[R] matching last argument in function

[Erik Iverson]

Yes that will work, that's exactly what I was getting at in my second 
paragraph.  I wrote a function that uses this idea, except the (single) 
unnamed argument can occur anywhere in the function (not necessarily 
last). It will stop if there is more than one unnamed argument.

test2 <- function(...) {
   dots <- list(...)

   if(sum(dots.missing.name <- names(dots) %in% "") > 1)
     stop("Only one argument should have missing name.")

   expr.ind <- ifelse(any(names(dots) == "expr"),
                      which(names(dots) == "expr"),
                      which(dots.missing.name))

   expr <- dots[[expr.ind]]
   opts <- dots[-expr.ind]
   opts
}


Gabor Csardi wrote:
> It should be possible i think. You just supply all the arguments via
> '...' and then cut off the last one. I don't see why this wouldn't work,
> but maybe i'm missing something.
> 
> Gabor
> 
> On Tue, Feb 12, 2008 at 12:58:25PM -0600, Erik Iverson wrote:
>> Alistair -
>>
>> I don't believe this is possible.  The only way formal arguments (like 
>> expr) can be matched after a '...' is with *exact* name matching.  Why 
>> do you want to avoid explicitly naming the expr argument?
>>
>> If you always want the expr argument last, you might be able to just use 
>> ... as the sole argument to your function, and strip off the last 
>> element inside the function as 'expr', and use all but the last element 
>> as your list of options.  This requires that expr always be given last 
>> though.  Probably best just to explicitly name the expr argument.
>>
>> Erik Iverson
>>
>> Alistair Gee 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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>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>