How to use for loop ?

Let's say I have a matrix A = zeros(1000,500). I have two another matrices of size 2 X 50 each for example B = [1 2 3 .... 50 ; 2 4 6 .... 100] and C = [3 6 9 .... 150 ; 4 8 12 ....200]. Let's say I have another matrix D of same size 2 X 50 with some numbers for example [2 4 2 ... 67 ; 6 9 3 .... 54]. Now I have add up all these random numbers in matrix D into the matrix A only at the positions dictated by the matrix B and C. For example the first row in matrix B and C as in [1 2 3 .... 50] and [3 6 9 .... 150] are the x and y positions/co-ordinate where the first row of random numbers as in [2 4 2 .... 67] are to be added with matrix A. The same thing goes for the second row also. Clearly, I have to add [2 4 2 .... 67] to the matrix A at the positions [(1,3) (2,6) (3,9) .... (50,150)] and similarly add up [6 9 3 .... 54] to the matrix A at the positions [(2,4) (4,8) (6,12) .... (100,200)].

How can I use for loop to code this ?

Answer 1

[Jan]

It does not matter, if B and C are matrices, but you care about the value of the elements only - correctly?

for k = 1:numel(B)
  A(B(k), C(K)) = A(B(k), C(K)) + D(k);
end

This seems to be very easy. Did you try anything else?

[Sudharsan Srinivasan]

Hi Jan, I used the same as technique as you have mentioned. But what does 'nume1' in your code mean ?

[Rik]

That isn't the number 1, it is the lowercase L. The function numel counts the number of elements of a matrix. It is therefore equivalent to prod(size(B)) or length(B(:))

Answer 2

[Rik]

Why insist on a loop?

A(sub2ind(size(A), B(:), C(:))) = D(:);

That should do the same thing.

Edit: the difference between my solution and Jan's is that mine doesn't work if the indices in B an C are non-unique combinations, which you could check with this:

length(unique([B(:) C(:)],'rows'))==numel(B)

If that is false, you can't use my solution.

if length(unique([B(:) C(:)],'rows'))==numel(B)
    %A(B(:),C(:))=D(:);
    A(sub2ind(size(A), B(:), C(:))) = D(:);
else
    for k = 1:numel(B)
        A(B(k), C(K)) = A(B(k), C(K)) + D(k);
    end
end

[Jan]

See this:

A = zeros(2, 2);
A([1,2], [1,2]) = 1;

This is not

A = [1, 0; ...
     0, 1];

but

A = [1, 1; ...
     1, 1];

So you need at least:

   A(sub2ind(size(A), B(:), C(:))) = D(:);

[Rik]

Good point. I'll edit my answer, thanks for the correction. Every time I make this mistake I think it's odd for Matlab to work in the way it does.

[Jan]

@Rik: I agree that it would be more intuitive to process the indices serially instead of accessing a rectangular sub-matrix.