How can id do this efficiently in R ? 1 0 0 0 2 0 0 0 3 rotate right 0 0 1 0 2 0 3 0 0 rotate left 0 0 3 0 2 0 1 0 0 What I want to do is described here: http://stackoverflow.com/questions/13049575/r-min-max-and-mean-of-off-diagonal-elements-in-a-matrix but I want do to it for all off-anti-diagonals. I am also concerned that the method described on stackoverflow isn't necessarily efficient. I work with matrices of size 3k * 3k. -- Witold Eryk Wolski
anti off diagonal min max mean - how to rotate matrix by 90 degree
2 messages · Witold E Wolski, Marc Schwartz
On Jul 17, 2013, at 3:04 PM, Witold E Wolski <wewolski at gmail.com> wrote:
How can id do this efficiently in R ? 1 0 0 0 2 0 0 0 3 rotate right 0 0 1 0 2 0 3 0 0 rotate left 0 0 3 0 2 0 1 0 0 What I want to do is described here: http://stackoverflow.com/questions/13049575/r-min-max-and-mean-of-off-diagonal-elements-in-a-matrix but I want do to it for all off-anti-diagonals. I am also concerned that the method described on stackoverflow isn't necessarily efficient. I work with matrices of size 3k * 3k.
I did not take the time to review in detail the solutions on SO, but here is one approach for rotation, albeit, not fully tested:
mat
[,1] [,2] [,3] [1,] 1 0 0 [2,] 0 2 0 [3,] 0 0 3 # Rotate right
t(mat)[, ncol(mat):1]
[,1] [,2] [,3] [1,] 0 0 1 [2,] 0 2 0 [3,] 3 0 0 # Rotate left
t(mat)[nrow(mat):1, ]
[,1] [,2] [,3] [1,] 0 0 3 [2,] 0 2 0 [3,] 1 0 0 big.mat <- matrix(rep(0, 3000 * 3000), ncol = 3000)
str(big.mat)
num [1:3000, 1:3000] 0 0 0 0 0 0 0 0 0 0 ...
system.time(mat.right <- t(big.mat)[, ncol(big.mat):1])
user system elapsed 0.273 0.000 0.273
system.time(mat.left <- t(big.mat)[nrow(big.mat):1], )
user system elapsed 0.094 0.000 0.099 Regards, Marc Schwartz