> diag(x[,x[,4]])
[1] 11 23 32
> diag(x[,x[,4]]) <- diag(x[,x[,4]])+5
> x
[,1] [,2] [,3] [,4]
[1,] 16 12 13 1
[2,] 21 22 28 3
[3,] 31 37 33 2
Cheers
Joris
On Fri, Jun 18, 2010 at 8:58 PM, Iuri Gavronski <i...@ufrgs.br> wrote:
> Hi,
>
> I would like to have an index for a column in a matrix encoded in a
> cell of the same matrix.
> For example:
> x = matrix(c(11,12,13,1,
> 21,22,23,3,
> 31,32,33,2),byrow=T,ncol=4)
>
> In this case, column 4 is the index. I then access the column
> specified in the index by:
>> for (i in 1:3) print(x[i,x[i,4]])
> [1] 11
> [1] 23
> [1] 32
>>
>> for (i in 1:3) {x[i,x[i,4]] <- x[i,x[i,4]] + 5}
>> x
> [,1] [,2] [,3] [,4]
> [1,] 16 12 13 1
> [2,] 21 22 28 3
> [3,] 31 37 33 2
>>
>
> Is there a way to get the same results without looping?
>
> Best,
>
> Iuri.
>
> ______________________________________________
> 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.
>
--
Joris Meys
Statistical consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
tel : +32 9 264 59 87
joris.m...@ugent.be
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