On Thu, Sep 3, 2009 at 4:34 PM, Henrique Dallazuanna <www...@gmail.com>wrote:
> Try this:
>
> #1
> lapply(my.array, '[', , 3)
>
>
this works! thank you a lot!
> #2
> newThirdColumn <- sample(3)
> lapply(my.array, replace, list = 7:9, values = newThirdColumn)
>
>
i did not understand this last line, so far i couldn't make it work.
would it be easier to replace the values (the third column of each matrix
in my.array) using an array like in #1?
thank you for your reply!
> 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
>
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