As a symptom of R being a functional language, it is possible to specify many control structures using pure functional notation. For example, an if statement:
作为R是函数式语言的症状，可以使用纯函数符号指定许多控制结构。例如，if语句：
> as.list((substitute(if(a == 1) 1 else 2)))
[[1]]
`if`
[[2]]
a == 1
[[3]]
[1] 1
[[4]]
[1] 2
> as.list(substitute(`if`(a == 1, 1, 2)))
[[1]]
`if`
[[2]]
a == 1
[[3]]
[1] 1
[[4]]
[1] 2
Out of curiosity I attempted to do the same for a function definition. Functions are typically constructed using the function(args) body syntax but there also exists a function named function in R. The problem I ran into is that the un-evaluated definition of a function will contain a pairlist:
出于好奇，我试图对函数定义做同样的事情。函数通常使用函数（args）body语法构造，但在R中也存在一个名为function的函数。我遇到的问题是函数的未评估定义将包含一个pairlist：
> substitute(function(x = 1, a) {x + a})[[2]]
$x
[1] 1
$a
[empty symbol]
The second argument is a pairlist between the names of the arguments and there default values, which are possibly empty symbols. As far as I know, it is impossible to pass a list or a pairlist as part of expression using only manual calls. Below are my attempts:
第二个参数是参数名称和默认值之间的一个pairlist，它们可能是空符号。据我所知，仅使用手动调用就不可能将列表或pairlist作为表达式的一部分传递。以下是我的尝试：
> substitute(`function`(x, {x + 1}))
Error: badly formed function expression
> substitute(`function`((x), {x + 1}))
function(`(`, x) {
x + 1
}
If I pass a single symbol as the second argument to `function`, an error is thrown (and this wouldn't work for a function with multiple arguments). When passing a call as the second argument, the call appears to be coerced to a list upon being parsed.
如果我将单个符号作为第二个参数传递给`function`，则会抛出一个错误（这对于具有多个参数的函数不起作用）。当将调用作为第二个参数传递时，调用似乎在被解析时被强制转换为列表。
This can be abused by using the first argument as the name of the call:
使用第一个参数作为调用名称可能会滥用此功能：
> substitute(`function`(a(x), {x + 1}))
function(a, x) {
x + 1
}
However, the call is not actually converted to a pairlist and only appears to be so. When evaluating this expression, an error is thrown. It would be ideal to somehow insert a list / pairlist into the result of a call to substitute. Does anyone know how to do so?
但是，呼叫实际上并没有转换成一个轮滑，只是看起来如此。在计算此表达式时，会引发错误。以某种方式将列表/ pairlist插入到替换调用的结果中是理想的。有谁知道怎么做？
1 个解决方案
#1
1
This will not be prompt answer but I just tried to focus on your function call.
这不会是快速回答，但我只是试着专注于你的函数调用。
Just try this code:
试试这段代码：
as.list((substitute(function(x = 1, a) {x + a})))
[[1]]
`function`
[[2]]
[[2]]$x
[1] 1
[[2]]$a
[[3]]
{
x + a
}
[[4]]
function(x = 1, a) {x + a}
#1
1
This will not be prompt answer but I just tried to focus on your function call.
这不会是快速回答，但我只是试着专注于你的函数调用。
Just try this code:
试试这段代码：
as.list((substitute(function(x = 1, a) {x + a})))
[[1]]
`function`
[[2]]
[[2]]$x
[1] 1
[[2]]$a
[[3]]
{
x + a
}
[[4]]
function(x = 1, a) {x + a}