[R] matching last argument in function

[Alistair Gee]

Hi Gabor,

That almost works ... but it fails when I nest with.options() within
another function:

with.options <- function(...) {
   L <- as.list(match.call())[-1]
   len <- length(L)
   old.options <- do.call(options, L[-len])
   on.exit(options(old.options))
   invisible(eval.parent(L[[len]]))
}

with.width <- function(w)
  with.options(width=w, print(1:25))

m.with.width(10)

> Error in function (...)  : object "w" not found

Enter a frame number, or 0 to exit

1: with.width(10)
2: with.options(width = w, print(1:25))
3: do.call(options, L[-len])
4: function (...)

I tried, unsuccessfully, to fix the problem by using eval.parent()
around do.call() and around L[-len].

This problem does not occur if I use my original implementation:

with.options <- function(..., expr) {
  options0 <- options(...)
  tryCatch(expr, finally=options(options0))
}

I realize that I can use my original implementation in this particular
case, but I'd like to have a single implementation that works
correctly, while not requiring explicitly naming the expr argument.

TIA

On Feb 12, 2008 12:43 PM, Gabor Grothendieck <ggrothendieck at gmail.com> wrote:
> Try this:
>
> with.options <- function(...) {
>     L <- as.list(match.call())[-1]
>     len <- length(L)
>     old.options <- do.call(options, L[-len])
>     on.exit(options(old.options))
>     invisible(eval.parent(L[[len]]))
> }
>
> > with.options(width = 40, print(1:25))
>  [1]  1  2  3  4  5  6  7  8  9 10 11 12
> [13] 13 14 15 16 17 18 19 20 21 22 23 24
> [25] 25
> > with.options(width = 80, print(1:25))
>  [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
>
>
>
>
>
> On Feb 12, 2008 12:45 PM, Alistair Gee <alistair.gee at gmail.com> wrote:
> > I often want to temporarily modify the options() options, e.g.
> >
> > a <- seq(10000001, 10000001 + 10) # some wide object
> >
> > with.options <- function(..., expr) {
> >  options0 <- options(...)
> >  tryCatch(expr, finally=options(options0))
> > }
> >
> > Then I can use:
> >
> > with.options(width=160, expr = print(a))
> >
> > But I'd like to avoid explicitly naming the expr argument, as in:
> >
> > with.options(width=160, print(a))
> >
> > How can I do this with R's argument matching? (I prefer the expr as
> > the last argument since it could be a long code block. Also, I'd like
> > with.options to take multiple options.)
> >
> > TIA
> >
>
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> >
>