So, to get back to mapply:
eval(mapply(function(x) substitute(function() z,list(z=x)), c("a", "b"))$a)()
or like this:
mapply(function(x) eval(substitute(function(i) z*i,list(z=x))), c(2,3))[[1]](2)
Am 03.05.2012 um 16:02 schrieb Jessica Streicher:
> Now.. i just tried around and this might be a bit strange way to do things..
>
> createFunc<-function(v){
> v_out<-NULL
> for(i in v){
> v_out[[i]]<-substitute(function(){x},list(x=i))
> }
> return(v_out)
> }
>
>> y<-createFunc(c("a","b"))
>> y
> $a
> function() {
> "a"
> }
>
> $b
> function() {
> "b"
> }
>
>> eval(y$a)()
> [1] "a"
>> eval(y$b)()
> [1] "b"
>
> Am 03.05.2012 um 12:31 schrieb Casper Ti. Vector:
>
>> As the title says, I want to apply a function (which itself returns
>> a function) to a list (or vector), and get a list (or vector) of
>> generated functions as the return value, but get unexpected result.
>>
>> Anyone with an idea about the reason of this phenomenon and a correct
>> way to implement the requirements? Thanks very much :)
>>
>> --
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>> from 2010 to 2013) from a key server.
>>
>> ______________________________________________
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>> and provide commented, minimal, self-contained, reproducible code.
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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