[R] Deparsing part of a list argument

Duncan Murdoch murdoch at stats.uwo.ca
Fri Oct 26 04:50:53 CEST 2007


On 25/10/2007 5:38 PM, Andrew Smith wrote:
> Here's a simple example of the type of function I'm trying to write,
> where the first argument is a list of functions:
> 
> myfun <- function(funlist, vec){
>      tmp <- lapply(funlist, function(x)do.call(x, args = list(vec)))
>      names(tmp) <- names(funlist)
>      tmp
> }
> 
>> myfun(list("Summation" = sum, prod, "Absolute value" = abs), c(1, 4, 6, 7))
> $Summation
> [1] 18
> 
> [[2]]
> [1] 168
> 
> $`Absolute value`
> [1] 1 4 6 7
> 
> 
> What I would like is for the names of the result to be the same as the
> names of funlist (as is the case in components 1 and 3 above), but if
> some components of funlist are unnamed, I would like the default name
> in the result to be the name of the actual function used.  So in the
> above example, I want the name of the second component to default to
> 'prod', since funlist[[2]] has no name.
> 
>  But the problem is that I can't get the function names corresponding
> to the unnamed elements of funlist by using the usual
> deparse(substitute)) trick, since they're part of a list which is
> treated as one big argument.  So deparse(substitute(funlist)) just
> gives me one big character string of all the functions, and I'm not
> sure how I can use that to get just the functions that aren't named.
> Any ideas?

I wouldn't do this, but you could use substitute to get the funlist 
argument to myfun.  It will be an expression which would deparse to 
list("Summation" = sum, prod, "Absolute value" = abs), but you could 
probably futz around and decompose it into its component parts.

But is this really worth it?

Duncan Murdoch



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