Filling a matrix using vectorization

Hello, I am trying to fill specific indices of a matrix, and I am using a for loop to do this, here is a small example code snippet:

r=2;
for s = 2:119
    nnum(r,s)=(r-1)*120+s;
    M(nnum(r,s),nnum(r,s)-1) = -1;
    M(nnum(r,s),nnum(r,s)+1) = -2;
    M(nnum(r,s),nnum(r,s)-120) = -3;
    M(nnum(r,s),nnum(r,s)+120) = -4;
    M(nnum(r,s),nnum(r,s))   =  5;
end

Now I have a big program with a lot of operations like the one above, and I wanted to reduce the computation time by using the vectorization to fill my matrices. Here is the code I tried:

r=2;
s = 2:119;
k=(r-1)*120+s;

F(k,k-1) = -1;
F(k,k+1) = -2;
F(k,k-120) = -3;
F(k,k+120) = -4;
F(k,k) = 5;

Now I thought that F and M would be equal but they are not. The thing is that I want only the first element of the vector in the first matrix index to couple with the first element of the vector in the second matrix index and so on. What is happening here is thet each element of the vector is coupling with all the elements from the other vector to form indices. Any ideas on how I can achieve the first code snippet without the use of a for loop?

[Nikita Agrawal]

These two are not same because the code runs in a sequential order and you are overwriting some cells in everystep in F. For eg: [k,k-1] & [k,k+1] could be same for many k when k is a variable.

[Haydar Dyab]

Yeah I knew the problem but couldn't find a way to solve it, thanks anw!

Answer 1

[Rik]

The problem is that this:

A([1 2],[1 3])

returns 4 positions, not two. If you want two instead, you will need to convert subindices to linear indices:

ind=sub2ind(size(A),[1 2],[1 3]);
A(ind)

[Haydar Dyab]

Yeah using linear indices works. Thanks!!