Try this:

diag(sapply(all.predicted.values, '[[', 'max.growth'))


On Mon, Oct 18, 2010 at 12:59 PM, Gregory Ryslik <rsaber@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
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>
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>



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

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