Matrix moving mean with overflow average

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!

Answer 1

[Matt J]

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

[Albert Zurita]

Thanks!! No, I don't want the wrap at the lower-right corner. Should I replace the numbers by NaN? In addition, for a rectangular window, the circshift I assume should change to accommodate the rows/columns definition of the window?

[Matt J]

Should I replace the numbers by NaN?

That's not for me to decide. However, you could do that and also omit nans from the mean calculation in this alternative forumulation:

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(end,end-1:end)=nan

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 NaN NaN

t=movmean(B,3,2,'omitnan','End','discard');
slidingMeans=movmean(t,3,1,'End','discard')

slidingMeans = 2×6

8.0000 9.0000 10.0000 11.0000 12.0000 13.0000 14.0000 15.0000 16.0000 17.0000 17.8333 18.6667

In addition, for a rectangular window, the circshift I assume should change to accommodate the rows/columns definition of the window?

Yes indeed.

[Albert Zurita]

Thanks for the tips. I have modified the code to adapt to a rectangular window as follows:

win = [3 5]; % rows, cols
A=reshape(1:24,[],4)'
B=[A,circshift(A(:,1:win(2)),-1)]
B(end,end-1:end)=nan
t=movmean(B,win(2),2,'omitnan','End','discard');
slidingMeans=movmean(t,win(1),1,'End','discard')

However, I am not fully convinced of the -1 in the circhift (if that is dependent on the window size), and with respect to the NaNs, I just want to stop averaging at the lower right border when I cannot complete the window box. To be illustrative I am exactly looking after obtaining the averages of the following boxes, for a window of [3 rows, 5 columns] and as a result obtain 8 numbers resulting from the 8 sequences below. Thanks!

[Matt J]

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]

This is excellent, and good methodology I learnt, thanks!