# Sum a matrix element using a window size of 4

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Rahul Gulia il 19 Giu 2023
Commentato: Rahul Gulia il 21 Giu 2023
Hi everyone,
I am trying to create a matrix(4,2) from a matrix (4,8), by adding 4 elements column wise.
For example:
A = [-2 -2 -2 2 -2 2 -2 -2
-2 2 -2 -2 -2 -2 -2 2
-2 -2 2 -2 -2 2 2 2
-2 2 2 2 -2 -2 2 -2]
should be converted to:
B = [-4 -4
-4 -4
-4 4
4 -4 ]
Looking forward to any kind of suggestion. I was trying a for loop, but that is kind of messy for a large matrix. Hoping to get some simpler solutions by directly using the sum() function. Kindly sugest.
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### Risposta accettata

Image Analyst il 19 Giu 2023
Try blockproc if you have the Image Processing Toolbox:
A = [-2 -2 -2 2 -2 2 -2 -2
-2 2 -2 -2 -2 -2 -2 2
-2 -2 2 -2 -2 2 2 2
-2 2 2 2 -2 -2 2 -2]
A = 4×8
-2 -2 -2 2 -2 2 -2 -2 -2 2 -2 -2 -2 -2 -2 2 -2 -2 2 -2 -2 2 2 2 -2 2 2 2 -2 -2 2 -2
sumFunction = @(theBlockStructure) sum(theBlockStructure.data(:));
B = blockproc(A, [1, 4], sumFunction)
B = 4×2
-4 -4 -4 -4 -4 4 4 -4
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Rahul Gulia il 21 Giu 2023
Thank you @Image Analyst.

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### Più risposte (1)

John D'Errico il 19 Giu 2023
Modificato: John D'Errico il 19 Giu 2023
Reshape the array, to be now of size 4x4x2
A = [-2 -2 -2 2 -2 2 -2 -2
-2 2 -2 -2 -2 -2 -2 2
-2 -2 2 -2 -2 2 2 2
-2 2 2 2 -2 -2 2 -2];
Think about the result. What will it look like?
A2 = reshape(A,[4,4,2])
A2 =
A2(:,:,1) = -2 -2 -2 2 -2 2 -2 -2 -2 -2 2 -2 -2 2 2 2 A2(:,:,2) = -2 2 -2 -2 -2 -2 -2 2 -2 2 2 2 -2 -2 2 -2
Now, can you sum that new array, along the SECOND dimension? TRY IT!
A2sum = sum(A2,2)
A2sum =
A2sum(:,:,1) = -4 -4 -4 4 A2sum(:,:,2) = -4 -4 4 -4
Are you getting close? I hope so. Next, all you need to do is to get rid of that pesky second, singleton dimension. For that, you could use either reshape, or squeeze. TRY IT!
The trick in all of these problems is to understand the sequence in which the elements in your array are stored , and to then visualize what you want in the end. Work towards that goal.
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Rahul Gulia il 20 Giu 2023
Thank you for your suggestion @John D'Errico. That is an interesting way to look at the problem. I would surely apply that for smaller size matrices.

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