On Apr 21, 2012, at 1:13 AM, Petr Savicky wrote:
On Fri, Apr 20, 2012 at 10:52:47AM -0400, David Winsemius wrote:
On Apr 20, 2012, at 7:05 AM, Petr Savicky wrote:
On Fri, Apr 20, 2012 at 03:03:40AM -0700, juliane0212 wrote:
I'm having some problems computing a matrix being symmetric on
both
diagonals.
Does anyone know a way to get from this matrix
M <- matrix(c(1,0,0,0,2,7,0,0,3,4,0,0,6,0,0,0), ncol=4)
to this one
M_final <- matrix(c(1,2,3,6,2,7,4,3,3,4,7,2,6,3,2,1),
ncol=4)
Hi.
Try the following.
M[row(M) > col(M)] <- t(M)[row(M) > col(M)]
n <- nrow(M)
M[row(M) + col(M) > n + 1] <- M[n:1, n:1][row(M) + col(M) > n + 1]
all(M == M_final)
[1] TRUE
How about?
M[3:4, ] <- rev(M[1:2,])
M
[,1] [,2] [,3] [,4]
[1,] 1 2 3 6
[2,] 2 7 4 3
[3,] 3 4 7 2
[4,] 6 3 2 1
Hi.
I am not sure, which matrix did you start from.
You are right. I misunderstood the requested task. (...and then used
the wrong starting matrix.) Sorry for the noise.
--
David.
If we start
from the original matrix, then we get
M <- matrix(c(1,0,0,0,2,7,0,0,3,4,0,0,6,0,0,0), ncol=4)
M[3:4, ] <- rev(M[1:2,])
M
[,1] [,2] [,3] [,4]
[1,] 1 2 3 6
[2,] 0 7 4 0
[3,] 0 4 7 0
[4,] 6 3 2 1
where the components 2 and 3 have two and not four copies.
Petr.
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David Winsemius, MD
West Hartford, CT
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