R caret package (rpart): constructing a classification tree

Question

I am struggling for several days to perform a classification tree using the caret package. The problem are my factor variables. I generate the tree, but when I try to use the best model to make predictions on the test sample, it fails, because the train function creates dummies for my factor variables and then the predict function cannot find these newly created dummies in the test set. How should I deal with this problem?

My code is as follows:

install.packages("caret", dependencies = c("Depends", "Suggests"))      
library(caret)                                      
db=data.frame(read.csv ("db.csv", head=TRUE, sep=";", na.strings ="?"))     
fix(db)
db$defaillance=factor(db$defaillance)
db$def=ifelse(db$defaillance==0,"No","Yes") 
db$def=factor(db$def)
db$defaillance=NULL
db$canal=factor(db$canal)
db$sect_isodev=factor(db$sect_isodev)
db$sect_risq=factor(db$sect_risq)       

#delete zero variance predictors                                
nzv <- nearZeroVar(db[,-78])
db_new <- db[,-nzv]

inTrain <- createDataPartition(y = db_new$def, p = .75, list = FALSE)                               
training <- db_new[inTrain,]
testing <- db_new[-inTrain,]
str(training)
str(testing)
dim(training)
dim(testing)

A sample o the str() function for training/testing is found below:

 $ FDR        : num  1305 211 162 131 143 ...
 $ FCYC       : num  0.269 0.18 0.154 0.119 0.139 ...
 $ BFDR       : num  803 164 108 72 76 63 100 152 188 80 ...
 $ TRES       : num  502 47 54 59 67 49 53 -7 -103 -109 ...
 $ sect_isodev: Factor w/ 9 levels "1","2","3","4",..: 4 3 3 3 3 3 3 3 3 3 ...
 $ sect_risq  : Factor w/ 6 levels "0","1","2","3",..: 6 6 6 6 6 6 6 6 6 6 ...
 $ def        : Factor w/ 2 levels "No","Yes": 1 1 1 1 1 1 1 1 1 1 ...
> dim(training)
[1] 14553    42
> dim(testing)
[1] 4850   42

Then my code goes like this:

fitControl <- trainControl(method = "repeatedcv",
                           number = 10,
                           repeats = 10,
                   classProbs = TRUE,
                   summaryFunction = twoClassSummary)

#CART1
set.seed(1234)
tree1 = train (def~.,
           training,
           method = "rpart",
           tuneLength=20,
           metric="ROC",
           trControl = fitControl)

A sample of

summary(tree1$finalModel)

is here

RNTB          38.397731
sect_isodev1   6.742289
sect_isodev3   4.005016
sect_isodev8   2.520850
sect_risq3     9.909127
sect_risq4     6.737908
sect_risq5     3.085714
SOLV          73.067539
TRES          47.906884
sect_isodev2   0.000000
sect_isodev4   0.000000
sect_isodev5   0.000000
sect_isodev6   0.000000
sect_isodev7   0.000000
sect_isodev9   0.000000
sect_risq0     0.000000
sect_risq1     0.000000
sect_risq2     0.000000

And here is the error:

model.tree1 <- predict(tree1$finalModel,testing) Error in eval(expr, envir, enclos) : object 'sect_isodev1' not found

I am curious yet about another thing. I have found in Max Kuhn's "Predictive Modelling with R" the following syntax:

predict(rpartTune$finalModel, newdata, type = "class")

where rpartTune$finalModel is a classification tree identical to mine (or mine identical to his). Now, R doesn't accept type="class". Only type="prob". I am troubled because of that.

Thank you in advance for your responses

Answer 1

Don't use predict.rpart with the train$finalModel unless you have a really good reason. The rpart object does;t know about anything that train did, including pre-process. It may not give you the correct answer. After all, you might be using train in order to avoid the minutia so let predict.train do the work.

Max

EDIT -

About the type = "class" and type = "prob" bit..

predict.rpart defaults to producing class probabilities. Although rpart is one of the earliest packages, that is atypical as most produce classes by default.

predict.train produces the classes by default and you have to use type = "prob" to get probabilities.