Help creating a symmetric matrix?

Sat, Dec 24, 2011 5:48 AM

On Sat, Dec 24, 2011 at 8:38 AM, William Revelle <lists at revelle.net> wrote:

I read the original question as *only* wanting the complete lower
triangle, with diagonal of 1 and 0 in the upper triangle.

If your interpretation is correct, there's also this convenience function:

library(ecodist)
z <- full(v)

Sarah

v<-c(0.33740, 0.26657, 0.23388, 0.23122, 0.21476, 0.20829, 0.20486,
0.19439, 0.19237,
0.18633, 0.17298, 0.17174, 0.16822, 0.16480, 0.15027)


z<-diag(6)
z[row(z) > col(z)] <- v
z <- z + t(z)
diag(z) <- 0

? ? ? ?[,1] ? ?[,2] ? ?[,3] ? ?[,4] ? ?[,5] ? ?[,6]
[1,] 0.00000 0.33740 0.26657 0.23388 0.23122 0.21476
[2,] 0.33740 0.00000 0.20829 0.20486 0.19439 0.19237
[3,] 0.26657 0.20829 0.00000 0.18633 0.17298 0.17174
[4,] 0.23388 0.20486 0.18633 0.00000 0.16822 0.16480
[5,] 0.23122 0.19439 0.17298 0.16822 0.00000 0.15027
[6,] 0.21476 0.19237 0.17174 0.16480 0.15027 0.00000


Bill


On Dec 24, 2011, at 6:04 AM, Sarah Goslee wrote:

Or the slightly shorter:

z<-diag(6)
z[row(z) > col(z)] <- v

which is what lower.tri() does,

and
z <- diag(6)
z[lower.tri(z)] <- v

also works.

Sarah

On Fri, Dec 23, 2011 at 9:31 PM, Rui Barradas <ruipbarradas at sapo.pt> wrote:

Matt Considine wrote

Hi,
I am trying to work with the output of the MINE analysis routine found at
? ?http://www.exploredata.net

Specifically, I am trying to read the results into a matrix (ideally an
n x n x 6 matrix, but I'll settle right now for getting one column into
a matrix.)

The problem I have is not knowing how to take what amounts to being one
half of a symmetric matrix - excluding the diagonal - and getting it
into a matrix. ?I have tried using "lower.tri" as found here
? ?https://stat.ethz.ch/pipermail/r-help/2008-September/174516.html
but it appears to only partially fill in the matrix. ?My code and an
example of the output is below. ?Can anyone point me to an example that
shows how to create a matrix with this sort of input?

Thank you in advance,
Matt

#v<-newx[,3]
#or, for the sake of this example
v<-c(0.33740, 0.26657, 0.23388, 0.23122, 0.21476, 0.20829, 0.20486,
0.19439, 0.19237,
0.18633, 0.17298, 0.17174, 0.16822, 0.16480, 0.15027)
z<-diag(6)
ind <- lower.tri(z)
z[ind] <- t(v)[ind]

z
? ? ? ? ?[,1] ? ?[,2] [,3] [,4] [,5] [,6]
[1,] 1.00000 0.00000 ? ?0 ? ?0 ? ?0 ? ?0
[2,] 0.26657 1.00000 ? ?0 ? ?0 ? ?0 ? ?0
[3,] 0.23388 0.19237 ? ?1 ? ?0 ? ?0 ? ?0
[4,] 0.23122 0.18633 ? NA ? ?1 ? ?0 ? ?0
[5,] 0.21476 0.17298 ? NA ? NA ? ?1 ? ?0
[6,] 0.20829 0.17174 ? NA ? NA ? NA ? ?1

Hello,

Aren't you complicating?

In the last line of your code, why use 'v[ind]' if 'ind' indexes the matrix,
not the vector?

z<-diag(6)
ind <- lower.tri(z)
z[ind] <- v ? ? ? ? ? ? ? ? ? ? ? ?#This works
z

Rui Barradas

--

Help creating a symmetric matrix?

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