2D sliding/moving/running average window

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Albert Zurita 2022-7-30

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评论： Bruno Luong 2022-8-25

Hello, I want to perform a matrix 'patch' moving average but I am not sure how to. So, for example, defining my moving window as 3 x 3. I would like, for each element in the matrix, to perform the average of the elements as in the sequence below (excerpt) for an exemplary 6 x 10 matrix. I know this can be done more efficiently through the use of conv2 or in frequency domain with fft2, using the window kernel, but I still want to do it with a for loop, as this will be later translated to another real time language code. In addition, I would like to be able to select the elements as a 1D array (a reshape of the n_rows, n_cols matrix). At the edges of the matrix I select the next row, so it probably makes more sense to do it as a 1D array. Finally, when reaching the end of the matrix I can only select a few elements in the corner, but in this case I will do the average just of those elements.

thanks!!

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Matt J 2022-7-30

I know this can be done more efficiently through the use of conv2 or in frequency domain with fft2, using the window kernel, but I still want to do it with a for loop

It sounds like you already know how you're going to do it, so there doesn't seem to be a question here.

Albert Zurita 2022-7-30

Yes, but I don't know how to select the elements, and especially given the 'corners' effect problem...

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Answer 1

Bruno Luong 2022-7-30

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编辑：Bruno Luong 2022-7-30

Just some hint if your want to implement efficiently using loop :

To do mean, you might do sliding sum onf the data, then divide by sliding sum on the array of size data but values are replaced with 1s (ones(size(data)).
The sliding 2D sum is "separable" meaning you just need to do sliding sum along one dimension, and apply the sliding sum along other dimension.
The sliding sum in 1D can be donne efficiently by simply adding sum of the previous step with the new entry element and substract the one that exits the window. If ther data is "quantified" such as multiples of the 2^power something, this method does not have cumulative error, otherwise you might have a small cumulative error compared to standard sum on the whole windows.

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Bruno Luong 2022-7-31

编辑：Bruno Luong 2022-7-31

在 MATLAB Online 中打开

I don't see yet you showing any effort to make your code/modification after many guidances, wonder how you translate the code to "realtime" language. But here we go :

% Create test data

A = peaks;

A = A + 0.1*randn(size(A));

win = [7 7];

B = slidingmean(A, win);

subplot(1,2,1)

surf(A)

title('Original')

subplot(1,2,2)

surf(B)

title('Sliding mean')

%%

function B = slidingmean(A, win)

B = slidingsum2(A, win) ./ slidingsum2(ones(size(A)), win);

end

%%

function B = slidingsum2(A, win)

if isscalar(win)

win = [win,win];

end

B = slidingsum(A, 1, win(1));

B = slidingsum(B, 2, win(2));

end

%%

function B = slidingsum(A, dim, win)

m = ndims(A);

szA = size(A);

nA = szA(dim);

wh = floor(win/2);

wt = win-wh-1;

B = zeros(size(A));

idx = repmat({':'},1,m);

szS = szA;

szS(dim) = 1;

S = zeros(szS);

for k=1-wh:nA+wt

in = k+wh;

if in <= nA

idx{dim} = in;

S = S + A(idx{:});

end

out = k-wt-1;

if out >= 1

idx{dim} = out;

S = S - A(idx{:});

end

if k>=1 && k<=nA

idx{dim} = k;

B(idx{:}) = S;

end

Bruno Luong 2022-8-1

编辑：Bruno Luong 2022-8-1

在 MATLAB Online 中打开

I see now that you want to do the wrap around with increment+1 in the row.

Instead of doing some kind of weird algorithm you could simply apply the normal algorithm on pad array on the right side by the left side:

All the specific wrap around is done in the way you pad your data, and leave the algorithm untouched.

A=zeros(4,6);
A(1:numel(A))=1:numel(A);
A
A = 4×6
     1     5     9    13    17    21
     2     6    10    14    18    22
     3     7    11    15    19    23
     4     8    12    16    20    24
win=3;
Apad=[A, [A(2:end,1:win-1); nan(1,win-1)]],
Apad = 4×8
     1     5     9    13    17    21     2     6
     2     6    10    14    18    22     3     7
     3     7    11    15    19    23     4     8
     4     8    12    16    20    24   NaN   NaN
% Doing average on Apad then remove the exceed
Bpad = slidingmean(Apad, win);
B = Bpad(:,1:size(A,2)),
B = 4×6
    3.5000    5.5000    9.5000   13.5000   17.5000   13.8333
    4.0000    6.0000   10.0000   14.0000   18.0000   14.3333
    5.0000    7.0000   11.0000   15.0000   19.0000   16.6250
    5.5000    7.5000   11.5000   15.5000   19.5000   18.0000

Note that if you want to discard NaN in the average you need to change the normalization from

... ./ slidingsum2(ones(size(A)), win)

to

... ./ slidingsum2(isfinite(A), win)

%%

function B = slidingmean(A, win)
valid = ~isnan(A);
A(~valid) = 0;
B = slidingsum2(A, win) ./ slidingsum2(valid, win);
end
%%
function B = slidingsum2(A, win)
if isscalar(win)
    win = [win,win];
end
B = slidingsum(A, 1, win(1));
B = slidingsum(B, 2, win(2));
end
%%
function B = slidingsum(A, dim, win)
m = ndims(A);
szA = size(A);
nA = szA(dim);
wh = floor(win/2);
wt = win-wh-1;
B = zeros(size(A));
idx = repmat({':'},1,m);
szS = szA;
szS(dim) = 1;
S = zeros(szS);
for k=1-wh:nA+wt
    in = k+wh;
    if in <= nA 
        idx{dim} = in;
        S = S + A(idx{:});
    end
    out = k-wt-1;
    if out >= 1
        idx{dim} = out;
        S = S - A(idx{:});
    end
    if k>=1 && k<=nA
        idx{dim} = k;
        B(idx{:}) = S;
    end
end
end

Bruno Luong 2022-8-1

编辑：Bruno Luong 2022-8-1

在 MATLAB Online 中打开

Cheese, I hope you should know by now what you want with this weird wrapping and make the Padding exactly like you want it to be. This is the LAST time I adapt the code for you.

I warn you, next question along special case on top level, this answer will be deleted, because I don't have any patient to follow you, sorry.

A=(1:10)+(0:10:50)'
A = 6×10
     1     2     3     4     5     6     7     8     9    10
    11    12    13    14    15    16    17    18    19    20
    21    22    23    24    25    26    27    28    29    30
    31    32    33    34    35    36    37    38    39    40
    41    42    43    44    45    46    47    48    49    50
    51    52    53    54    55    56    57    58    59    60
win = 3;
Apad = [[nan(1,win-1); A(1:end,end-win+2:end)], [A; nan(1,size(A,2))], [A(2:end,1:win-1); nan(2,win-1)]]
Apad = 7×14
   NaN   NaN     1     2     3     4     5     6     7     8     9    10    11    12
     9    10    11    12    13    14    15    16    17    18    19    20    21    22
    19    20    21    22    23    24    25    26    27    28    29    30    31    32
    29    30    31    32    33    34    35    36    37    38    39    40    41    42
    39    40    41    42    43    44    45    46    47    48    49    50    51    52
    49    50    51    52    53    54    55    56    57    58    59    60   NaN   NaN
    59    60   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN
Bpad = slidingmean(Apad, win);
B = Bpad(1:end-1,win:size(A,2)+win-1)
B = 6×10
    7.2000    7.0000    8.0000    9.0000   10.0000   11.0000   12.0000   13.0000   14.0000   15.0000
   12.3750   12.0000   13.0000   14.0000   15.0000   16.0000   17.0000   18.0000   19.0000   20.0000
   21.0000   22.0000   23.0000   24.0000   25.0000   26.0000   27.0000   28.0000   29.0000   30.0000
   31.0000   32.0000   33.0000   34.0000   35.0000   36.0000   37.0000   38.0000   39.0000   40.0000
   41.0000   42.0000   43.0000   44.0000   45.0000   46.0000   47.0000   48.0000   49.0000   48.6250
   48.0000   47.0000   48.0000   49.0000   50.0000   51.0000   52.0000   53.0000   54.0000   53.8000
function B = slidingmean(A, win)
valid = ~isnan(A);
A(~valid) = 0;
B = slidingsum2(A, win) ./ slidingsum2(valid, win);
end
%%
function B = slidingsum2(A, win)
if isscalar(win)
    win = [win,win];
end
B = slidingsum(A, 1, win(1));
B = slidingsum(B, 2, win(2));
end
%%
function B = slidingsum(A, dim, win)
m = ndims(A);
szA = size(A);
nA = szA(dim);
wh = floor(win/2);
wt = win-wh-1;
B = zeros(size(A));
idx = repmat({':'},1,m);
szS = szA;
szS(dim) = 1;
S = zeros(szS);
for k=1-wh:nA+wt
    in = k+wh;
    if in <= nA 
        idx{dim} = in;
        S = S + A(idx{:});
    end
    out = k-wt-1;
    if out >= 1
        idx{dim} = out;
        S = S - A(idx{:});
    end
    if k>=1 && k<=nA
        idx{dim} = k;
        B(idx{:}) = S;
    end
end
end

Bruno Luong 2022-8-25

编辑：Bruno Luong 2022-8-25

在 MATLAB Online 中打开

"Apad matrix does not seem to pad the number of columns correctly."

Sorry but You still get it wrong again. The correct is

Apad = [[nan(1,win(2)-1); ...
         A(1:end,end-win(2)+2:end)], ...
         [A; nan(1,size(A,2))], ...
       [A(2:end,1:win(2)-1); ...
        nan(2,win(2)-1)]];
Bpad = slidingmean(Apad, win);
B = Bpad(1:end-1,win(2):size(A,2)+win(2)-1);

Up to know you only specify the wrap around on the horizontal direction, so only win(2) matter.

You might expect something different now, and I again insist that you must pad according whatever the weird wrap around you want. It's up to you to change the pad if you change the wrap shiftting that beside you nobody know.

Bruno Luong 2022-8-25

在 MATLAB Online 中打开

Also you might change some of the hard-code first index

nan(1, ...
A(2:...
Bpad(1:end-1 ...

with expression using win(1)

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Answer 2

David Hill 2022-7-30

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https://ww2.mathworks.cn/matlabcentral/answers/1771140-2d-sliding-moving-running-average-window#answer_1018220

在 MATLAB Online 中打开

a=randi(100,15);%whatever initial size of your matrix
b=size(a,1);
m=zeros(b);
for x=1:b^2
  idx=[x-b-1,x-1,x+b-1;x-b,x,x+b;x-b+1,x+1,x+b+1];
  if idx(2,3)>b^2
      idx(:,3)=[];
  end
  if mod(idx(3,2),b)==1
      idx(3,:)=[];
  end
  if idx(2,1)<1
      idx(:,1)=[];
  end
  if mod(idx(1,2),b)==0
      idx(1,:)=[];
  end
  m(x)=mean(a(idx),'all');
end

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Image Analyst 2022-8-24

Albert, did you even see my answer below? Please read it. It does what you asked. You can specify the image file and window size and is 100% manual (non-vectorized). At least give it a try even if you prefer Bruno's answer (which has vectorized indexing).

Albert Zurita 2022-8-25

Yes, please see my comment to your reply below

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Answer 3

Image Analyst 2022-7-31

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https://ww2.mathworks.cn/matlabcentral/answers/1771140-2d-sliding-moving-running-average-window#answer_1018700

编辑：Image Analyst 2022-8-24

See my attached "manual" convolution demo. It lets you pick a standard demo image (converts to gray scale if necessary) and then asks you for the number of rows and columns in the scanning filter window (kernel). Then it has a 4-nested for loop where it sums the image pixel times the kernel value. It also counts the number of pixels in the kernel that overlap the image so in essence it shrinks the window when it "leaves the image" by ignoring those pixels. So if the 3x3 kernel is centered over (1,1) only 4 pixels are considered rather than the full 9 because 5 pixels are off the image and don't overlap. The input image and output image are displayed side by side. It does not use any MATLAB convolution functions or vectorized indexing to get any of the pixels in the window so it's completely 100% manual.

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Albert Zurita 2022-8-25

Hi, indeed your proposal looks very good. The only issue is the 'circular smooth' that Bruno has implemented as per my original post. Since there is some sort of similarity between the end of one row and the start of the next one Bruno creates that padded matrix. Perhaps your code could be adapted to work with the padded matrix, but the indexing should be changed.

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