提取传递给函数参数的函数的参数值

Question

我可以访问传递给结构为命名列表的函数的参数值

compile_name <- function(first_name, last_name){
  paste(first_name, last_name)
}

create_project <- function(project_name,...) {
  print(as.list(match.call()))
}

create_project(project_name = compile_name(first_name = "first", last_name = "last"), more_stuff = "more")


[[1]]
create_project

$project_name
compile_name(first_name = "first", last_name = "last")

$more_stuff 
[1] more

有没有办法将传入参数（此处为 project_name）的函数的参数解析为列表。在上面的示例中，我最终在 project_name 参数的列表点中以字符串的形式调用函数。像这样：

 [[1]]
create_project

$project_name
  $first_name 
    [1] "first"
  $last_name
    [1] "last"

$more_stuff 
[1] "more"

另外，有没有办法检查一个参数是否包含一个函数来对函数内部的 this 做出反应？

Answer 1

compile_name <- function(first_name, last_name){
  paste(first_name, last_name)
}

convert_call_to_list <- function(x) {
  if (is.call(x)) as.list(x) else x
}

create_project <- function(project_name,...) {
  first_pass <- as.list(match.call())
  second_pass <- lapply(first_pass, convert_call_to_list)
  setNames(second_pass, names(first_pass))
}

create_project(project_name = compile_name(first_name = "first", last_name = "last"), more_stuff = "more")
#> [[1]]
#> create_project
#> 
#> $project_name
#> $project_name[[1]]
#> compile_name
#> 
#> $project_name$first_name
#> [1] "first"
#> 
#> $project_name$last_name
#> [1] "last"
#> 
#> 
#> $more_stuff
#> [1] "more"

^{由reprex package (v0.2.1) 于 2019 年 2 月 27 日创建}

Answer 2

您只是在寻找list(...)吗？

create_project = function (project_name, ...) {
    lst = list(...)
    # do something with `lst`.
}

Answer 3

这是一个函数，它检查它的任何参数是否属于“语言”类型（即表达式），如果是，则进一步检查是否有任何 那些 参数是函数：

my_function <- function(a, b) {
  args <- as.list(match.call())[-1]

  if (is.language(args$b)) {
    sub_args <- as.list(match.call(call = substitute(b)))
  }

  if (any(sapply(sub_args, is.symbol))) {
    print('found a sub-function!')
    print(sub_args)
  }
}

my_function(a = 10, b = my_function(1, mean))

[1] "found a sub-function!"

[[1]]
my_function

$a
[1] 1

$b
mean