Binning regions of a 2D matrix slices (and extending this into the third dimension)

Hello,

I am having an issue binning regions of a 3D matrix (but I want the binning to occur in the X-Y dimension for each Z slice). Let's say I have a matrix of 512x128x1200, each 2D slice has dimension 512x128, but say I want to bin the pixel counts such that we create a 2D sliced 32x32 matrix (each element being the sum of 16x4 regions within the original larger matrix, is this possible? Then if we can perform this binning in 2D I want to extend this to each element in the Z-direction

It also would be nice then in future to have control of the X/Y bin 'width' so that the matrix can be reshaped into other square orientations, I have just chosen 32x32 as an arbritary starting point!

Hopefully that makes sense!

Cheers

Answer 1

[Bruno Luong]

This assumes the first/second dimensions of your matrix are divisible by 32

% Test matrix
A=rand(512,128,1200);

binsz = 32;
[m,n,p] = size(A);
mr = m/binsz;
nr = n/binsz;
B = reshape(A,mr,binsz,nr,binsz,p);
B = sum(B,[1 3]);
B = reshape(B,binsz,binsz,p);

[Eunan McShane]

Thank you! Excellent, and can adapt to take into account some NaN data within the matrix

Cheers!

[Bruno Luong]

See argument 'omitnan'of SUM.

Alternatively you can replace NaN by 0

B(isnan(B))=0

before the sum

[Matt J]

Bruno's technique is pre-implemented for you in sepblockfun

https://www.mathworks.com/matlabcentral/fileexchange/48089-separable-block-wise-operations

B=sepblockfun(A,[16,4,1],@(x,d) sum(x,d,'omitnan'))

Answer 2

[Alan Moses]

Hi Eunan,

You can use the convn function to help you achieve this. Please look at the below example where I have taken ‘X’ as the large matrix and reshaped it into a smaller matrix ‘Z’:

X = 2*ones(512,128,3);
x_width = 16;
y_width = 4;

Y = convn(X,ones(x_width,y_width),'valid');
Z = Y(1:x_width:end,1:y_width:end,:); %Z dimensions – 32x32x3

Hope this helps!

[Eunan McShane]

Thank you! This does work, however (I should have said) there is a portion of the data which is NaN and so I want to ignore these entries in the sum, the other answer allows me to use nansum to this end!

Cheers though this is a really useful function which I can use for other applications!