CODEWITHSUNDEEP
rclassificationprediction
Simon
Simon

Reputation: 141

Classification - Usage of factor levels

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!

Upvotes: 14

Views: 35711

Answers (5)

Seyma Kalay
Seyma Kalay

Reputation: 2861

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)

Upvotes: 0

Salma Elshahawy
Salma Elshahawy

Reputation: 1190

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

Upvotes: 1

Agile Bean
Agile Bean

Reputation: 7161

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))))

Upvotes: 0

Dbercules
Dbercules

Reputation: 719

How about this base function:

 make.names(churn) ~ .,

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

Source

Upvotes: 9

einar
einar

Reputation: 473

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.

Upvotes: 36

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