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rdecision-treer-caretc5.0

How to plot final c50 decision tree model (library C50) from caret::train object


I trained Decision Tree model using train function from caret library:

gr = expand.grid(trials = c(1, 10, 20), model = c("tree", "rules"), winnow = c(TRUE, FALSE))
dt = train(y ~ ., data = train, method = "C5.0", trControl = trainControl(method = 'cv', number = 10), tuneGrid = gr)

Now I would like to plot Decision Tree for the final model. But this command doesn't work:

plot(dt$finalModel)

Error in data.frame(eval(parse(text = paste(obj$call)[xspot])), eval(parse(text = paste(obj$call)[yspot])),  : 
  arguments imply differing number of rows: 4160, 208, 0

Someone already asked about it here: topic

Suggested solution was to use bestTune from the fitted train object to define the relevant c5.0 model manually. And then plot that c5.0 model normally:

c5model = C5.0(x = x, y = y, trials = dt$bestTune$trials, rules = dt$bestTune$model == "rules", control = C5.0Control(winnow = dt$bestTune$winnow))
plot(c5model)

I tried to do so. Yes, it makes possible to plot c5.0 model, BUT predicted probabilities from train object and manually recreated c5.0 model don't match.

So, my question is: is it possible to extract final c5.0 model from caret::train object and plot this Decision Tree?


Solution

  • The predicted probabilities should be the same, see below:

    library(MASS)
    library(caret)
    library(C50)
    library(partykit)
    
    traindata = Pima.tr
    testdata = Pima.te
    
    gr = expand.grid(trials = c(1, 2), 
    model = c("tree"), winnow = c(TRUE, FALSE))
    
    dt = train(x = traindata[,-ncol(testdata)], y = traindata[,ncol(testdata)], 
    method = "C5.0",trControl = trainControl(method = 'cv', number=3),tuneGrid=gr)
    
    c5model = C5.0.default(x = traindata[,-ncol(testdata)], y = traindata[,ncol(testdata)], 
    trials = dt$bestTune$trials, rules = dt$bestTune$model == "rules", 
    control = C5.0Control(winnow = dt$bestTune$winnow))
    
    all.equal(predict(c5model,testdata[,-ncol(testdata)],type="prob"),
    predict(dt$finalModel,testdata[,-ncol(testdata)],type="prob"))
    [1] TRUE
    

    So I would suggest you double check whether the predictions are the same.

    The error you see plotting the final model from caret comes from what is stored under $call which is weird, we can replace it with a call that would work for the plotting:

    plot(c5model)
    

    enter image description here

    finalMod = dt$finalModel
    finalMod$call = c5model$call
    plot(finalMod)
    

    enter image description here

    Or you can rewrite it like you would with the results from your training but you can see it gets a bit complication with the expression (or at least I am not very good with it):

    newcall = substitute(C5.0.default(x = X, y = Y, trials = ntrials, rules = RULES, control = C5.0Control(winnow = WINNOW)),
    list(
    X = quote(traindata[, -ncol(traindata)]),
    Y = quote(traindata[, ncol(traindata)]),
    RULES = dt$bestTune$model == "rules",
    ntrials = dt$bestTune$trials,
    WINNOW = dt$bestTune$winnow)
    )
    
    finalMod = dt$finalModel
    finalMod$call = newcall