If I use do.call() to run a model with parameters supplied as a list, the "call" returned with the model lists all entries within any data frames of the parameters. This prints an extremely long model output for large datasets.
library(randomForest)
data(iris)
do.call(randomForest, list(Species ~ ., data=iris))
#Call:
# randomForest(formula = Species ~ ., data = structure(list(Sepal.Length = c(5.1, 4.9, 4.7,
#4.6, 5, 5.4, 4.6, 5, 4.4, 4.9, 5.4, 4.8,...
Is it possible to prevent printing the data frame entries so the output matches a normal model call, for example, to randomForest?
randomForest(Species ~ ., data=iris)
#Call:
# randomForest(formula = Species ~ ., data = iris)
I could try to reconstruct and replace the "call" slot in the model object after assigning it, or set it to NULL, but this seems like a bad solution.
mod <- do.call(randomForest, list(Species ~ ., data=iris))
mod$call <- 'randomForest(formula = Species ~ ., data = iris)'
mod
#Call:
# "randomForest(formula = Species ~ ., data = iris)"
I'm sure there is a better and simpler solution, but I cannot find it. Thanks in advance for any help.
Use quote :
do.call("randomForest", list(Species ~ ., data = quote(iris)))
giving:
Call:
randomForest(formula = Species ~ ., data = iris)
Type of random forest: classification
Number of trees: 500
No. of variables tried at each split: 2
OOB estimate of error rate: 4.67%
Confusion matrix:
setosa versicolor virginica class.error
setosa 50 0 0 0.00
versicolor 0 47 3 0.06
virginica 0 4 46 0.08