How to update all diagonal values in Matrex matrix

Question

I work with Elixir Matrex (https://hexdocs.pm/matrex/Matrex.html) and want to update or set all diagonals elements to zeros (and later to ones).

I tried:

matrix_size = 5
randommatrix = Matrex.random(matrix_size)

for index_i <- 1 .. matrix_size, do: randommatrix = Matrex.update(randommatrix, index_i, index_i, fn a -> 0 end)

#subsequent work with matrix that has all diagonal elements zero
#...

This approach updates randommatrix only "inside" comprehension and the subsequent code sees the randommatrix unchanged.

I think that I can do what I need through recursive function, however, is there any other efficient method (minimal computer time)? Thanks

Answer 1

If you don't want to use recursion, you could use the apply/2 method. This allows you to apply a function to every element of a matrix. The update function gets the respective value and index. This way we can easily update all values on the diagonal.

Example: For a 5x5 matrix the following indices are to be updated: [1, 7, 13, 19, 25]. So if we take the index modulo 6 (matrix_size + 1), the remainder is always 1.

matrix_size = 5
matrix = Matrex.random(matrix_size)
# Update diagonal to zero
Matrex.apply(matrix, fn val, index -> if rem(index, matrix_size+1) == 1 do 0 else val end)

Answer 2

Why don't you want to make it a recursion? I think that would be an elegant way to solve this problem.

Something like this (code is not tested):

update(matrix, 0), do: matrix
update(matrix, index) do
    update(
        Matrex.update(matrix, index, index, fn a -> 0 end),
        index-1
    )
end

Example call: udpate(matrix, 5)

First the elment at position (5,5) is updated. Thereby a new matrix is created. With this new matrix we go into recursion. Next, the element at position (4,4) is updated - and so on.

The termination condition is done with pattern matching on the second argument. So when counting down to 0, nothing more is done than returning the finished matrix.