future_lapply(X, FUN = with, a) looks for 'a' in the wrong environment

[HenrikBengtsson]

> X <- list(list(a = 1), list(a = 2))
> y <- lapply(X, FUN = with, a)
> X <- list(list(a = 1), list(a = 2))
> y0 <- lapply(X, FUN = with, a)
> y1 <- future.apply::future_lapply(X, FUN = with, a)
Error: object 'a' not found

Enter a frame number, or 0 to exit   

1: future.apply::future_lapply(X, FUN = with, a)
2: future_xapply(FUN = FUN, nX = nX, chunk_args = X, args = list(...), get_chu
3: getGlobalsAndPackagesXApply(FUN = FUN, args = args, MoreArgs = MoreArgs, en
4: (function () 
{
    replicate(sink.number(), sink(NULL))
    if (interactive()

Selection: 0

[HenrikBengtsson]

A lower-level reproducible example:

X <- list(list(a = 1))
future.apply:::future_xapply(nX = length(X), chunk_args = X, FUN = with, a, future.stdout = FALSE, future.seed = FALSE, future.scheduling = 1.0, future.chunk.size = 1L, future.label = "", debug = FALSE)
## Error: object 'a' not found

Enter a frame number, or 0 to exit   

1: future.apply:::future_xapply(nX = length(X), chunk_args = X, FUN = with, a,
2: getGlobalsAndPackagesXApply(FUN = FUN, args = args, MoreArgs = MoreArgs, en
3: (function () 
{
    replicate(sink.number(), sink(NULL))
    if (interactive()

This happens because a is a symbol and future.apply:::getGlobalsAndPackagesXApply() calls:

future.apply/R/globals.R

Line 4 in 81bfef5

use_args <- !is.null(args)

as the first thing.

[HenrikBengtsson]

This has to do with the globals package and a bit with the future package. I doubt there's anything to be solved in this package.