You probably need mapply since you have 2 list of arguments which you want to 
use "in sync"

mapply(function(x1,x2)x1[[x2]],all.predicted.values,max.growth)

might be what you want.



On Oct 18, 2010, at 5:17 PM, Gregory Ryslik wrote:

> Unfortunately, that gives me null everywhere. Here's the data I have for 
> all.predicted.values and max.growth. Perhaps this will help. Thus I want 
> all.predicted.values[[1]][[4]] then all.predicted.values[[2]][3]] and then 
> all.predicted.values[[3]][[4]].
> 
> I've attached what your statement outputs at the end.
> 
> Thanks again!
> 
> Browse[2]> max.growth
> [[1]]
> [1] 4
> 
> [[2]]
> [1] 3
> 
> [[3]]
> [1] 4
> 
> Browse[2]> all.predicted.values
> [[1]]
> [[1]][[1]]
>  [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
> 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
> [55] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
> 0 0 0 0 0 0 0 0 0 0
> Levels: 0 1 2
> 
> [[1]][[2]]
>  [1] 2 2 2 0 2 0 2 2 2 2 2 2 2 2 0 2 2 2 2 2 2 2 2 2 2 0 0 2 2 2 2 0 0 0 2 2 
> 0 0 2 2 0 2 2 2 2 2 0 2 2 2 0 2 2 0
> [55] 0 0 2 0 2 0 0 0 0 2 2 2 2 0 2 2 2 0 2 2 0 0 2 2 2 2 2 2 2 0 0 0 2 0 2 2 
> 2 2 0 2 2 2 0 2 0 0
> Levels: 0 1 2
> 
> [[1]][[3]]
>  [1] 0 0 0 0 2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 2 0 2 2 2 0 0 0 2 0 0 2 0 0 0 0 0 
> 0 0 2 0 0 0 0 0 2 2 0 0 0 2 0 0 0 0
> [55] 0 0 2 0 2 0 0 0 0 2 2 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 
> 2 0 0 0 0 0 0 2 0 0
> Levels: 0 1 2
> 
> [[1]][[4]]
>  [1] 0 0 0 0 2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 2 0 2 2 2 0 0 0 2 0 0 2 0 0 0 0 0 
> 0 0 2 0 0 0 0 0 2 2 0 0 0 2 0 0 0 0
> [55] 0 0 2 0 2 0 0 0 0 2 2 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 
> 2 0 0 0 0 0 0 2 0 0
> Levels: 0 1 2
> 
> 
> [[2]]
> [[2]][[1]]
>  [1] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 
> 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
> [55] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 
> 2 2 2 2 2 2 2 2 2 2
> Levels: 0 1 2
> 
> [[2]][[2]]
>  [1] 2 2 2 2 1 2 2 2 2 2 1 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 1 2 
> 2 2 1 2 2 1 1 2 2 2 2 2 2 2 2 1 2 2
> [55] 2 2 2 2 1 2 2 2 2 1 2 2 1 1 1 2 2 2 1 2 1 2 1 2 1 2 2 2 1 1 2 2 1 2 2 1 
> 1 2 1 1 1 2 2 1 2 2
> Levels: 0 1 2
> 
> [[2]][[3]]
>  [1] 2 2 2 0 1 2 2 2 2 2 1 2 2 2 0 1 2 1 2 2 2 2 2 2 2 0 0 2 1 2 2 2 0 0 1 2 
> 0 0 1 2 0 1 1 2 2 2 0 2 2 2 0 1 2 2
> [55] 0 2 2 2 1 0 0 0 0 1 2 2 1 1 1 2 2 0 1 2 1 0 1 2 1 2 2 2 1 1 2 2 1 2 2 1 
> 1 2 1 1 1 2 2 1 0 2
> Levels: 0 1 2
> 
> 
> [[3]]
> [[3]][[1]]
>  [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
> 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
> [55] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
> 0 0 0 0 0 0 0 0 0 0
> Levels: 0 1 2
> 
> [[3]][[2]]
>  [1] 2 2 2 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 0 0 2 2 2 2 2 0 0 2 2 
> 2 0 2 2 0 2 2 2 2 2 0 2 2 2 0 2 2 2
> [55] 0 2 2 2 2 2 0 0 2 2 2 2 2 2 2 2 2 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 
> 2 2 2 2 2 2 2 2 2 2
> Levels: 0 1 2
> 
> [[3]][[3]]
>  [1] 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 
> 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0
> [55] 0 0 0 0 1 0 0 0 0 1 0 0 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 1 1 0 0 1 0 0 1 
> 1 0 1 1 1 0 0 1 1 0
> Levels: 0 1 2
> 
> [[3]][[4]]
>  [1] 2 2 2 0 1 0 2 2 0 2 1 2 2 0 0 1 1 1 1 0 2 0 0 0 2 0 0 0 1 2 0 0 0 0 1 2 
> 0 0 1 2 0 1 1 2 0 0 0 2 2 0 0 1 2 0
> [55] 0 0 0 0 1 0 0 0 0 1 0 2 1 1 1 2 0 0 1 2 1 1 1 2 1 2 2 2 1 1 0 0 1 0 2 1 
> 1 2 1 1 1 2 0 1 1 0
> Levels: 0 1 2
> 
> 
> Browse[2]>    
> predicted.values.for.max.growth<-diag(sapply(all.predicted.values,'[[','max.growth'))
> Browse[2]> predicted.values.for.max.growth
> [[1]]
> NULL
> 
> [[2]]
> [1] 0
> 
> [[3]]
> [1] 0
> 
> [[4]]
> [1] 0
> 
> [[5]]
> NULL
> 
> [[6]]
> [1] 0
> 
> [[7]]
> [1] 0
> 
> [[8]]
> [1] 0
> 
> [[9]]
> NULL
> 
> 
> 
> On Oct 18, 2010, at 11:08 AM, Henrique Dallazuanna wrote:
> 
>> Try this:
>> 
>> diag(sapply(all.predicted.values, '[[', 'max.growth'))
>> 
>> 
>> On Mon, Oct 18, 2010 at 12:59 PM, Gregory Ryslik <rsa...@comcast.net> wrote:
>> Hi,
>> 
>> I have a list of n items and the ith element has m_i elements within it.
>> 
>> I want to do something like:
>> 
>> predicted.values<- lapply(all.predicted.values,'[[',max.growth[[i]])
>> 
>> Where max.growth[[i]] is the element I want to extract from each of the ith 
>> predicted elements. Thus, for example, I want to extract the max.growth[[1]] 
>> element from  all.predicted.values[[1]] (which is itself a list). Then I 
>> want to extract max.growth[[2]] element from all.predicted.values[[2]].
>> 
>> I realize I can do this with a for loop but then if I can do this as one 
>> line that would be preferable.
>> 
>> Thanks!
>> 
>> Greg
>>       [[alternative HTML version deleted]]
>> 
>> ______________________________________________
>> 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.
>> 
>> 
>> 
>> -- 
>> Henrique Dallazuanna
>> Curitiba-Paraná-Brasil
>> 25° 25' 40" S 49° 16' 22" O
> 
> 
>       [[alternative HTML version deleted]]
> 
> ______________________________________________
> 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.

--
Erich Neuwirth
Didactic Center for Computer Science and Institute for Scientific Computing
University of Vienna





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