Classification - Usage of factor levels

user_8040310 0
Viewed 29069

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.

DATA OVERVIEW

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!

prediction classification r
5 Answers

How about this base function:

 make.names(churn) ~ .,

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

Source

user_55686750

Adding to the correct answer of @einar, here's the dplyr syntax of converting the factor levels:

training.dt  %>% 
  mutate(churn = factor(churn, 
          levels = make.names(levels(churn))))

I slightly prefer to change only the labels of the factor levels, as the levels change the underlying data, like this:

training.dt  %>% 
  mutate(churn = factor(churn, 
          labels = make.names(levels(churn))))
user_77690760

I had the same issue and fixed it by setting classProbs = FALSE in the trainControl() this solved the issue and kept the level 0 and 1

user_81110130

I got the same problem,

class(iris$Species); levels(iris$Species)
iris.lvls <- factor(iris, levels = c("1", "2", "3"))
class(iris.lvls); levels(iris.lvls)
user_92092030
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