library(AppliedPredictiveModeling)
data(segmentationOriginal)
library(caret)
library(randomForest)
Load the vowel.train and vowel.test data sets:
library(ElemStatLearn)
data(vowel.train)
data(vowel.test)
Set the variable y to be a factor variable in both the training and test set. Then set the seed to 33833. Fit (1) a random forest predictor relating the factor variable y to the remaining variables and (2) a boosted predictor using the “gbm” method. Fit these both with the train() command in the caret package.
What are the accuracies for the two approaches on the test data set? What is the accuracy among the test set samples where the two methods agree?
set.seed(33833)
vowel.train$y <- factor(vowel.train$y)
vowel.test$y <- factor(vowel.test$y)
fit.rf <- randomForest(y~., data=vowel.train)
fit.gbm <- train(y~., method="gbm", data=vowel.train, verbose = F)
predict.rf <- predict(fit.rf, vowel.test)
predict.gbm <- predict(fit.gbm, vowel.test)
confusionMatrix(predict.rf, vowel.test$y)$overall[1]
## Accuracy
## 0.5800866
confusionMatrix(predict.gbm, vowel.test$y)$overall[1]
## Accuracy
## 0.517316
confusionMatrix(predict.gbm, predict.rf)$overall[1]
## Accuracy
## 0.6904762
set.seed(3433)
library(AppliedPredictiveModeling)
data(AlzheimerDisease)
adData = data.frame(diagnosis,predictors)
inTrain = createDataPartition(adData$diagnosis, p = 3/4)[[1]]
training = adData[ inTrain,]
testing = adData[-inTrain,]
Set the seed to 62433 and predict diagnosis with all the other variables using a random forest (“rf”), boosted trees (“gbm”) and linear discriminant analysis (“lda”) model. Stack the predictions together using random forests (“rf”). What is the resulting accuracy on the test set? Is it better or worse than each of the individual predictions?
set.seed(62433)
ad.fit.rf <- train(diagnosis~., method="rf", data=training)
ad.fit.gbm <- train(diagnosis~., method="gbm", data=training, verbose = F)
ad.fit.lda <- train(diagnosis~., method="lda", data=training)
## Warning in lda.default(x, grouping, ...): variables are collinear
## Warning in lda.default(x, grouping, ...): variables are collinear
## Warning in lda.default(x, grouping, ...): variables are collinear
## Warning in lda.default(x, grouping, ...): variables are collinear
## Warning in lda.default(x, grouping, ...): variables are collinear
## Warning in lda.default(x, grouping, ...): variables are collinear
## Warning in lda.default(x, grouping, ...): variables are collinear
## Warning in lda.default(x, grouping, ...): variables are collinear
ad.pdt.rf <- predict(ad.fit.rf, testing)
ad.pdt.gbm <- predict(ad.fit.gbm, testing)
ad.pdt.lda <- predict(ad.fit.lda, testing)
Stacked <- data.frame(rf = ad.pdt.rf, gbm = ad.pdt.gbm, lda = ad.pdt.lda, diagnosis = testing$diagnosis)
stack.fit <- train(diagnosis ~., method="rf", data=Stacked)
## note: only 2 unique complexity parameters in default grid. Truncating the grid to 2 .
stack.pdt <- predict(stack.fit, Stacked)
confusionMatrix(ad.pdt.rf, testing$diagnosis)$overall[1]
## Accuracy
## 0.7682927
confusionMatrix(ad.pdt.gbm, testing$diagnosis)$overall[1]
## Accuracy
## 0.8170732
confusionMatrix(ad.pdt.lda, testing$diagnosis)$overall[1]
## Accuracy
## 0.7682927
confusionMatrix(stack.pdt, testing$diagnosis)$overall[1]
## Accuracy
## 0.8170732
set.seed(3523)
library(AppliedPredictiveModeling)
data(concrete)
inTrain = createDataPartition(concrete$CompressiveStrength, p = 3/4)[[1]]
training = concrete[ inTrain,]
testing = concrete[-inTrain,]
Set the seed to 233 and fit a lasso model to predict Compressive Strength. Which variable is the last coefficient to be set to zero as the penalty increases? (Hint: it may be useful to look up ?plot.enet).
set.seed(233)
lassoFit <- train(CompressiveStrength ~., data = training,
method = "lasso")
lassoFit$finalModel
## $call
## elasticnet::enet(x = as.matrix(x), y = y, lambda = 0)
##
## $actions
## $actions[[1]]
## Cement
## 1
##
## $actions[[2]]
## Superplasticizer
## 5
##
## $actions[[3]]
## Age
## 8
##
## $actions[[4]]
## BlastFurnaceSlag
## 2
##
## $actions[[5]]
## Water
## 4
##
## $actions[[6]]
## FineAggregate
## 7
##
## $actions[[7]]
## FlyAsh
## 3
##
## $actions[[8]]
## FineAggregate
## -7
##
## $actions[[9]]
## FineAggregate
## 7
##
## $actions[[10]]
## CoarseAggregate
## 6
##
## $actions[[11]]
## [1] 11
##
##
## $allset
## [1] 1 2 3 4 5 6 7 8
##
## $beta.pure
## Cement BlastFurnaceSlag FlyAsh Water Superplasticizer
## 0 0.00000000 0.00000000 0.00000000 0.0000000 0.0000000
## 1 0.01802441 0.00000000 0.00000000 0.0000000 0.0000000
## 2 0.02729231 0.00000000 0.00000000 0.0000000 0.1559023
## 3 0.03847584 0.00000000 0.00000000 0.0000000 0.4343197
## 4 0.04599070 0.01006844 0.00000000 0.0000000 0.5510774
## 5 0.06499050 0.03893552 0.00000000 -0.1174308 0.5737881
## 6 0.06678431 0.04187677 0.00000000 -0.1401379 0.5739760
## 7 0.08502706 0.06390658 0.03312720 -0.1768114 0.4070855
## 8 0.10150634 0.08388323 0.06377012 -0.2190396 0.2504283
## 9 0.10301008 0.08567182 0.06624407 -0.2188092 0.2387808
## 10 0.11351343 0.09842089 0.08097549 -0.1794448 0.2567532
## CoarseAggregate FineAggregate Age
## 0 0.00000000 0.000000000 0.00000000
## 1 0.00000000 0.000000000 0.00000000
## 2 0.00000000 0.000000000 0.00000000
## 3 0.00000000 0.000000000 0.02765634
## 4 0.00000000 0.000000000 0.04042040
## 5 0.00000000 0.000000000 0.07968720
## 6 0.00000000 -0.003483276 0.08499933
## 7 0.00000000 0.000000000 0.09810130
## 8 0.00000000 0.000000000 0.11088807
## 9 0.00000000 0.001412076 0.11162794
## 10 0.01130084 0.014480343 0.11346049
## attr(,"scaled:scale")
## [1] 2831.2770 2416.1259 1767.8101 595.3663 168.3104 2133.9390 2234.6943
## [8] 1756.7394
##
## $vn
## [1] "Cement" "BlastFurnaceSlag" "FlyAsh"
## [4] "Water" "Superplasticizer" "CoarseAggregate"
## [7] "FineAggregate" "Age"
##
## $mu
## [1] 35.83765
##
## $normx
## [1] 2831.2770 2416.1259 1767.8101 595.3663 168.3104 2133.9390 2234.6943
## [8] 1756.7394
##
## $meanx
## Cement BlastFurnaceSlag FlyAsh Water
## 281.108398 75.020284 53.444057 181.400904
## Superplasticizer CoarseAggregate FineAggregate Age
## 6.321318 972.082300 774.158527 44.727390
##
## $lambda
## [1] 0
##
## $L1norm
## [1] 0.00000 51.03211 103.51208 230.62128 318.29920 584.55773
## [7] 627.40970 799.82678 970.15842 985.46860 1108.17911
##
## $penalty
## [1] 4.532905e+02 3.512262e+02 2.914402e+02 2.078625e+02 1.686523e+02
## [6] 7.241288e+01 5.892304e+01 3.131011e+01 3.697331e+00 2.375469e+00
## [11] 1.313172e-12
##
## $df
## [1] 1 2 3 4 5 6 7 7 7 8 9
##
## $Cp
## 0 1 2 3 4 5
## 1201.603156 1014.917230 861.914214 572.237452 422.520823 129.536499
## 6 7 8 9 10
## 105.671332 34.171119 6.766949 8.339652 9.000000
##
## $sigma2
## 10
## 108.795
##
## $xNames
## [1] "Cement" "BlastFurnaceSlag" "FlyAsh"
## [4] "Water" "Superplasticizer" "CoarseAggregate"
## [7] "FineAggregate" "Age"
##
## $problemType
## [1] "Regression"
##
## $tuneValue
## fraction
## 3 0.9
##
## $obsLevels
## [1] NA
##
## $param
## list()
##
## attr(,"class")
## [1] "enet"
suppressMessages(library(lubridate)) # For year() function below
## Warning: package 'lubridate' was built under R version 3.4.3
dat = read.csv("gaData.csv")
training = dat[year(dat$date) < 2012,]
testing = dat[(year(dat$date)) > 2011,]
tstrain = ts(training$visitsTumblr)
Fit a model using the bats() function in the forecast package to the training time series. Then forecast this model for the remaining time points. For how many of the testing points is the true value within the 95% prediction interval bounds?
suppressMessages(library(forecast))
## Warning: package 'forecast' was built under R version 3.4.3
visit.fit <- bats(training[,3])
visit.pdt <- forecast(visit.fit, h=length(testing[,3]))
inbounds <- sum((testing$visitsTumblr <= visit.pdt$upper[,2]) & (testing$visitsTumblr >= visit.pdt$lower[,2]))
inbounds/length(testing[,3])
## [1] 0.9617021
set.seed(3523)
library(AppliedPredictiveModeling)
data(concrete)
inTrain = createDataPartition(concrete$CompressiveStrength, p = 3/4)[[1]]
training = concrete[ inTrain,]
testing = concrete[-inTrain,]
Set the seed to 325 and fit a support vector machine using the e1071 package to predict Compressive Strength using the default settings. Predict on the testing set. What is the RMSE?
set.seed(325)
suppressMessages(library(e1071))
## Warning: package 'e1071' was built under R version 3.4.3
fit.svm <- svm(CompressiveStrength~., data=training)
pdt.svm <- predict(fit.svm, testing)
RMSE(testing$CompressiveStrength, pdt.svm)
## [1] 6.715009