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

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?

Thanks in advance!

Babak

[Stephen23]

"Of course, I can solve this in a for-loop which is time conssumming"

I doubt that. Why do you think that?

"I wonder if this is the fastest approach?"

Probably a loop. Lets compare against using SUB2IND (which itself is a perfectly reasonable approach):

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 = c; % preallocate
for k = 1:numel(v)
    v(k) = A(r(k),c(k));
end
toc

Elapsed time is 0.003391 seconds.

tic
w = A(sub2ind(size(A),r,c));
toc

Elapsed time is 0.004914 seconds.

isequal(v,w)

ans = logical
   1

[Torsten]

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]

I suspect sub2ind might just apply the little string of code I devised in my answer, or something very similar to it.

Answer 1

[Torsten]

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]

Hi Torsten,

Beautiful!

Answer 2

[Davide Masiello]

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]

Nice! Thanks Davide!