In R, I specify a model with no intercept as follows:
data(iris)
lmFit <- lm(Sepal.Length ~ 0 + Petal.Length + Petal.Width, data=iris)
> round(coef(lmFit),2)
Petal.Length Petal.Width
2.86 -4.48
However, if I fit the same model with caret, the resulting model includes an intercept:
library(caret)
caret_lmFit <- train(Sepal.Length~0+Petal.Length+Petal.Width, data=iris, "lm")
> round(coef(caret_lmFit$finalModel),2)
(Intercept) Petal.Length Petal.Width
4.19 0.54 -0.32
How do I tell caret::train to exclude the intercept term?
As discussed in a linked SO question https://stackoverflow.com/a/41731117/7613376, this works in caret v6.0.76 (And the trace answer above no longer seems to work with code refactoring in caret):
caret_lmFit <- train(Sepal.Length~0+Petal.Length+Petal.Width, data=iris, "lm",
tuneGrid = expand.grid(intercept = FALSE))
> caret_lmFit$finalModel
Call:
lm(formula = .outcome ~ 0 + ., data = dat)
Coefficients:
Petal.Length Petal.Width
2.856 -4.479
@rcs already told you which line in which function you need to change.
Just use trace to modify that function:
trace(caret::createModel,
quote(modFormula <- as.formula(".outcome ~ .-1")), at=5, print=FALSE)
caret_lmFit <- train(Sepal.Length~0+Petal.Length+Petal.Width, data=iris, "lm")
round(coef(caret_lmFit$finalModel),2)
#Petal.Length Petal.Width
# 2.86 -4.48
untrace(caret::createModel)
However, I don't use caret. There might be unforeseen consequences. It's also often not a good idea to exclude the intercept from the model.
