R：当函数调用为3点时，无法使用Deparse和Replace完全捕获所有参数

Question

I am trying to capture all arguments of a function call so deparse and substitute are used. It has worked well until I want to extend the usage of the function using 3 dots as arguments; turns out only the first argument is captured. A simplified version of the function is shown below.

func_3dot <- function(...){
  tmp <- deparse(substitute(...))
  print(tmp)
}

For the example below [1] "mod_1" "mod_2" are expected. Is there any way I could still capture all input arguments without changing the above function much? Thanks!!

mod_1 <- lm(mpg ~ wt, data = mtcars)
mod_2 <- update(mod_1, . ~ . + hp)

> func_3dot(mod_1, mod_2)
[1] "mod_1"

Answer 1

Use list(...) instead of just ... , and deparse the entries separately. For example,

func_3dot <- function(...){
  tmp <- substitute(list(...))
  sapply(tmp, deparse)[-1]
}

The [-1] removes the call to list from the deparsed results.

Answer 2

You can use match.call :

func_3dot <- function(...){
  tmp <- vapply(as.list((match.call()[-1])), deparse, FUN.VALUE = character(1))
  print(tmp)
}

func_3dot(mod_1, mod_2)
#[1] "mod_1" "mod_2"

Answer 3

One option with purrr and rlang

library(rlang)
library(purrr)
func_3dot <- function(...){
    unname(map_chr(quos(...), quo_name))
  }

func_3dot(mod_1, mod_2)
#[1] "mod_1" "mod_2"

---

Or using base R

func_3dot <- function(...){
   sapply(as.list(substitute(...())), deparse)
 }

func_3dot(mod_1, mod_2)
#[1] "mod_1" "mod_2"