Hi,
Try:
indx<- combn(ncol(uu),2)
sapply(seq_len(ncol(indx)),function(i) rho(uu[,indx[1,i]],uu[,indx[2,i]]))
# [1] 0.9746318 0.9594119 0.9512583 0.9462555 0.9981909 0.9961499 0.9946367
# [8] 0.9996186 0.9990560 0.9998746
A.K.
On Sunday, January 26, 2014 10:04 AM, Kathryn Lord <kathryn.lord2...@gmail.com>
wrote:
Dear R users,
I'd like to compute "rho"(looks like a correlation matrix) with every two
columns of "uu" matrix below.
Toy example,
> uu <- matrix(1:15, nr=3, nc=5)
> uu
[,1] [,2] [,3] [,4] [,5]
[1,] 1 4 7 10 13
[2,] 2 5 8 11 14
[3,] 3 6 9 12 15
rho <- function(x,y)
{
sum(x*y)/(sqrt(sum(x^2))*sqrt(sum(y^2)))
}
rho_12 <- rho(uu[,1],uu[,2])
rho_13 <- rho(uu[,1],uu[,3])
rho_14 <- rho(uu[,1],uu[,4])
rho_15 <- rho(uu[,1],uu[,5])
rho_23 <- rho(uu[,2],uu[,3])
rho_24 <- rho(uu[,2],uu[,4])
rho_24 <- rho(uu[,2],uu[,5])
rho_34 <- rho(uu[,3],uu[,4])
rho_35 <- rho(uu[,3],uu[,5])
rho_45 <- rho(uu[,4],uu[,5])
Actually, the matrix uu is huge, 20*1000. Would you plz tell me how to
calculate "rho", more efficiently??
Any suggestion will be greatly appreciated.
Best,
Kathryn Lord
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