I’m working on a ml project. I trained a xgboost model through the caret library, and I want to calculate the AIC for this model. Since there’s no “straightforward” way to find it ( like AIC(xgb)), I’d like to know how I could do it.

I trained the model with these specifics:

ctrl_xgb <- trainControl(method="cv", number=10, search="grid", summaryFunction = twoClassSummary, classProbs = TRUE)

param_grid_xgb <- expand.grid(nrounds=500, max_depth = c(3, 6, 9),eta = c(0.01, 0.1, 0.3),
    gamma = c(0, 0.2, 0.4), subsample = c(0.8, 0.9, 1), colsample_bytree = c(0.8, 0.9,    1),min_child_weight=c(1, 5, 10))
         
xgb<-train(Target~., data=under, method="xgbTree", metric="ROC", tuneGrid=param_grid_xgb, 
           trControl=ctrl_xgb,verbosity=0)

and the best paramters are:

nrounds max_depth  eta   gamma colsample_bytree min_child_weight subsample  
  500         9    0.01   0.4              0.8                1       0.9

I know there’s this formula for AIC:

AIC=2k−2ln(L)

with k being the number of parameters of the model.
For the log-likelihood I used this code:

ll<- sum(log(ifelse(aicdf$Target == 1, aicdf$xgb_true, 1 - aicdf$xgb_true)))

with aicdf$xgb_true being the prediction of the model on a validation dataset. Correct me if I’m wrong.

If this is a correct procedure to calculate AIC, in my case, what would the number of parameters be equal to?

Thank you!