构造内部函数调用glms的step/step step

Question

我遇到了一个与能见度/环境有关的问题。简而言之，构造内部函数的glms不能使用step/stepAIC进行简化：

foo = function(model) {
  m = glm(y~x, family=model$family, data = dframe)
  return(m)
}

y = rbinom(100, 1, 0.5)
x = y*rnorm(100) + rnorm(100)
dframe = data.frame(y, x)
m = glm(y~x, family='binomial', data = dframe)
m2 = foo(m)
library(MASS)
summary(m2)
print(m2$family)
m3 = stepAIC(m2, k = 2)

这将导致以下错误：

Error in glm(formula = y ~ 1, family = model$family, data = dframe) : 
  object 'model' not found

这一点，尽管m2看起来很适合它和家庭的定义。如果这个例子有点人为的话，很抱歉。

Answer 1

找到了解决方案--最初的glm需要用do.call构建。

foo = function(model) {
  form.1<-as.formula(y ~ x)
  dat = model$data
  fam = model$family
  m <- do.call("glm", list(form.1, data=dat, family=fam))
  ##m = glm(y~x, family='binomial', data = model$dframe)
  return(m)
}

y = rbinom(100, 1, 0.5)
x = y*rnorm(100) + rnorm(100)
dframe = data.frame(y, x)
m = glm(y~x, family='binomial', data = dframe)
m2 = foo(m)
library(MASS)
summary(m2)
print(m2$family)
m3 = stepAIC(m2, k = 2)