How do I compute the maxpool of a image? Let us say stride of 2,2 on a mxn matrix?

Asked by Kannan U V on 5 Jul 2018

Latest activity Edited by Matt J on 6 Jul 2018

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If I were to implement just the max pooling operation on an image as mentioned in the following page https://www.quora.com/What-is-max-pooling-in-convolutional-neural-networks What is the most efficient way of computing it without going into for loops

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3 Answers

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Answer by Kannan U V on 6 Jul 2018

Accepted Answer

The following does the trick

fun = @(block_struct) max(block_struct.data(:));
b = blockproc (a, [X Y], fun);

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Matt J on 6 Jul 2018

But it is not very efficient. Compare:

a=rand(5000);
X=4; Y=4; %window sizes
tic
fun = @(block_struct) max(block_struct.data(:));
b = blockproc (a, [X Y], fun);
toc
%Elapsed time is 19.764354 seconds.
tic
b=sepblockfun(a,[X,Y],'max');
toc
%Elapsed time is 0.092457 seconds.

It is probably in fact the least efficient approach you could use. Even a double for-loop is faster:

    tic;
         [m,n]=size(a);
         ex=ones(1,m/X)*X;
         ey=ones(1,n/Y)*Y;
         ac=mat2cell(a,ex,ey);
         for i=1:m/X
             for j=1:n/Y
                 ac{i,j}=max(ac{i,j}(:));
             end 
         end
         b=cell2mat(ac);
    toc
    %Elapsed time is 6.203763 seconds.

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Answer by Matt J on 6 Jul 2018

Edited by Matt J on 6 Jul 2018

What is the most efficient way of computing it without going into for loops

The most efficient way in the entire universe is to use SEPBLOCKFUN (Download) as follows,

X=2; Y=2; %window sizes
maxpool=sepblockfun(yourImage,[X,Y],'max');

This assumes the image dimensions m,n are evenly divisible by X,Y respectively. Otherwise, you must pad the image to make it so.

1 Comment

Kannan U V on 6 Jul 2018

Thanks for your time and answer. It looks like the following does the trick

fun = @(block_struct) max(block_struct.data(:));
b = blockproc (a, [X Y], fun);

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Answer by Anton Semechko on 5 Jul 2018

Edited by Anton Semechko on 5 Jul 2018

Here is an example:

% Sample image
im=imread('cameraman.tif'); % sample image
% 4 pixels comprising non-overlapping 2-by-2 neighbourhoods  
im_nw=im(1:2:end,1:2:end);
im_sw=im(2:2:end,1:2:end);
im_se=im(2:2:end,2:2:end);
im_ne=im(1:2:end,2:2:end);
% Select pixel with maximum intensity
im_max=max(cat(3,im_nw,im_sw,im_se,im_ne),[],3);
% Visualize
figure('color','w')
ha=subplot(1,2,1);
imshow(im,imref2d(size(im)))
title(ha,'original','FontSize',20)
ha=subplot(1,2,2);
imshow(im_max,imref2d(size(im_max)))
title(ha,'2x2 max-pool','FontSize',20)

Note that even though two images appear to have the same size when visualized using 'imshow', the dimensions of im_max are half that of im. Recursive application of 2-by-2 max-pool will result in downsampled images with sizes 1/2, 1/4, 1/8, etc. of the original image.

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Kannan U V on 5 Jul 2018

Thanks for your time and answer. I am looking for variable X, Y window sizes too.

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