[R] Extracting elements from a nested list

[jim holtman]

Try posting your data using 'dput' so it is easily read for testing.

On Mon, Oct 18, 2010 at 11:17 AM, Gregory Ryslik <rsaber at comcast.net> 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 <rsaber at 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 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.
>>
>>
>>
>> --
>> Henrique Dallazuanna
>> Curitiba-Paraná-Brasil
>> 25° 25' 40" S 49° 16' 22" O
>
>
>        [[alternative HTML version deleted]]
>
>
> ______________________________________________
> 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.
>
>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?