如何正确地用点为 lm() 编写包装器？ （错误 ..3 在不正确的上下文中使用）

Question

我正在尝试为lm() 构建一个带有点... 的包装器，但取得了部分成功。它仅适用于前两个 formula= 和 data= 参数。但是对于随后的每个参数，它似乎都不起作用并引发错误。

我对函数中lm() 的结果所做的不是问题的原因。它似乎只与使用点有关。所以我能够在这个 MWE 中找出问题：

lm_fun <- function(...) {
  lm(...)
  # <some other stuff>
}

lm_fun(formula=mpg ~ hp, data=mtcars)
## works fine

mtcars$w <- runif(nrow(mtcars))  ## adding weights
lm_fun(formula=mpg ~ hp, data=mtcars, weights=w)
# Error in eval(substitute(subset), data, env) : 
#   ..3 used in an incorrect context, no ... to look in

lm_fun(formula=mpg ~ hp, data=mtcars, subset=am == 1)
# Error in eval(substitute(subset), data, env) : 
#   ..3 used in an incorrect context, no ... to look in

lm_fun(formula=mpg ~ hp, data=mtcars, subset=am == 1, weights=w)
# Error in eval(extras, data, env) : 
#   ..4 used in an incorrect context, no ... to look in

这似乎是与that question 不同的问题，因为我在函数调用中提供了完整的参数集，并且没有在函数中添加任何内容。

我错过了什么吗？这种行为的原因是什么，我怎样才能让它正常工作？

编辑

@user63230 提供了一个nice workaround below。但是，它不适用于 lapply，这对我的需求很重要。

lm_fun2 <- function(...) eval(substitute(lm(...)))

fo <- c(mpg ~ hp, mpg ~ am)

lapply(fo, lm_fun, data=mtcars)
# works

lapply(fo, lm_fun2, data=mtcars)
# Error in stats::model.frame(formula = X[[i]], data = mtcars, drop.unused.levels = TRUE) : 
#   object 'X' not found 

Answer 1

我认为问题在于搜索w等的环境，因此如果将其包装在eval(substitute(中，则以下示例有效：

lm_fun <- function(...) {
  result <- eval(substitute(lm(...)))
  result
}
lm_fun(formula = mpg ~ hp, data = mtcars)
# Call:
# lm(formula = mpg ~ hp, data = mtcars)

# Coefficients:
# (Intercept)           hp  
#    30.09886     -0.06823 

lm_fun(formula = mpg ~ hp, data = mtcars, weights = w)
# Call:
# lm(formula = mpg ~ hp, data = mtcars, weights = w)

# Coefficients:
# (Intercept)           hp  
#    31.41313     -0.07391  

lm_fun(formula = mpg ~ hp, data = mtcars, weights = w, na.action = na.omit)
# Call:
# lm(formula = mpg ~ hp, data = mtcars, weights = w, na.action = na.omit)

# Coefficients:
# (Intercept)           hp  
#    31.41313     -0.07391 

lm_fun(formula = mpg ~ hp, data = mtcars, weights = w, na.action = na.omit, 
  singular.ok = F)
# Call:
# lm(formula = mpg ~ hp, data = mtcars, weights = w, na.action = na.omit, 
#     singular.ok = F)

# Coefficients:
# (Intercept)           hp  
#    31.41313     -0.07391 

lm_fun(formula=mpg ~ hp, data=mtcars, subset=am == 1, weights=w)
# Call:
# lm(formula = mpg ~ hp, data = mtcars, subset = am == 1, weights = w)

# Coefficients:
# (Intercept)           hp  
#    32.96650     -0.07072

也见here。该公式自动与df 关联，但weights 可能在df 之外。