Replace diagonal elements in an array

Question

Does anyone know a neat / efficient way to replace diagonal elements in an array, similar to using diag(x) <- value for a matrix? In other words, something like this:

 > m<-array(1:27,c(3,3,3)) > for(k in 1:3){ + diag(m[,,k])<-5 + } > m , , 1 [,1] [,2] [,3] [1,] 5 4 7 [2,] 2 5 8 [3,] 3 6 5 , , 2 [,1] [,2] [,3] [1,] 5 13 16 [2,] 11 5 17 [3,] 12 15 5 , , 3 [,1] [,2] [,3] [1,] 5 22 25 [2,] 20 5 26 [3,] 21 24 5 

but without using a for loop (my arrays are quite large, and this manipulation will already be in the loop).

Many thanks.

Answer 1

Try the following:

 with(expand.grid(a = 1:3, b = 1:3), replace(m, cbind(a, a, b), 5)) 

EDIT:

The question asked is neat / efficient, but of course this is not the same thing. One liner here is compact and without a loop, but if you are looking for speed, I think you will find that the loop in the question is actually the fastest of all the answers.

Answer 2

For this you can use the following function if you have only 3 dimensions in your array. You can generalize to more dimensions based on this code, but I'm too lazy to do this for you; -)

 `arraydiag<-` <- function(x,value){ dims <- dim(x) id <- seq_len(dims[1]) + dims[2]*(seq_len(dims[2])-1) id <- outer(id,(seq_len(dims[3])-1)*prod(dims[1:2]),`+`) x[id] <- value dim(x) <- dims x } 

This works like:

 m<-array(1:36,c(3,3,4)) arraydiag(m)<-NA m 

Note that contrary to the diag () function, this function cannot deal with matrices that are not square. You can look at the diag () source code to find out how to adapt this code so that it does.

Answer 3

 diagArr <- function (dim) { n <- dim[2] if(dim[1] != n) stop("expecting first two dimensions to be equal") d <- seq(1, n*n, by=n+1) as.vector(outer(d, seq(0, by=n*n, length=prod(dim[-1:-2])), "+")) } m[diagArr(dim(m))] <- 5 

This is written with the intention that it works for measurements above 3, but I have not tested it in this case. However, it should be good.