在 R 中使用 deparse 进行强制评估

Question

I have a function that produces a function:

fun1 <- function (x)
    return (function() x)

By the time a function is produced, the value of x is already basically a constant:

fun2 <- fun1 ("a")

However, when I have fun2 printed, it won't show the actual value, it only shows "x" even though it is by now obsolete:

> fun2
function() x
<environment: 0x55a9f94f9fc8>

How can I force the evaluation of x so that fun2 gets printed as

function() "a"

or however it was produced?

Answer 1

There's no need to resort to deparse here. You can write the evaluated value of x into the body of the newly created function:

fun1 <- function (x){
  f <- function() x
  body(f) <- x
  f
}

So that:

fun1("a")
#> function () 
#> "a"
#> <environment: 0x00000146d0449dd8>

and

f <- fun1("a")
f()
#> [1] "a"

EDIT

If you wanted f to take an argument you could do:

fun1 <- function (x){
  f <- function(y) y + x
  body(f)[[3]] <- x
  f
}

so that

fun1(3)
#> function (y) 
#> y + 3
#> <environment: 0x00000146d4b702a0>

Answer 2

This substitutes all instances of x in the body of fun . In the example fun1 has one argument but it would continue to work with multiple fun1 arguments used in fun substituting them all into fun . We have also set the environment of fun to the environment in the caller of fun1 allowing the caller to change that by specifying envir if needed.

fun1 <- function (x, envir = parent.frame()) {
    fun <- function() x
    body(fun) <- do.call("substitute", list(body(fun)))
    environment(fun) <- envir
    fun
}

fun2 <- fun1("a")
fun2
## function () 
## "a"

environment(fun2)
## <environment: R_GlobalEnv>