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
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
sorry, my previous answer was not correct; you need:
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(t(x == 1), arr.ind = TRUE)[, "row", drop = FALSE]
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
you may try a matrix multiplication, which has a very high performance in R x%*%1:ncol(x) hth. stefan.petersson at inizio.se schrieb:> 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. >-- Eik Vettorazzi Institut f?r Medizinische Biometrie und Epidemiologie Universit?tsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/42803-8243 F ++49/40/42803-7790
stefan.petersson wrote:> > > 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 > >max.col(x) -- View this message in context: http://www.nabble.com/Matrix-transformation-problem-tp17773270p17779431.html Sent from the R help mailing list archive at Nabble.com.