[R] Matrix transformation problem
Dimitris Rizopoulos
dimitris.rizopoulos at med.kuleuven.be
Wed Jun 11 11:17:08 CEST 2008
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
>
> ______________________________________________
> R-help at 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.
>
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
More information about the R-help
mailing list