# Find elements having a specific row and column numbers in a matrix

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Mohammad Shojaei Arani on 21 Dec 2022
Commented: Davide Masiello on 21 Dec 2022
Hello,
I have a simple question but could not find a single command which can do the job. I explain via an example. Consider the following matrix
A = [ 85 45 15 28 11 4 59
79 81 87 32 94 10 85
90 79 64 62 93 25 62
29 41 82 66 100 16 72
37 33 72 59 22 41 98];
Assume that r = [2 5 2 2 5 2 1], c = [3 7 7 1 2 1 6]; What I want, are 7 elements of A whose first component comes from vector r but whose second component comes from vector c. I mean, A(2,3), A(5,7), A(2,7), A(2,1),A(5,2), A(2,1), A(1,6). Of course, I can solve this in a for-loop which is time conssumming. In Matlab, if I write A(r,c) then I get something else which I do not want. The answer to my question is actually the diagonal elements of this matric, that is to say,
diag(A(r,c))
but, I wonder if this is the fastest approach?
Do you know a better way?
Babak
##### 3 CommentsShow 1 older commentHide 1 older comment
Torsten on 21 Dec 2022
Interestingly, the solution from @Davide Masiello seems to be the fastest:
A = [85,45,15,28,11,4,59;79,81,87,32,94,10,85;90,79,64,62,93,25,62;29,41,82,66,100,16,72;37,33,72,59,22,41,98];
r = [2,5,2,2,5,2,1];
c = [3,7,7,1,2,1,6];
tic
v = A(size(A,1)*(c-1)+r);
toc
Elapsed time is 0.001910 seconds.
Davide Masiello on 21 Dec 2022
I suspect sub2ind might just apply the little string of code I devised in my answer, or something very similar to it.

Torsten on 21 Dec 2022
A = [ 85 45 15 28 11 4 59
79 81 87 32 94 10 85
90 79 64 62 93 25 62
29 41 82 66 100 16 72
37 33 72 59 22 41 98];
r = [2 5 2 2 5 2 1];
c = [3 7 7 1 2 1 6];
A(sub2ind(size(A),r,c))
ans = 1×7
87 98 85 79 33 79 4
Mohammad Shojaei Arani on 21 Dec 2022
Hi Torsten,
Beautiful!

Davide Masiello on 21 Dec 2022
If the answer is the diagonal, then I'd use the diag function.
However, if you are still interested in generalizing the code, you could do
A = [ 85 45 15 28 11 4 59
79 81 87 32 94 10 85
90 79 64 62 93 25 62
29 41 82 66 100 16 72
37 33 72 59 22 41 98];
r = [2 5 2 2 5 2 1];
c = [3 7 7 1 2 1 6];
A(size(A,1)*(c-1)+r)
ans = 1×7
87 98 85 79 33 79 4
Mohammad Shojaei Arani on 21 Dec 2022
Nice! Thanks Davide!