Blockwise matrix addition without using more than 2 dimensions and cycles

2 Ansichten (letzte 30 Tage)
Dear All,
I am looking for a solution of blockwise matrix addition. To give a simple example, let us have
A = [1,2,3,4,5,6,7,8; 9,10,11,12,13,14,15,16];
I would like to get the result:
B = [1+5,2+6,3+7,4+8; 9+13,10+14,11+15,12+16];
The evident solution which comes to mind would be:
sum(reshape(A,2,4,2), 3);
However, I am using a third-party tool with special variables for which 3D arrays are not implemented. My next idea was to use:
B = zeros(2,4);
for i = 1:4
B(:,i) = sum(A(:, i:4:end), 2);
end
If possible I am looking for functions that can solve this without using a cycle as in my application the number of columns will reach hundreds and thousands. I thank for in advance for anyone's help.

Akzeptierte Antwort

the cyclist
the cyclist am 22 Nov. 2022
This is awkward, but it does what you want. I think there is probably a better way.
% Inputs
A = [1,2,3,4,5,6,7,8; 9,10,11,12,13,14,15,16];
groupSize = 4;
% Output
B = movsum(A,[0,groupSize],2) - movsum(A,[0,groupSize-1],2) + A;
B = B(:,1:groupSize)
B = 2×4
6 8 10 12 22 24 26 28
  1 Kommentar
Bence Cseppento
Bence Cseppento am 23 Nov. 2022
Thanks for your effort, it does look good but I'll have to test it for some other cases.
In the mean time I managed to solve my problem by managing a conversion between types and using the original 3D version, but I did not know about the movsum() function before so you have given me food for thought, thank you!

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Weitere Antworten (1)

the cyclist
the cyclist am 21 Nov. 2022
A = [1,2,3,4,5,6,7,8; 9,10,11,12,13,14,15,16];
B = A(:,1:end/2) + A(:,(end/2)+1:end)
B = 2×4
6 8 10 12 22 24 26 28
  1 Kommentar
Bence Cseppento
Bence Cseppento am 21 Nov. 2022
Thank you, I was a bit over-complicating it because of the context of my problem which I have not shared. However, I am still looking for a more general solution, e.g. I want to get a 2x20 matrix from a 2x2000 matrix, by summing the (:,1:20), (:,21:40) ... (:,1981:2000) submatrices without the use of a cycle.

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