Julia - How to efficiently turn to zero the diagonal of a matrix?

Question

In Julia, what would be an efficient way of turning the diagonal of a matrix to zero?

Answer 1

Assuming m is your matrix of size N x N this could be done as:

setindex!.(Ref(m), 0.0, 1:N, 1:N)

Another option:

using LinearAlgebra
m[diagind(m)] .= 0.0

And some performance tests:

julia> using LinearAlgebra, BenchmarkTools

julia> m=rand(20,20);

julia> @btime setindex!.(Ref($m), 0.0, 1:20, 1:20);
  55.533 ns (1 allocation: 240 bytes)

julia> @btime $m[diagind($m)] .= 0.0;
  75.386 ns (2 allocations: 80 bytes)

Answer 2

Performance wise, simple loop is faster (and more explicit, but it is taste dependent)

julia> @btime foreach(i -> $m[i, i] = 0, 1:20)
  11.881 ns (0 allocations: 0 bytes)

julia> @btime setindex!.(Ref($m), 0.0, 1:20, 1:20);
  50.923 ns (1 allocation: 240 bytes)

And it is faster then diagind version, but not by much

julia> m = rand(1000, 1000);

julia> @btime foreach(i -> $m[i, i] = 0.0, 1:1000)
  1.456 μs (0 allocations: 0 bytes)

julia> @btime foreach(i -> @inbounds($m[i, i] = 0.0), 1:1000)
  1.338 μs (0 allocations: 0 bytes)

julia> @btime $m[diagind($m)] .= 0.0;
  1.495 μs (2 allocations: 80 bytes)

Answer 3

Przemyslaw Szufel's solutions bench-marked for 1_000 x 1_000 matrix size show that diagind performs best:

julia> @btime setindex!.(Ref($m), 0.0, 1:1_000, 1:1_000);
  2.222 μs (1 allocation: 7.94 KiB)

julia> @btime $m[diagind($m)] .= 0.0;
  1.280 μs (2 allocations: 80 bytes)

Answer 4

Here is a more general way how to performantly use setindex! on an Array by accessing custom indices: Using an array for in array indexing

Linear indexing is best performing, that's why diagind runs better than the Cartesian Indices.