r - 当有下标时从公式中提取变量

Question

有几篇文章与获取 R 回归公式中的变量列表相关 - 基本答案是使用 all.vars。例如，

> all.vars(log(resp) ~ treat + factor(dose))
[1] "resp"  "treat" "dose"

这很好，因为它删除了所有函数和运算符(以及重复项，未显示)。但是，当公式包含 $ 运算符或下标(例如 in

)时，就会出现问题

> form = log(cows$weight) ~ factor(bulls[[3]]) * herd$breed
> all.vars(form)
[1] "cows"   "weight" "bulls"  "herd"   "breed"

这里将数据框名称cows、bulls、herd标识为变量，与实际变量的名称解耦或丢失。相反，我真正想要的是这样的结果:

> mystery.fcn(form)
[1] "cows$weight" "bulls[[3]]"  "herd$breed"

最优雅的方法是什么？我有一个提案将作为答案发布，但也许有人有更优雅的解决方案，并且会赢得更多选票!

Answer 1

一种有效的方法，虽然有点乏味，是将运算符 $ 等替换为变量名的合法字符，将字符串转回公式，应用 all。 vars，并取消破坏结果:

All.vars = function(expr, retain = c("\\$", "\\[\\[", "\\]\\]"), ...) {
    # replace operators with unlikely patterns _Av1_, _Av2_, ...
    repl = paste("_Av", seq_along(retain), "_", sep = "")
    for (i in seq_along(retain))
        expr = gsub(retain[i], repl[i], expr)
    # piece things back together in the right order, and call all.vars
    subs = switch(length(expr), 1, c(1,2), c(2,1,3))
    vars = all.vars(as.formula(paste(expr[subs], collapse = "")), ...)
    # reverse the mangling of names
    retain = gsub("\\\\", "", retain)  # un-escape the patterns
    for (i in seq_along(retain))
        vars = gsub(repl[i], retain[i], vars)
    vars
}

使用retain参数来指定我们希望保留而不是视为运算符的模式。默认值是 $、[[ 和 ]](全部都经过适当转义)以下是一些结果:

> form = log(cows$weight) ~ factor(bulls[[3]]) * herd$breed
> All.vars(form)
[1] "cows$weight" "bulls[[3]]"  "herd$breed" 

将 retain 更改为还包括 ( 和 ):

> All.vars(form, retain = c("\\$", "\\(", "\\)", "\\[\\[", "\\]\\]"))
[1] "log(cows$weight)"   "factor(bulls[[3]])" "herd$breed"

这些点被传递给 all.vars，它实际上与 all.names 相同，但具有不同的默认值。这样我们也可以获得retain中没有的函数和运算符:

> All.vars(form, functions = TRUE)
[1] "~"           "log"         "cows$weight" "*"          
[5] "factor"      "bulls[[3]]"  "herd$breed"