Education 2.95721774 54.469563
## Urban -0.22123191 8.959947
## US 1.51594678 6.052909varImpPlot(bag.car)
It looks like Shelveloc and Price 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.
set.seed(123)
tree.car<-randomForest(Sales ~ ., data = Carseats,
subset = train , mtry = 5, importance = TRUE)
yhat.bag <- predict(tree.car, newdata = carseats.test)
mean((yhat.bag - y.test)^2)## [1] 3.01707importance(tree.car)## %IncMSE IncNodePurity
## CompPrice 15.6874802 152.97700
## Income 6.1782093 104.83134
## Advertising 7.4822778 89.84070
## Population 0.1519680 74.70762
## Price 35.6880806 345.59391
## ShelveLoc 42.1336259 333.55620
## Age 18.3059283 204.29930
## Education 1.8670124 63.31604
## Urban -0.1467837 11.01253
## US 0.3849865 11.47038varImpPlot(tree.car)
This appears to show a lower test MSE as compared to the earlier
models.
- Now analyze the data using BART, and report your results.
library(BART)## Warning: package 'BART' was built under R version 4.3.2## Loading required package: nlme## Loading required package: nnet## Loading required package: survivalx <- Carseats[, 2:11]
y <- Carseats[, "Sales"]
xtrain <- x[train, ]
ytrain <- y[train]
xtest <- x[-train, ]
ytest <- y[-train]set.seed (1)
bartfit<- gbart(xtrain , ytrain , x.test = xtest)## *****Calling gbart: type=1
## *****Data:
## data:n,p,np: 200, 14, 200
## y1,yn: 3.230000, 3.070000
## x1,x[n*p]: 104.000000, 1.000000
## xp1,xp[np*p]: 138.000000, 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: 2,0.95,0.260569,3,0.191523,7.43
## *****sigma: 0.991574
## *****w (weights): 1.000000 ... 1.000000
## *****Dirichlet:sparse,theta,omega,a,b,rho,augment: 0,0,1,0.5,1,14,0
## *****printevery: 100
##
## MCMC
## done 0 (out of 1100)
## done 100 (out of 1100)
## done 200 (out of 1100)
## done 300 (out of 1100)
## done 400 (out of 1100)
## done 500 (out of 1100)
## done 600 (out of 1100)
## done 700 (out of 1100)
## done 800 (out of 1100)
## done 900 (out of 1100)
## done 1000 (out of 1100)
## time: 6s
## trcnt,tecnt: 1000,1000yhat.bart <- bartfit$yhat.test.mean
mean (( ytest - yhat.bart)^2)## [1] 1.621989BART appears to indicate the lowest test MSE of 1.62.
- 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.
set.seed(123)
train_indices<- sample(1:nrow(OJ), 800)
train_oj <- OJ[train_indices, ]
test_oj<-OJ[-train_indices, ]- 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 = train_oj)
summary(oj_tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = train_oj)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff"
## Number of terminal nodes: 8
## Residual mean deviance: 0.7625 = 603.9 / 792
## Misclassification error rate: 0.165 = 132 / 800The output indicates a training error rate of .16, with 8 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 1071.00 CH ( 0.60875 0.39125 )
## 2) LoyalCH < 0.5036 350 415.10 MM ( 0.28000 0.72000 )
## 4) LoyalCH < 0.276142 170 131.00 MM ( 0.12941 0.87059 )
## 8) LoyalCH < 0.0356415 56 10.03 MM ( 0.01786 0.98214 ) *
## 9) LoyalCH > 0.0356415 114 108.90 MM ( 0.18421 0.81579 ) *
## 5) LoyalCH > 0.276142 180 245.20 MM ( 0.42222 0.57778 )
## 10) PriceDiff < 0.05 74 74.61 MM ( 0.20270 0.79730 ) *
## 11) PriceDiff > 0.05 106 144.50 CH ( 0.57547 0.42453 ) *
## 3) LoyalCH > 0.5036 450 357.10 CH ( 0.86444 0.13556 )
## 6) PriceDiff < -0.39 27 32.82 MM ( 0.29630 0.70370 ) *
## 7) PriceDiff > -0.39 423 273.70 CH ( 0.90071 0.09929 )
## 14) LoyalCH < 0.705326 130 135.50 CH ( 0.78462 0.21538 )
## 28) PriceDiff < 0.145 43 58.47 CH ( 0.58140 0.41860 ) *
## 29) PriceDiff > 0.145 87 62.07 CH ( 0.88506 0.11494 ) *
## 15) LoyalCH > 0.705326 293 112.50 CH ( 0.95222 0.04778 ) *Node 4 is the 3rd split at LoyalCH < 0.276142. There are 170 observations within the node with a deviance 131. It appears that roughly .13% of the observations were misclassified as CH, with the remaining 87% being correctly identified as MM.
- Create a plot of the tree, and interpret the results.
plot(oj_tree)
text(oj_tree, pretty=0)
The splits indicate that LoyaCH and PriceDiff are the two most important
variables in this model.
- 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, test_oj,
type = "class")
table(oj_pred, test_oj$Purchase)##
## oj_pred CH MM
## CH 150 34
## MM 16 70(150+70)/270## [1] 0.8148148It appears the test error rate is around 19%
- Apply the cv.tree() function to the training set in order to determine the optimal tree size.
set.seed (7)
cv.oj<-cv.tree(oj_tree, FUN = prune.misclass)
cv.oj## $size
## [1] 8 5 3 2 1
##
## $dev
## [1] 140 140 160 166 313
##
## $k
## [1] -Inf 0 8 11 154
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"Both 8 and 5 are tied for lowest dev.
- Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.’
plot(cv.oj$size, cv.oj$dev, type = 'b')
(h) Which tree size corresponds to the lowest cross-validated
classification error rate?
Both 5 and 8
- Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
prune.oj<- prune.misclass(oj_tree, best = 5)
summary(prune.oj)##
## Classification tree:
## snip.tree(tree = oj_tree, nodes = c(4L, 7L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff"
## Number of terminal nodes: 5
## Residual mean deviance: 0.826 = 656.6 / 795
## Misclassification error rate: 0.165 = 132 / 800plot(prune.oj)
text(prune.oj, pretty = 0)
(j) Compare the training error rates between the pruned and unpruned
trees. Which is higher?
The training error was quite comparable .16
- Compare the test error rates between the pruned and unpruned trees. Which is higher?
prune_pred <- predict(prune.oj, test_oj,
type = "class")
table(prune_pred, test_oj$Purchase)##
## prune_pred CH MM
## CH 150 34
## MM 16 70(150+70)/270## [1] 0.8148148Comparable test error rates.
- We now use boosting to predict Salary in the Hitters data set.
- Remove the observations for whom the salary information is unknown, and then log-transform the salaries.
Hitters = na.omit(Hitters)
Hitters$Salary = log(Hitters$Salary)- Create a training set consisting of the first 200 observations, and a test set consisting of the remaining observations.
set.seed(123)
train_indices<- sample(1:nrow(Hitters), 200)
train_hit<- Hitters[train_indices, ]
test_hit<-Hitters[-train_indices, ]- Perform boosting on the training set with 1,000 trees for a range of values of the shrinkage parameter λ. Produce a plot with different shrinkage values on the x-axis and the corresponding training set MSE on the y-axis.
set.seed (1)
shrinkage_values <- seq(0.01, 0.2, by = 0.01)
mse_values <- numeric(length(shrinkage_values))
for (i in seq_along(shrinkage_values)) {
hit_boost<- gbm(
formula = Salary ~ .,
distribution = "gaussian",
data = train_hit,
n.trees = 1000,
shrinkage = shrinkage_values[i],)
predictions_train<- predict(hit_boost, newdata = train_hit, n.trees = 1000)
mse_values[i]<-mean((predictions_train - train_hit$Salary)^2)}
plot(shrinkage_values, mse_values, type = "b", xlab = "Shrinkage Parameter (λ)", ylab = "Training Set MSE", main = "Boosting: Shrinkage vs. Training Set MSE")
(d) Produce a plot with different shrinkage values on the x-axis and the
corresponding test set MSE on the y-axis.
set.seed (1)
shrinkage_values<- seq(0.01, 0.2, by = 0.01)
mse_test<- numeric(length(shrinkage_values))
for (i in seq_along(shrinkage_values)) {
hit_boost<- gbm(
formula = Salary ~ .,
distribution = "gaussian",
data = train_hit,
n.trees = 1000,
shrinkage = shrinkage_values[i],)
predictions_train<- predict(hit_boost, newdata = test_hit, n.trees = 1000)
mse_test[i]<-mean((predictions_train - test_hit$Salary)^2)}
plot(shrinkage_values, mse_values, type = "b", xlab = "Shrinkage Parameter (λ)", ylab = "Training Set MSE", main = "Boosting: Shrinkage vs. Training Set MSE")
- Compare the test MSE of boosting to the test MSE that results from applying two of the regression approaches seen in Chapters 3 and 6.
Lets first get the test error for the previous model
which.min(mse_test)## [1] 5mse_test[5]## [1] 0.182951shrinkage_values[5]## [1] 0.05lambda value 0.05 produces the lowest test error of 18
ml_fit<-lm(Salary ~ ., data = train_hit)
pred<-predict(ml_fit, test_hit)
mean((pred - test_hit$Salary)^2)## [1] 0.4636651x <- model.matrix(Salary ~ ., Hitters)[, -1]Lets try ridge
set.seed(123)
y<-Hitters$Salary
x1_train<-x[1:200,]
x1_test<-x[201:263,]
y1_train<-y[1:200]
y1_test<-y[201:263]
ridge.mod<-glmnet(x1_train, y1_train, alpha = 0)
ridge.pred<- predict(ridge.mod, newx = x1_test)
mean ((ridge.pred - y1_test)^2)## [1] 0.5145349The method of boosting results in a quite significant reduction in test error, as compared to LM and ridge regression.
- Which variables appear to be the most important predictors in the boosted model?
summary(hit_boost)
## var rel.inf
## CAtBat CAtBat 18.7349427
## CRBI CRBI 9.7748552
## Assists Assists 7.5894815
## Walks Walks 6.7376674
## PutOuts PutOuts 6.6569733
## CHmRun CHmRun 6.4233218
## CRuns CRuns 5.6908153
## AtBat AtBat 5.5493044
## RBI RBI 5.3605685
## Runs Runs 5.3319071
## HmRun HmRun 4.3953566
## Errors Errors 3.9842625
## Years Years 3.7912156
## Hits Hits 2.9571114
## CHits CHits 2.7213600
## CWalks CWalks 2.6797075
## League League 0.7050832
## Division Division 0.5579991
## NewLeague NewLeague 0.3580668CAtBat followed by CRBI are the two most important.
- Now apply bagging to the training set. What is the test set MSE for this approach?
set.seed (1)
bag.hit<-randomForest(Salary ~ ., data = train_hit, mtry = 19, importance = TRUE)
yhat.bag <- predict(bag.hit, newdata = test_hit)
mean((yhat.bag - test_hit$Salary)^2)## [1] 0.1674674importance(bag.hit)## %IncMSE IncNodePurity
## AtBat 9.0942582 7.3808624
## Hits 4.5646002 6.1348510
## HmRun 0.8793614 1.4728781
## Runs 6.3357764 2.9861875
## RBI 8.5466292 6.3313359
## Walks 10.8810629 7.9538743
## Years 6.2213577 1.5016902
## CAtBat 20.1976310 49.1244534
## CHits 9.9028060 15.3977528
## CHmRun 6.7311147 4.7713401
## CRuns 18.6762434 42.7473717
## CRBI 9.7795240 7.2005027
## CWalks 5.4507949 5.4216838
## League -1.7684417 0.1456641
## Division -0.9509188 0.1200144
## PutOuts 0.5207492 2.2833580
## Assists 1.9050328 1.3871213
## Errors 2.1942694 1.4765472
## NewLeague -0.6469658 0.2690485varImpPlot(bag.hit)
Not bad at all! This models test error is lower than that of boosting at
.16. It indicates the same most important variables as boosting.
- This question uses the Caravan data set.
- Create a training set consisting of the first 1,000 observations, and a test set consisting of the remaining observations.
Caravan$Purchase<-ifelse(Caravan$Purchase == "Yes", 1, 0)
train_indices<-1:1000
train_car<- Caravan[train_indices, ]
test_car<-Caravan[-train_indices, ]- Fit a boosting model to the training set with Purchase as the response and the other variables as predictors. Use 1,000 trees, and a shrinkage value of 0.01. Which predictors appear to be the most important?
set.seed(123)
boost.car<- gbm(Purchase ~ ., data = train_car, distribution = "gaussian", n.trees = 1000, shrinkage = 0.1)## Warning in gbm.fit(x = x, y = y, offset = offset, distribution = distribution,
## : variable 50: PVRAAUT has no variation.## Warning in gbm.fit(x = x, y = y, offset = offset, distribution = distribution,
## : variable 71: AVRAAUT has no variation.summary(boost.car)
## var rel.inf
## MOPLHOOG MOPLHOOG 5.84547588
## MKOOPKLA MKOOPKLA 5.13986245
## MOSTYPE MOSTYPE 5.03245809
## MBERHOOG MBERHOOG 4.90296559
## MSKA MSKA 4.38878354
## MGODGE MGODGE 4.18422333
## MAUT1 MAUT1 4.09888626
## MSKC MSKC 4.09832492
## MINK3045 MINK3045 3.98730910
## MBERMIDD MBERMIDD 3.48940487
## ABRAND ABRAND 3.36540347
## PPERSAUT PPERSAUT 3.19411639
## MFWEKIND MFWEKIND 2.76553266
## PBRAND PBRAND 2.74024602
## MGODPR MGODPR 2.62913019
## MAUT2 MAUT2 2.49000540
## MRELGE MRELGE 2.24093534
## MBERARBG MBERARBG 1.96804295
## APERSAUT APERSAUT 1.91085248
## MINKM30 MINKM30 1.88666994
## MHHUUR MHHUUR 1.83311135
## MSKB1 MSKB1 1.66511872
## MINK4575 MINK4575 1.66403284
## MFGEKIND MFGEKIND 1.52363199
## MZFONDS MZFONDS 1.51940089
## MAUT0 MAUT0 1.48393368
## MFALLEEN MFALLEEN 1.42959139
## MOPLMIDD MOPLMIDD 1.35712582
## MRELOV MRELOV 1.28841041
## MZPART MZPART 1.25929968
## MINKGEM MINKGEM 1.22869319
## PWAPART PWAPART 1.12908530
## MSKB2 MSKB2 1.10051328
## MINK7512 MINK7512 1.08680660
## MGODRK MGODRK 1.06714585
## MBERARBO MBERARBO 1.04217903
## MOPLLAAG MOPLLAAG 0.98962937
## MSKD MSKD 0.82949400
## MGODOV MGODOV 0.73230725
## PBYSTAND PBYSTAND 0.70677359
## MGEMOMV MGEMOMV 0.64226934
## MBERBOER MBERBOER 0.53154042
## MOSHOOFD MOSHOOFD 0.49436570
## MHKOOP MHKOOP 0.47951406
## PLEVEN PLEVEN 0.46311789
## MAANTHUI MAANTHUI 0.41468831
## MRELSA MRELSA 0.37218441
## MINK123M MINK123M 0.36565223
## PMOTSCO PMOTSCO 0.33226742
## MGEMLEEF MGEMLEEF 0.30293182
## MBERZELF MBERZELF 0.23642419
## ALEVEN ALEVEN 0.07013109
## PWABEDR PWABEDR 0.00000000
## PWALAND PWALAND 0.00000000
## PBESAUT PBESAUT 0.00000000
## PVRAAUT PVRAAUT 0.00000000
## PAANHANG PAANHANG 0.00000000
## PTRACTOR PTRACTOR 0.00000000
## PWERKT PWERKT 0.00000000
## PBROM PBROM 0.00000000
## PPERSONG PPERSONG 0.00000000
## PGEZONG PGEZONG 0.00000000
## PWAOREG PWAOREG 0.00000000
## PZEILPL PZEILPL 0.00000000
## PPLEZIER PPLEZIER 0.00000000
## PFIETS PFIETS 0.00000000
## PINBOED PINBOED 0.00000000
## AWAPART AWAPART 0.00000000
## AWABEDR AWABEDR 0.00000000
## AWALAND AWALAND 0.00000000
## ABESAUT ABESAUT 0.00000000
## AMOTSCO AMOTSCO 0.00000000
## AVRAAUT AVRAAUT 0.00000000
## AAANHANG AAANHANG 0.00000000
## ATRACTOR ATRACTOR 0.00000000
## AWERKT AWERKT 0.00000000
## ABROM ABROM 0.00000000
## APERSONG APERSONG 0.00000000
## AGEZONG AGEZONG 0.00000000
## AWAOREG AWAOREG 0.00000000
## AZEILPL AZEILPL 0.00000000
## APLEZIER APLEZIER 0.00000000
## AFIETS AFIETS 0.00000000
## AINBOED AINBOED 0.00000000
## ABYSTAND ABYSTAND 0.00000000MOSTYPE and PPERSAUT are the two most important variables fromt his model.
- Use the boosting model to predict the response on the test data. Predict that a person will make a purchase if the estimated probability of purchase is greater than 20 %. Form a confusion matrix. What fraction of the people predicted to make a purchase do in fact make one? How does this compare with the results obtained from applying KNN or logistic regression to this data set?
probs_test<-predict(boost.car, test_car, n.trees = 1000, type = "response")
pred_test<-ifelse(probs_test > 0.2, 1, 0)
table(test_car$Purchase, pred_test)## pred_test
## 0 1
## 0 4206 327
## 1 241 48Roughly 13%