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How to divide a 2d array into blocks and create an array with the mean value of each block?

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Here's an 4x8 array.Three of its values are NaNs:
B=zeros(4,8);
B(1,:) = [1 3 4 3 10 0 3 4];
B(2,:) = [5 7 2 NaN 8 2 3 4];
B(3,:) = [8 4 6 2 NaN 3 5 7];
B(4,:) = [2 2 5 5 1 NaN 4 4];
I want to divide the array into 2x2 blocks and form a new array 'C' that is the mean of each block. If the block contains one or more NaNs, I want to find the average of the numbers that are not NaNs. In this example, I would like the following output:
C =
4.0000 3.0000 5.0000 3.5000
4.0000 4.5000 2.0000 5.0000
I tried the code below but it just returns a NaN if any of the block elements are NaN. How should I amend this, please?
C=blockproc(B, [2 2], @(block_struct) mean(block_struct.data(:)))
C =
4.0000 NaN 5.0000 3.5000
4.0000 4.5000 NaN 5.0000

Risposta accettata

Jan
Jan il 26 Mag 2022
Modificato: Jan il 26 Mag 2022
B = [1 3 4 3 10 0 3 4; ...
5 7 2 NaN 8 2 3 4; ...
8 4 6 2 NaN 3 5 7; ...
2 2 5 5 1 NaN 4 4];
B = repmat(reshape(B, 1, 4, 8), 10, 1, 1); % The test data
% Code without BLOCKPROC:
N = ~isnan(B); % Mask for the NaNs
S = size(B);
T = [S(1), 2, S(2)/2, 2, S(3)/2];
BB = reshape(B .* N, T); % Set NaNs to 0
NN = reshape(N, T);
R = sum(BB, [2, 4]) ./ sum(NN, [2, 4]); % Sum over 2x2 blocks
R = reshape(R, T([1,3,5])); % Remove singelton dimensions
  2 Commenti
Steve Francis
Steve Francis il 26 Mag 2022
Thanks, Jan. Is the advantage of this method (a) speed and (b) for users without the image processing toolbox? Is there any other disadvantage in using blockproc?
Jan
Jan il 27 Mag 2022
You are welcome. I do not have the image processing toolbox, so I use solutions without blockproc.
The method shown above is fast and more flexibel, if the inputs are no 2D or 3D RGB arrays and the blocks are built on the 1st 2 dmensions. Instead of ~isnan() other masks can be defined. blockproc is more powerful, when specific functions are applied to the blocks, but power costs processing time.
See also: https://www.mathworks.com/matlabcentral/fileexchange/24812-blockmean (no masking of NaNs impelemented, frist 2 dimensions processed also, dimensions do not need to be a multiple of the window size)

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Più risposte (1)

David Hill
David Hill il 26 Mag 2022
C=blockproc(B, [2 2], @(block_struct) mean(block_struct.data(:),'omitnan'))
  2 Commenti
Steve Francis
Steve Francis il 26 Mag 2022
Thanks for this @David Hill. Just to widen the question, if B was a 10x4x8 array (i.e. 10 'layers' of 4x8 arrays), I could create a corresponding C (10 layers of 2x4 arrays) using the following code.
C = zeros(10, 2, 4);
for n=1:10
C(n,:,:) = blockproc(squeeze(B(n,:,:)), [2 2], @(block_struct) mean(block_struct.data(:),'omitnan'));
end
Is there a way to do it without the 'for' loop?
Steve Francis
Steve Francis il 27 Mag 2022
David's solution directly answered my original question. Jan's answer introduced me to an intriguing alternative. Thanks to both.

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