Classification - Usage of factor levels

Question

I am currently working on a predictive model for a churn problem.
Whenever I try to run the following model, I get this error: At least one of the class levels is not a valid R variable name. This will cause errors when class probabilities are generated because the variables names will be converted to X0, X1. Please use factor levels that can be used as valid R variable names.

fivestats <- function(...) c( twoClassSummary(...), defaultSummary(...))
fitControl.default    <- trainControl( 
    method  = "repeatedcv"
  , number  = 10
  , repeats = 1 
  , verboseIter = TRUE
  , summaryFunction  = fivestats
  , classProbs = TRUE
  , allowParallel = TRUE)
set.seed(1984)

rpartGrid             <-  expand.grid(cp = seq(from = 0, to = 0.1, by = 0.001))
rparttree.fit.roc <- train( 
    churn ~ .
  , data      = training.dt  
  , method    = "rpart"
  , trControl = fitControl.default
  , tuneGrid  = rpartGrid
  , metric = 'ROC'
  , maximize = TRUE
)

In the attached picture you see my data, I already transformed some data from chr to factor variable.

I do not get what my problem is, if I would transform the entire data into factors, then for instance the variable total_airtime_out will probably have around 9000 factors.

Thanks for any kind of help!

Answer 1

It's not exactly possible for me to reproduce your error, but my educated guess is that the error message tells you everything you need to know:

At least one of the class levels is not a valid R variable name. This will cause errors when class probabilities are generated because the variables names will be converted to X0, X1. Please use factor levels that can be used as valid R variable names.

Emphasis mine. Looking at your response variable, its levels are "0" and "1", these aren't valid variable names in R (you can't do 0 <- "my value"). Presumably this problem will go away if you rename the levels of the response variable with something like

levels(training.dt$churn) <- c("first_class", "second_class")

as per this Q.

Answer 2

How about this base function:

 make.names(churn) ~ .,

to "make syntactically valid names out of character vectors"?

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