Hi,
I think it's a side effect of lazy evaluation, where you should
probably use the ?force like a jedi,
lapply(1:5,function(k){force(k) ; function(){k}})[[2]]()
HTH,
baptiste
On 18 March 2011 07:01, jamie.f.olson <inspired2apa...@gmail.com> wrote:
> So, I've been confused by this for a while. If I want to create functions in
> an apply, it only uses the desired value for the variable if I create a new
> local variable:
>
>> lapply(1:5,function(h){k=h;function(){k}})[[1]]()
> [1] 1
>> lapply(1:5,function(k){function(){k}})[[1]]()
> [1] 5
>>
>
> Normally, a function will use values for a variable if the variable is local
> to the definition of the function:
>
>> a = function(x){function(){x}}
>> a(5)
> function(){x}
>
>> a(5)()
> [1] 5
>> a(6)()
> [1] 6
>
>
> So why doesn't this work for apply? Does apply work more like loops, with a
> shared variable that changes values so that the first function defined
> actually uses the last value of the variable?
>
>> a = list();for (k in 1:5){a = c(a,function(){k})}
>> a[[1]]()
> [1] 5
>>
>
>
> Or is there something else entirely?
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Scope-and-apply-type-functions-tp3385230p3385230.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help@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.
>
______________________________________________
R-help@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.
