how to determine if a function's result is invisible
On 10/29/2006 8:03 AM, Gabor Grothendieck wrote:
On 10/28/06, Duncan Murdoch <murdoch at stats.uwo.ca> wrote:
On 10/28/2006 6:03 PM, Philippe Grosjean wrote:
Duncan Murdoch wrote: [...]
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
Excellent, many thanks... but I am afraid I cannot use this function
because you force evaluation on parent.frame(), where I need to evaluate
it in .GlobalEnv (which is NOT equal to parent.frame() in my context).
Would it be possible to change it to:
withVisible <- function(x, env = parent.frame()) {
x <- substitute(x)
v <- .Internal(eval.with.vis(x, env, baseenv()))
v
}
...so that we got additional flexibility?
As I said, that's not needed. Use evalq(withVisible(x), envir=.GlobalEnv).
Even if its not strictly necessary in terms of minimality it still might be convenient and consistent with other eval-style functions which do tend to provide an env= or in the case of lm-style functions a data= argument. Also its very easy to do and the underlying internal function supports it.
I agree with Luke here. It's a bad design to make every function do everything. This function reveals the "R_visible" flag. It doesn't need to do anything else. Duncan Murdoch