在用航速定义对象时，无法在函数中找到对象

Question

我使用speedglm将GLM与数据相匹配。当我直接调用函数时，代码按预期工作，但是当我创建一个符合模型的函数时，我会得到一个错误，即找不到参数。

变量(在下面的示例中是w)显然存在于函数的范围内，但似乎该变量只在速度函数中稍后才计算，其中w不再可用，所以我认为。这就是我开始质疑我目前对R.

在创建函数时，是否犯了错误，是否使用了一些奇怪的技巧来对打破常规(?)的变量(源代码这里)进行范围分析？逻辑还是我对R函数是如何工作的有错误的理解？

我正在尝试理解这种行为，并修复我的train_glm函数，使它能够与速度和重量一起工作。

米维

library(speedglm)

# works as expected
m1 <- speedglm(wt ~ cyl, data = mtcars, weights = mtcars$wt)

# define a small helper function that just forwards its arguments
train_glm <- function(f, d, w) {
  speedglm(formula = f, data = d, weights = w)
}
# does not work
m <- train_glm(wt ~ cyl, d = mtcars, w = mtcars$wt)
#> Error in eval(extras, data, env) : object 'w' not found

更奇怪的是，如果我更改代码，我发现

# removing the weights as a base case -> WORKS
train_glm3 <- function(f, d) {
  speedglm(formula = f, data = d)
}
m3 <- train_glm3(wt ~ cyl, d = mtcars)
# works


# hardcoding the weights inside the function -> BREAKS
train_glm4 <- function(f, d) {
  speedglm(formula = f, data = d, weights = d$wt)
}
m4 <- train_glm4(wt ~ cyl, d = mtcars)
# Error in eval(extras, data, env) : object 'd' not found


# creating a new dataset and hardcoding the weights inside the function
# but using the name of the dataset at the highest environment -> WORKS
train_glm5 <- function(f, d) {
  speedglm(formula = f, data = d, weights = mtcars2$wt)
}
mtcars2 <- mtcars
m5 <- train_glm5(wt ~ cyl, d = mtcars2)
# works

Answer 1

解决方案(感谢@Mike的提示)是通过使用这个答案提供的解决方案或使用类似于这样的do.call来计算代码：

library(speedglm)

train_glm_docall <- function(f, d, w) {
  do.call(
    speedglm,
    list(
      formula = f,
      data = d,
      weights = w
    )
  )
}

m2 <- train_glm_docall(f = wt ~ cyl, d = mtcars, w = mtcars$wt)
class(m2)
#> [1] "speedglm" "speedlm"