[R] Function that works within a package and not when copied in global environment. Why?
Marc Girondot
marc_grt at yahoo.fr
Thu Feb 2 17:34:08 CET 2017
Thanks Bert for the explanation about identical.
For the vectorize.args, note that vectorize.args is not a function but
an variable that is unknown in the namespace nlWaldTest.
> nlWaldTest::vectorize.args
Erreur : 'vectorize.args' n'est pas un objet exporté depuis
'namespace:nlWaldTest'
Furthermore, if the function is created from a copy of the original one:
smartsub <- getFromNamespace(".smartsub", ns="nlWaldTest")
or if it is created manually: by copy-paste of the code:
smartsub2 <- function (pat, repl, x)
{
args <- lapply(as.list(match.call())[-1L], eval, parent.frame())
names <- if (is.null(names(args)))
character(length(args))
else names(args)
dovec <- names %in% vectorize.args
do.call("mapply", c(FUN = FUN, args[dovec], MoreArgs =
list(args[!dovec]),
SIMPLIFY = SIMPLIFY, USE.NAMES = USE.NAMES))
}
Both are defined in the global env, but the first one works and not the
second one.
I am surprised and don't understand how it is possible.
Sincerely
Marc Girondot
> 2. You need to review how namespaces work. From the "Writing R
> extensions" manual:
>
> "The namespace controls the search strategy for variables used by
> **functions in the package**. If not found locally, R searches the
> package namespace first, then the imports, then the base namespace and
> then the normal search path."
>
> So if vectorize.args() is among the package functions, it will be
> found by package functions but not by those you write unless
> specifically qualified by :: or ::: depending on whether it is
> exported.
>
> Cheers,
> Bert
>
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Thu, Feb 2, 2017 at 6:30 AM, Marc Girondot via R-help
> <r-help at r-project.org> wrote:
>> Dear experts,
>>
>> In the package nlWaldTest, there is an hidden function : .smartsub
>>
>> I can use it, for example:
>>
>>> getFromNamespace(".smartsub", ns="nlWaldTest")(pat="x", repl="b" ,
>> x="essai(b[1], b[2], x[1])")
>> [1] "essai(b[1], b[2], b[1])"
>>
>> Now I try to create this function in my global environment:
>> smartsub <- getFromNamespace(".smartsub", ns="nlWaldTest")
>>
>> It works also:
>>> smartsub(pat="x", repl="b" , x="essai(b[1], b[2], x[1])")
>> [1] "essai(b[1], b[2], b[1])"
>>
>> But if I create the function manually:
>>> smartsub2 <- function (pat, repl, x)
>> {
>> args <- lapply(as.list(match.call())[-1L], eval, parent.frame())
>> names <- if (is.null(names(args)))
>> character(length(args))
>> else names(args)
>> dovec <- names %in% vectorize.args
>> do.call("mapply", c(FUN = FUN, args[dovec], MoreArgs =
>> list(args[!dovec]),
>> SIMPLIFY = SIMPLIFY, USE.NAMES = USE.NAMES))
>> }
>>> smartsub2(pat="x", repl="b" , x="essai(b[1], b[2], x[1])")
>> Error in names %in% vectorize.args : objet 'vectorize.args' introuvable
>>
>> It fails because vectorize.args is unknown
>>
>> Indeed smartsub2 is different from smartsub.
>>> identical(smartsub, smartsub2)
>> [1] FALSE
>>
>> 1/ Why are they different? They are just a copy of each other.
>>
>> 2/ Second question, vectorize.args is indeed not defined before to be used
>> in the function. Why no error is produced in original function?
>>
>> Thanks a lot
>>
>> Marc
>>
>>
>> --
>>
>> __________________________________________________________
>> Marc Girondot, Pr
>>
>> Laboratoire Ecologie, Systématique et Evolution
>> Equipe de Conservation des Populations et des Communautés
>> CNRS, AgroParisTech et Université Paris-Sud 11 , UMR 8079
>> Bâtiment 362
>> 91405 Orsay Cedex, France
>>
>> Tel: 33 1 (0)1.69.15.72.30 Fax: 33 1 (0)1.69.15.73.53
>> e-mail: marc.girondot at u-psud.fr
>> Web: http://www.ese.u-psud.fr/epc/conservation/Marc.html
>> Skype: girondot
>>
>> ______________________________________________
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