Matrix moving mean with overflow average

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Albert Zurita
Albert Zurita 2022-11-13
评论： Albert Zurita 2022-11-15
I am wondering if there is a (simply) way when using movmean for controlling row overflow and for specifying a rectangular window instead of a square window. The idea would be to keep averaging the patch corresponding to each number in red until the end of row is found, and simply jump to the next row. Thanks!

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Matt J
Matt J 2022-11-13
编辑：Matt J 2022-11-13
Ran in:
Here's one way. I assumed here you want the same wrap-around to occur in the lower-right corner of the matrix as well.
A=reshape(1:24,[],4)'
A = 4×6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
B=[A,circshift(A(:,1:2),-1) ]
B = 4×8
1 2 3 4 5 6 7 8 7 8 9 10 11 12 13 14 13 14 15 16 17 18 19 20 19 20 21 22 23 24 1 2
slidingMeans=conv2(B,ones(3)/9,'valid')
slidingMeans = 2×6
8.0000 9.0000 10.0000 11.0000 12.0000 13.0000 14.0000 15.0000 16.0000 17.0000 15.3333 13.6667
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Matt J
Matt J 2022-11-15
Ran in:
win = [3 5]; % rows, cols
A=reshape(1:24,[],4)'
A = 4×6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
B=[A,circshift(A(:,1:win(2)-1),-1)];
B(end,end-win(2)+2:end)=nan
B = 4×10
1 2 3 4 5 6 7 8 9 10 7 8 9 10 11 12 13 14 15 16 13 14 15 16 17 18 19 20 21 22 19 20 21 22 23 24 NaN NaN NaN NaN
slidingMeans=conv2(B,ones(win)/prod(win),'valid' )
slidingMeans = 2×6
9.0000 10.0000 11.0000 12.0000 13.0000 14.0000 15.0000 16.0000 NaN NaN NaN NaN
Albert Zurita
Albert Zurita 2022-11-15
This is excellent, and good methodology I learnt, thanks!

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