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Combination of rows of two different matrices

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J AI
J AI on 23 Jul 2020
Commented: Fangjun Jiang on 24 Jul 2020
Sorry if this is a repeated question but I failed to find an answer myself. I have two matrices:
A = [-0.6, -0.2;
-60, 2;
6, -20];
B = [-0.4, -0.8;
-40, 8;
4, -80];
I want to find all the possible combinations of sum of each row (sum of individual elements of a row) of A with each row of B, i.e., my desired result is (order does not matter):
ans = [-1, -1;
-40.6, 7.8;
3.4, -80.2;
-60.4, 1.2;
-100, 10;
-56, -78;
5.6, -20.8;
-34, -12;
10, -100];
which is a matrix resulting from possible combinations of A and B.
Thanks in advance.
(Please no for loops. It is pretty trivial then.)
EDIT: I have used 2 columns and 3 rows as an example. Looking for more general solution, i.e., for n number of columns and m number of rows.

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Accepted Answer

Fangjun Jiang
Fangjun Jiang on 23 Jul 2020
Edited: Fangjun Jiang on 23 Jul 2020
Feels non-ideal. Any better solution?
>> C=A(:,1)+B(:,1)';
D=A(:,2)+B(:,2)';
reshape([C(:),D(:)],[],2)
ans =
-1.0000 -1.0000
-60.4000 1.2000
5.6000 -20.8000
-40.6000 7.8000
-100.0000 10.0000
-34.0000 -12.0000
3.4000 -80.2000
-56.0000 -78.0000
10.0000 -100.0000
better one
>> m=size(A,1);
ind1=repmat(1:m,1,m);
ind2=repelem(1:m,m);
A(ind1,:)+B(ind2,:)
ans =
-1.0000 -1.0000
-60.4000 1.2000
5.6000 -20.8000
-40.6000 7.8000
-100.0000 10.0000
-34.0000 -12.0000
3.4000 -80.2000
-56.0000 -78.0000
10.0000 -100.0000

  2 Comments

J AI
J AI on 23 Jul 2020
This works, but I should have been a bit more specific with my question - the number of columns I have used is 2, however, it may vary. So I was looking for a more general solution. Thanks anyway!

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More Answers (1)

Bruno Luong
Bruno Luong on 23 Jul 2020
Edited: Bruno Luong on 23 Jul 2020
A = [-0.6, -0.2;
-60, 2;
6, -20];
B = [-0.4, -0.8;
-40, 8;
4, -80];
Single statement
reshape(permute(A,[3 1 2])+permute(B,[1 3 2]),[],size(A,2))
or a variation
reshape(reshape(A,1,size(A,1),[])+reshape(B,size(B,1),1,[]),[],size(A,2))
Gives
ans =
-1.0000 -1.0000
-40.6000 7.8000
3.4000 -80.2000
-60.4000 1.2000
-100.0000 10.0000
-56.0000 -78.0000
5.6000 -20.8000
-34.0000 -12.0000
10.0000 -100.0000
>>

  3 Comments

J AI
J AI on 23 Jul 2020
These ones are actually faster for larger data - the variation being the fastest. Thanks a lot for your contribution!
Bruno Luong
Bruno Luong on 23 Jul 2020
Yes the variation version does no more no less than the required combination sums and puts at the results at the right place. Not a hair uneccesary arithmetic or memory moving (first version).
Fangjun Jiang
Fangjun Jiang on 24 Jul 2020
Brilliant, Bruno Luong! I learned implicit expansion in a whole new dimension.

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