Sliding max window and its average for multi-dimensional arrays

Question

I have a 60 x 21 x 700 matrix, where the 60 x 21 represent a pressure output x number of frames. I want to find the 2 x

Answer 1

One other approach using im2col

%// getting each 2x2 sliding submatrices as columns
cols = im2col(A,[2 2],'sliding'); 

%// getting the mean & reshaping it to 2D matrix
C = reshape(mean(cols),size(A,1)-1,[]); 

%// to find the maximum of the sub-matrix-means and its corresponding index.
[B,i] = max(mean(cols)); 

%// reshaping the corresponding sub-matrix to 2D matrix
mat = reshape(cols(:,i),2,2); 

Results:

>> mat

mat =

 6    10
 8     9

>> C

C =

1.5000    1.5000    1.5000    1.5000
2.0000    2.2500    2.7500    2.7500
2.0000    3.7500    6.0000    5.5000
1.7500    4.5000    8.2500    7.2500

>> B

B =

8.2500

Answer 2

2D Arrays: For the given 2D array input, you can use 2D convolution -

%// Perform 2D convolution with a kernel of `2 x 2` size with all ones
conv2_out = conv2(A,ones(2,2),'same')

%// Find starting row-col indices of the window that has the maximum conv value
[~,idx] = max(conv2_out(:))
[R,C] = ind2sub(size(A),idx)

%// Get the window with max convolution value
max_window = A(R:R+1,C:C+1)

%// Get the average of the max window
out =  mean2(max_window)

Sample step-by-step run of the code -

A =
     1     2     2     1     1
     2     1     1     2     2
     2     3     4     4     3
     1     2     6    10     5
     2     2     8     9     5
conv2_out =
     6     6     6     6     3
     8     9    11    11     5
     8    15    24    22     8
     7    18    33    29    10
     4    10    17    14     5
idx =
    14
R =
     4
C =
     3
max_window =
     6    10
     8     9
out =
         8.25

---

Multi-dimensional Arrays: For multi-dimensional array case, you need to perform ND convolution -

%// Perform ND convolution with a kernel of 2 x 2 size with all ONES
conv_out = convn(A,ones(2,2),'same')

%// Get the average for all max windows in all frames/slices  
[~,idx] = max(reshape(conv_out,[],size(conv_out,3)),[],1)
max_avg_vals = conv_out([0:size(A,3)-1]*numel(A(:,:,1)) + idx)/4

%// If needed, get the max windows across all dim3 slices/frames
nrows = size(A,1)
start_idx = [0:size(A,3)-1]*numel(A(:,:,1)) + idx
all_idx = bsxfun(@plus,permute(start_idx(:),[3 2 1]),[0 nrows;1 nrows+1])
max_window = A(all_idx)

Sample input, output -

>> A
A(:,:,1) =
     4     1     9     9
     3     7     5     5
     9     6     1     6
     7     1     1     5
     4     2     2     1
A(:,:,2) =
     9     4     2     2
     3     6     4     5
     3     9     1     1
     6     6     8     8
     5     3     6     4
A(:,:,3) =
     5     5     7     7
     6     1     9     9
     7     7     5     4
     4     1     3     7
     1     9     3     1
>> max_window
max_window(:,:,1) =
     9     9
     5     5
max_window(:,:,2) =
     8     8
     6     4
max_window(:,:,3) =
     7     7
     9     9
>> max_avg_vals
max_avg_vals =
            7          6.5            8

Answer 3

A moving average can be done with a simple convolution. It has to be 2D in your case, so:

A = [01 02 02 01 01
     02 01 01 02 02
     02 03 04 04 03
     01 02 06 10 05
     02 02 08 09 05];

B = [1 1;1 1] / 4 ; %// prepare moving average filter [2x2]

C = conv2(A,B,'valid') ; %// perform 2D moving average

Produces:

C =
    1.5     1.5     1.5     1.5
    2       2.25    2.75    2.75
    2       3.75    6       5.5
    1.75    4.5     8.25    7.25

Which is exactly the average of each of your [2x2] areas.