"RSNNS" package doesn't allow me to use CARET's train function

Question

So far I was using the CARET package with RandomForest for my training.

I use CARET's train function with cross validation and all is working well.

That is until I wanted to try using neural network and uploaded the RSNNS package. Now, whenever I'm trying to use train (with my old rf algorithms) I get the following error:

Error in UseMethod("train") : no applicable method for 'train' applied to an object of class "c('tbl_df', 'tbl', 'data.frame')"

Is that bug? Why RSNNS causes that?

Answer 1

The problem is that RSNNS::train() is masking caret::train() because the RSNNS version was loaded after caret. Fix the problem by calling caret::train() with the packageName::function() syntax.

library(caret)
library(RSNNS)

library(mlbench)
data(Sonar)

inTraining <- createDataPartition(Sonar$Class, p = .75, list=FALSE)
training <- Sonar[inTraining,]
testing <- Sonar[-inTraining,]
fitControl <- trainControl(method = "cv",
                           number = 3)
# error because RSNNS::train does not work like caret::train()
system.time(fit <- train(Class ~ ., method="rf",data=Sonar,trControl = fitControl))
# correct by calling caret::train()
system.time(fit <- caret::train(Class ~ ., method="rf",data=Sonar,trControl = fitControl))
fit

...and the output:

> system.time(fit <- train(Cx=Sonar[,-61],y=Sonar[,61], method="rf",data=Sonar,trControl = fitControl))
Error in UseMethod("train") : 
  no applicable method for 'train' applied to an object of class "data.frame"
Timing stopped at: 0.033 0 0.034
> # correct by calling caret::train()
> system.time(fit <- caret::train(x=Sonar[,-61],y=Sonar[,61], method="rf",data=Sonar,trControl = fitControl))
   user  system elapsed 
  3.888   0.069   3.981 
> fit
Random Forest 

208 samples
 60 predictor
  2 classes: 'M', 'R' 

No pre-processing
Resampling: Cross-Validated (3 fold) 
Summary of sample sizes: 139, 138, 139 
Resampling results across tuning parameters:

  mtry  Accuracy   Kappa    
   2    0.8175983  0.6292393
  31    0.7645963  0.5249374
  60    0.7694272  0.5336925

Accuracy was used to select the optimal model using the largest value.
The final value used for the model was mtry = 2.
>