插入符号中的逻辑回归-是否没有截获?

问题描述

使用插入符号包在R中执行逻辑回归，并尝试强制执行零拦截，以使x = 0处的概率为0.5.在其他形式的回归中，似乎可以使用tunegrid关闭截距，但是它没有逻辑回归的功能.有什么想法吗?

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)

是的，我知道"在x = 0处的概率应为0.5，因此试图强加它.

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

推荐答案

有一个渐近有关如何为插入符号设置自定义模型的信息.因此，在下面的解决方案中，您还可以看到拦截仍然存在的原因:

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]]

如果您看合适的话，那一行会显示:

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)
...

因此，默认情况下存在拦截.您可以更改此行并在修改后的模型上运行插入符号:

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
                  }

我们拟合了模型:

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

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