Sum of antidiagonal of a matrix

Question

I'm trying to sum the elements along the antidiagonal (secondary diagonal, minor diagonal) of a matrix.

So, if I have a matrix m:

m <- matrix(c(2, 3, 1, 4, 2, 5, 1, 3, 7), 3)
m

     [,1] [,2] [,3]
[1,]    2    4    1
[2,]    3    2    3
[3,]    1    5    7

I'm looking for the sum m[3, 1] + m[2, 2] + m[1, 3], i.e. 1 + 2 + 1

I can't figure out how to set up an iteration. As far as I know there is no function for this (like diag() for the other diagonal).

Answer 1

Using

m <- matrix(c(2, 3, 1, 4, 2, 5, 1, 3, 7), 3)

1) Reverse the rows as shown (or the columns - not shown), take the diagonal and sum:

sum(diag(m[nrow(m):1, ]))
## [1] 4

2) or use row and col like this:

sum(m[c(row(m) + col(m) - nrow(m) == 1)])
## [1] 4

This generalizes to other anti-diagonals since row(m) + col(m) - nrow(m) is constant along all anti-diagonals. For such a generalization it might be more convenient to write the part within c(...) as row(m) + col(m) - nrow(m) - 1 == 0 since then replacing 0 with -1 uses the superdiagonal and with +1 uses the subdiagonal. -2 and 2 use the second superdiagonal and subdiagonal respectively and so on.

3) or use this sequence of indexes:

n <- nrow(m)
sum(m[seq(n, by = n-1, length = n)])
## [1] 4

4) or use outer like this:

n <- nrow(m)
sum(m[!c(outer(1:n, n:1, "-"))])
## [1] 4

This one generalizes nicely to other anti-diagonals too as outer(1:n, n:1, "-") is constant along anti-diagonals. We can write m[outer(1:n, n:1) == 0] and if we replace 0 with -1 we get the super anti-diagonal and with +1 we get the sub anti-diagonal. -2 and 2 give the super super and sub sub antidiagonals. For example sum(m[c(outer(1:n, n:1, "-") == 1)]) is the sum of the sub anti-diagonal.