Logistic Regression in Caret - No Intercept?

问题

Performing logistic regression in R using the caret package and trying to force a zero intercept such that probability at x=0 is .5. In other forms of regression, it seems like you can turn the intercept off using tunegrid, but that has no functionality for logistic regression. Any ideas?

model <- train(y ~ 0+ x, data = data, method = "glm", family = binomial(link="probit"),
               trControl = train.control)

And yes, I "know" that the probability at x=0 should be .5, and thus trying to force it.

回答1:

There's a vignette on how to set up a custom model for caret. So in the solution below, you can also see why the intercept persist:

library(caret)
glm_wo_intercept = getModelInfo("glm",regex=FALSE)[[1]]

if you look at the fit, there's a line that does:

glm_wo_intercept$fit

....
modelArgs <- c(list(formula = as.formula(".outcome ~ ."), data = dat), theDots)
...

So the intercept is there by default. You can change this line and run caret on this modified model:

glm_wo_intercept$fit = function(x, y, wts, param, lev, last, classProbs, ...) {
  dat <- if(is.data.frame(x)) x else as.data.frame(x)
  dat$.outcome <- y
  if(length(levels(y)) > 2) stop("glm models can only use 2-class outcomes")

  theDots <- list(...)
  if(!any(names(theDots) == "family"))
        {
    theDots$family <- if(is.factor(y)) binomial() else gaussian()
                    }
  if(!is.null(wts)) theDots$weights <- wts
  # change the model here
  modelArgs <- c(list(formula = as.formula(".outcome ~ 0+."), data = dat), theDots)

  out <- do.call("glm", modelArgs)
  out$call <- NULL
  out
                  }

We fit the model:

data = data.frame(y=factor(runif(100)>0.5),x=rnorm(100))
model <- train(y ~ 0+ x, data = data, method = glm_wo_intercept, 
family = binomial(),trControl = trainControl(method = "cv",number=3))

predict(model,data.frame(x=0),type="prob")
  FALSE TRUE
1   0.5  0.5

来源：https://stackoverflow.com/questions/59539412/logistic-regression-in-caret-no-intercept