Dear anna,
Unless the original matrix has a massive number of columns, why not just use
loops? R programmers often have an unnecessary phobia of loops, and will puzzle
over a programming problem for hours that can be solved by loops in seconds.
You don't say what specifically you want to do, but, for example:
> f <- function(X){
+ nc <- ncol(X)
+ Y <- matrix(0, nc, nc)
+ for (i in 1:(nc - 1)){
+ for (j in (i+1):nc){
+ Y[i, j] <- sum(X[, i] * X[, j])
+ }
+ }
+ return(Y)
+ }
>
> (A <- matrix(1:12, 4, 3))
[,1] [,2] [,3]
[1,] 1 5 9
[2,] 2 6 10
[3,] 3 7 11
[4,] 4 8 12
> f(A)
[,1] [,2] [,3]
[1,] 0 70 110
[2,] 0 0 278
[3,] 0 0 0
The example is artificial, since it just computes part of the matrix product,
> upper.tri(diag(3)) * (t(A) %*% A)
[,1] [,2] [,3]
[1,] 0 70 110
[2,] 0 0 278
[3,] 0 0 0
>
but it has the structure that you outlined (if I understand it correctly).
For a moderate number of columns (you don't say how many you have), the
computation isn't prohibitively slow:
> B <- matrix(rnorm(1e5), 100, 1000)
> system.time(f(B))
user system elapsed
4.12 0.01 4.15
I hope this helps,
John
------------------------------------------------
John Fox, Professor
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/
On Wed, 27 Aug 2014 04:54:44 +0100
anna pannas <anna_pan...@yahoo.co.uk> wrote:
> hello
>
> i want to fill a matrix by its upper off diagonal elements
>
> specifically I want to take the first and second column of the matrix and I
> apply a function to then that returns a single number which I want to place
> in the (1,2) entry of the matrix, then I want to take the first and third
> column of the matrix and apply the same function, get the single number and
> place it to (2,3) entry of the matrix and so on
>
> how can i do it?
>
> thanks
> anna
>
> [[alternative HTML version deleted]]
>
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