Sum a matrix element using a window size of 4

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.

Answer 1

[Image Analyst]

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

[Rahul Gulia]

Thank you @Image Analyst.

Answer 2

[John D'Errico]

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.

[Rahul Gulia]

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.