From ggrothendieck at gmail.com  Sat Oct 28 19:39:43 2006
From: ggrothendieck at gmail.com (Gabor Grothendieck)
Date: Sat, 28 Oct 2006 13:39:43 -0400
Subject: [Rd] how to determine if a function's result is invisible
In-Reply-To: <45438F9B.7000905@stats.uwo.ca>
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	<45438F9B.7000905@stats.uwo.ca>
Message-ID: <971536df0610281039g702780bbge339ea5423e38ad1@mail.gmail.com>

Thanks!

On 10/28/06, Duncan Murdoch <murdoch at stats.uwo.ca> wrote:
> On 10/26/2006 5:26 AM, Gabor Grothendieck wrote:
> > On 10/26/06, Duncan Murdoch <murdoch at stats.uwo.ca> wrote:
> >> On 10/25/2006 11:02 PM, Gabor Grothendieck wrote:
> >>> On 10/25/06, Duncan Murdoch <murdoch at stats.uwo.ca> wrote:
> >>>> On 10/25/2006 8:14 PM, Gabor Grothendieck wrote:
> >>>>> Suppose we have a function such as the following
> >>>>>
> >>>>> F <- function(f, x) f(x)+1
> >>>>>
> >>>>> which runs function f and then transforms it.  I would like the
> >>>>> corresponding function which works the same except that
> >>>>> unlike F returns an invisible result if and only if f does.
> >>>>>
> >>>>> Is there some way of determining whether f returns
> >>>>> an invisible result or not?
> >>>>>
> >>>>> Thus we want this:
> >>>>>
> >>>>> f <- function(x) x
> >>>>> g <- function(x) invisible(x)
> >>>>>
> >>>>>> F(f, 1)
> >>>>> 2
> >>>>>
> >>>>>> F(g, 1)
> >>>> I don't think there's a way to do that.  Internally there's a global
> >>>> flag called R_Visible; if it is set to zero, the value won't print.  But
> >>>>  it gets reset to 1 very easily (e.g. by adding 1 to the result of an
> >>>> invisible function), and it's not available in the API for you to write
> >>>> C code to look at it.
> >>>>
> >>>> I think you'll just have to do require the user of your F to tell you
> >>>> that they want the result to be invisible.
> >>>>
> >>>> Duncan Murdoch
> >>>>
> >>> Perhaps R_Visible be made available at the R level in the future.
> >>> It would be helpful in situations where you are transforming a
> >>> function but want to keep aspects of it such as whether the
> >>> return result is invisible.
> >> Actually, there is a way, but it's undocumented (i.e., use at your own
> >> risk).  It's the eval.with.vis function.  This is an internal function
> >> that is used within source() and capture.output(); you'll have to guess
> >> from the usage there what the args are.   But here's an F that does
> >> something close to what you want:
> >>
> >>  > fix(F)
> >>  > f <- function() 1
> >>  > g <- function() invisible(1)
> >>  >
> >>  > F <- function (expr)
> >> + {
> >> +     expr <- substitute(expr)
> >> +     pf <- parent.frame()
> >> +     tmp <- .Internal(eval.with.vis(expr, pf,
> >> +         baseenv()))
> >> +     tmp
> >> + }
> >>  > F(f())
> >> $value
> >> [1] 1
> >>
> >> $visible
> >> [1] TRUE
> >>
> >>  > F(g())
> >> $value
> >> [1] 1
> >>
> >> $visible
> >> [1] FALSE
> >>
> >>
> >
> >
> > Perfect.  Thanks!!!!
>
> I've just added this function to R-devel (to become 2.5.0 next spring):
>
> withVisible <- function(x) {
>     x <- substitute(x)
>     v <- .Internal(eval.with.vis(x, parent.frame(), baseenv()))
>     v
> }
>
> Luke Tierney suggested simplifying the interface (no need to duplicate
> the 3 parameter eval interface, you can just wrap this in evalq() if you
> need that flexibility); the name "with.vis" was suggested, but it looks
> like an S3 method for the with() generic, so I renamed it.
>
> Duncan Murdoch
>




[Rd] how to determine if a function's result is invisible
Gabor Grothendieck