Remove the loop from a sum of index expression

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Consider the following simple example:
% Example setup
n = 5;
m = 10;
M = randn( m, m );
J = randi( n, m, m );
% Code to simplify
V = zeros( n, 1 );
for i = 1 : n
V( i ) = sum( M( J == i ) );
end
Is it possible to replace the loop by a one line expression?
For the sake of answering this, assume that I do not particularly care about either efficiency, or readability. I really just want a one line solution.
The actual code does not have random M and J, so I am also not interested in solutions based on these particular distributions.
  2 Comments
Jiri Hajek
Jiri Hajek on 10 Nov 2022
Hi, your example code produces a vector in each step of the loop, is that a typo? Please clarify, what is the desired result.
Tom Holden
Tom Holden on 10 Nov 2022
No it doesn't. The sum collapses the vector.

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Accepted Answer

Stephen23
Stephen23 on 10 Nov 2022
Edited: Stephen23 on 10 Nov 2022
Hiding the loop even more (really everything uses loops, even vectorized code):
% Example setup
n = 5;
m = 10;
M = randn( m, m );
J = randi( n, m, m );
% Code to simplify
V = zeros( n, 1 );
for k = 1:n
V(k) = sum(M(J==k));
end
V
V = 5×1
-8.9895 -0.9206 -1.5965 0.6393 -3.7407
% Simpler code:
V = accumarray(J(:),M(:),[n,1])
V = 5×1
-8.9895 -0.9206 -1.5965 0.6393 -3.7407
  1 Comment
Tom Holden
Tom Holden on 10 Nov 2022
Somehow I have never encountered "accumarray" before. But it is perfect here.

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More Answers (1)

Torsten
Torsten on 10 Nov 2022
Here is a loop in disguise:
n = 5;
m = 10;
M = randn( m, m );
J = randi( n, m, m );
V = arrayfun(@(i) sum( M( J == i ) ),1:n)
  1 Comment
Tom Holden
Tom Holden on 10 Nov 2022
A fair answer! It still feels like there ought to be a solution that is less explicitly a loop than this.

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