1+outer(1:9, 1:9, "+")%%3
In particular the following are equal to your jj1 and jj2:
jj1. <- 1+outer(0:8, rep(0:2, e=3), "+")%%3
jj2. <- 1+outer(0:8, c(1,2,1,3,1,3,2,3,2)-1, "+")%%3
I couldn't figure out jj3, but this system may not work so easily for that. hope this helps. spencer graves
Robin Hankin wrote:
Hello everybody.
I am trying to reproduce a particular matrix in an elegant way. If I have
jj1 <- structure(c(1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2, 3,1,2,3,1,2,3,1,2,3,2,3,1,2,3,1,2,3,1,2,3, 1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,3,1,2,3,1, 2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1, 2),.Dim = as.integer(c(9,9)))
[ image(jj1) is good here ] then I can get this with
kronecker(matrix(1,3,1),kronecker(1+outer(0:2,0:2,"+")%%3,matrix(1,1,3)))
I want to reproduce the following matrices in an equivalent way:
jj2 <- matrix(c(1,2,3,1,2,3,1,2,3,2,3,1,2,3,1,2,3,1, 1,2,3,1,2,3,1,2,3,3,1,2,3,1,2,3,1,2,1,2,3,1,2, 3,1,2,3,3,1,2,3,1,2,3,1,2,2,3,1,2,3,1,2,3,1,3, 1,2,3,1,2,3,1,2,2,3,1,2,3,1,2,3,1),9,9)
jj3 <- structure(c(1,2,3,2,3,1,3,1,2,1,2,1,2,3,2,3,1, 3,1,3,1,2,1,2,3,2,3,2,3,1,3,1,2,1,2,3,2,3, 2,3,1,3,1,2,1,2,1,2,3,2,3,1,3,1,3,1,2,1,2, 3,2,3,1,3,1,3,1,2,1,2,3,2,3,2,3,1,3,1,2,1, 2),.Dim = as.integer(c(9,9)))
[ note that jj1-jj3 each have precisely 3 occurrences of A, B, and C along each row, column and (broken) diagonal ].
Can anyone give me a nice elegant way of creating jj2 and jj3 please?
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