Pass missing argument to function using do.call

Question

How do I pass a missing argument to a function using do.call? The following does not work:

x <- array(1:9, c(3, 3))
do.call(`[`, list(x, , 1))

I expect this to do the same calculation as

x[,1]

Answer 1

you can use quote(expr=)

x <- array(1:9, c(3, 3))

do.call(`[`, list(x, quote(expr=), 1))
#> [1] 1 2 3

identical(x[,1], do.call(`[`, list(x, quote(expr=), 1)))
#> [1] TRUE

^{Created on 2022-04-20 by the reprex package (v2.0.1)}

do.call(`[`, alist(x,, 1)) also works.

So does do.call(`[`, list(x, TRUE, 1)) for this specific case, since subsetting rows with TRUE keeps all of them.

Answer 2

Update: Thanks to @moodymudskipper: ("is_missing() is always TRUE so you're just subsetting rows and keeping everything. We could do do.call([, list(x, TRUE, 1)). This solves this specific issue (perhaps better in fact), but not the general one of dealing with missing arguments.")

I meant rlangs missing_arg():

do.call(`[`, list(x, rlang::missing_arg() , 1))

First answer: not good -> see comments moodymudskipper:

We could use rlangs is_missing(): See here https://rlang.r-lib.org/reference/missing_arg.html

"rlang::is_missing() simplifies these rules by never treating default arguments as missing, even in internal contexts:"

do.call(`[`, list(x, rlang::is_missing() , 1))