Try this:
#1
lapply(my.array, '[', , 3)
#2
newThirdColumn <- sample(3)
lapply(my.array, replace, list = 7:9, values = newThirdColumn)
On Thu, Sep 3, 2009 at 11:16 AM, Carlos Hernandez <carlos.u...@gmail.com>wrote:
> Dear All,
> I created a list (of length Z) in the following way:
>
> my.array <- vector("list", Z)
>
> then i assigned a matrix (of T rows by N columns) in each of the elements
> of
> the list my.array in the following way:
>
> my.array[[i]] <- matrix.data ##( matrix.data has dimensions TxN, and i
> repeated this command for i from 1 to Z, the matrix.data contains only
> numeric data)
>
> and
> 1. i would like to extract all the third columns of each of the Z matrices
> stored in my.array (such that i get a new list only with the 3rd columns of
> each matrix in the elements of a new list)
>
> 2. i would like to know how could i replace all the 3rd columns of each
> matrix in my.array if i have a second matrix (size ZxT) with these columns.
>
> is there a simple way to do these tasks? i appreciate any hints or advice.
>
> Carlos
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
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> 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]]
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and provide commented, minimal, self-contained, reproducible code.
