Thank you also for your help
Michel
Le 20/11/2013 19:04, Dennis Murphy a écrit :
Hi:
which(m == 1L, arr.ind = TRUE)
Dennis
On Wed, Nov 20, 2013 at 2:28 AM, Arnaud Michel wrote:
Hi
I have the following problem
I would like to build, from a matrix filled with 0 and with 1, a matrix
or a data.f
> indx <- arrayInd(which(m>0), .dim=c(5, 5))
> indx
[,1] [,2]
[1,]41
[2,]23
[3,]43
[4,]24
[5,]15
# If you want the result sorted
> indx[order(indx[,1], indx[,2]),]
[,1] [,2]
[1,]15
[2,]23
[3,]24
[4,]41
[5,]43
Thank you Pascal
Its fine
Michel
Le 20/11/2013 11:55, Pascal Oettli a écrit :
Hello,
One approach is:
m <- structure(c(0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0,
0, 0, 1, 0, 0, 0, 0), .Dim = c(5L, 5L))
out <- which(m==1, arr.ind=TRUE)
out[order(out[,1]),]
Regards,
Pascal
On 20 Nove
Hello,
One approach is:
m <- structure(c(0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0,
0, 0, 1, 0, 0, 0, 0), .Dim = c(5L, 5L))
out <- which(m==1, arr.ind=TRUE)
out[order(out[,1]),]
Regards,
Pascal
On 20 November 2013 19:28, Arnaud Michel wrote:
> Hi
> I have the following problem
> I w
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