[R] Matrix transformation problem

[Dimitris Rizopoulos]

try this:

x <- matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0),
    ncol = 3, byrow = TRUE)
which(x == 1, arr.ind = TRUE)[, "col", drop = FALSE]


I hope it helps.

Best,
Dimitris

----
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
     http://www.student.kuleuven.be/~m0390867/dimitris.htm


----- Original Message ----- 
From: <stefan.petersson at inizio.se>
To: <r-help at r-project.org>
Sent: Wednesday, June 11, 2008 10:10 AM
Subject: [R] Matrix transformation problem


>
> ng,
>
> I have a matrix (x) with binary content. Each row of the matrix 
> holds exactly one 1, and the rest of the row is zeros. The thing is 
> that I need to 'collapse' the matrix to one column where each row 
> holds the original column index of the 1's (y). Sometimes, the 
> matrix is quite large, so I have a perfomance problem.
>
> x <- matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 
> 1,0,0),ncol=3,byrow=T)
> x
>     [,1] [,2] [,3]
> [1,]    1    0    0
> [2,]    0    0    1
> [3,]    0    1    0
> [4,]    0    0    1
> [5,]    0    1    0
> [6,]    1    0    0
>
> In the matrix above, on the first row, the 1 is in column 1, hence 
> '1' on the first row in the matrix below. On the second row in the 
> matrix above, the 1 is in column 3, hence the '3' on the second row 
> in the matrix below. And so on...
>
> y
>     [,1]
> [1,]    1
> [2,]    3
> [3,]    2
> [4,]    3
> [5,]    2
> [6,]    1
>
> ______________________________________________
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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.
> 


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