I want to create a function that returns another function with a parameter from the former, e.g.
foo1 <- function(y) function(x) x+y
(Edit: I wrote foo1 <- function(y) function(x+y) originally, changed this. Also, see the answer of @Konrad Rudolph as to why this is a bad definition.)
but! I would like to be able to see what parameter value ‘y’ was chosen. The above prints like
foo1(1.2)
#> function(x) x + y
#> <environment: 0x000001dd24275a90>
Now, I could achieve this through
foo2 <- function(y){
eval(parse(text = sprintf("function(x) x + %s", y)))
}
Then printing ‘foo2’ shows what parameter value ‘y’ was used in its creation
foo2(1.2)
#> function(x) x + 1.2
#> <environment: 0x000001dd242604a0>
However, I suspect there is a more elegant solution.
I thought something like
foo3 <- function(y){
eval(substitute(function(x) x+dummy, env = list(dummy=y)))
}
could work, but no. That will print
foo3(1.2)
#> function(x) x+dummy
#> <environment: 0x000001dd23e137d0>
Does anyone have a more elegant solution than my ‘foo2’?
First of all, beware that your original function is broken (beyond the syntax error):
foo1 <- function(y) function(x) x+y
a <- 10
myfoo <- foo1(a)
a <- 20
myfoo(1)
# [1] 21
You probably wanted the last call to return 11, not 21.
To fix this, we need to force eager evaluation of y – otherwise it will be lazily evaluated when it’s used, and not before:
foo1 <- function (y) {
force(y)
function (x) x + y
}
Next, to answer your actual question: your foo3 almost works! R just tricks you, because it stores the “source code” of the function in an attribute (srcref) of the function. And when displaying the function definition via print, it prints that srcref (if available) instead of the actual body. We can delete the attribute to fix that:
foo <- function (y) {
fun <- eval(substitute(function(x) x + dummy, env = list(dummy = y)))
attr(fun, 'srcref') <- NULL
fun
}
foo(1.2)
# x + 1.2
# <environment: 0x108952138>
(Note that we don’t need to explicitly force evaluation of y in this definition of foo(), since passing it to list() (and then to substitute()) will cause it to be evaluated anyway.)
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