Overcome for loops for fast processing

I have the following part of MATLAB code

skin_rgb = zeros(256.^3,1);
for i = [2000 : 2003,2008,2010:2011,2013:2015,2017:2018,2021,2024,2027:2041,2043,2045:2059,2061:2063]
    I = imread(['skin_img',num2str(i),'.tif']);
         [x,y,~]= size(I);
         for l = 1:x
             for j = 1:y
                  if (I(l,j,1) ~= 255 && I(l,j,2) ~= 255 && I(l,j,3) ~= 255)
                     aa = double([I(l,j,1) , I(l,j,2) , I(l,j,3)]);
                     [~,ee1] = ismember(aa,rgb_hist,'rows');
                     skin_rgb(ee1) = skin_rgb(ee1) + 1; 
                     b1 = b1 + 1;
                 end
             end
         end

end

The output of the code is perfect. The issue is it is very slow. Inside the if loop the size of rgb_hist variable is 255.^3 x 3. Is there any way to get rid of the two nested for loops? As I am doing Image processing on high resolution image the variable x and y (different for each "i" iteration) have a big number in it.

Thank you

[Walter Roberson]

Is every possible combination in rgbhist?

[Muhammad Farhan Mughal]

Yes, every possible combination is in rgb_hist.

Answer 1

[Bruno Luong]

skin_rgb = 0;
for i = [2000 : 2003,2008,2010:2011,2013:2015,2017:2018,2021,2024,2027:2041,2043,2045:2059,2061:2063]
    I = imread(['skin_img',num2str(i),'.tif']);
    RGB = reshape(double(I),[],3)+1;
    skin_rgb = skin_rgb + accumarray(RGB,1,256+[0 0 0]);
end
skin_rgb = permute(skin_rgb,[3 2 1]);
skin_rgb = skin_rgb(:);

[Muhammad Farhan Mughal]

Thanks a lot.

Answer 2

[Walter Roberson]

skin_rgb = zeros(256.^3,1);
for i = [2000 : 2003, 2008, 2010:2011, 2013:2015, 2017:2018, 2021, 2024, 2027:2041, 2043, 2045:2059, 2061:2063]
    I = imread(['skin_img',num2str(i),'.tif']);
    r = 1 + double(I(:,:,1)); g = 1 + double(I(:,:,2)); b = 1 + double(I(:,:,3));
    skin_rgb = skin_rgb + accumarray([r(:), g(:), b(:)], 1, [256.^3, 1]);
end

[Muhammad Farhan Mughal]

Thank you for the kind response, but MATLAB giving an error that "

   Error using accumarray
  Third input SZ must be a full row vector with one element for each column of SUBS.

Error in hist_RGB (line 33)
    skin_rgb = skin_rgb + accumarray([r(:), g(:), b(:)], 1, [256.^3, 1]);

Waiting for your kind response.

[Walter Roberson]

 skin_rgb = skin_rgb + accumarray([r(:), g(:), b(:)], 1, [256, 256, 256]);

And after the loop

skin_rgb = skin_rgb(:) ;

[Muhammad Farhan Mughal]

Thanks a lot.

[Bruno Luong]

You'll notice my answer correct these two errors

[Muhammad Farhan Mughal]

I have tested it many times but found that

skin_rgb = skin_rgb(:) ;

is not working fine. Do I need to reshape after skin_rgb after 3 columns?

[Bruno Luong]

256^3 = 16777216 is obviously not divisible by 3. So on earth you want to reshape in 3 columns???

[Muhammad Farhan Mughal]

Sorry for that, but skin_rgb = skin_rgb(:) is not arranged in order as asked in the question. I just wanted to solve that issue.

[Bruno Luong]

The order depends on entirely on the content of rgb_hist that you never specify what's in and in which order. Question never correctly formulated.

[Muhammad Farhan Mughal]

I accept my mistake of asking question. rgb_hist has the following order

rgb_hist = [1 1 1 ; 1 1 2 ; .... ; 1 1 256; 1 2 1 ; 1 2 2; ,.... ;1 256 256 ; ..... ; 256 256 256];

Its a 256.^3 x 3 matrix. Starting from 1,1,1 to 256,256,256.

[Bruno Luong]

Usually RGB values are in [0:255] range not [1:256]. Assuming the above rgb_hist is to be substracted by 1, on this case just change

skin_rgb = skin_rgb(:);

to

skin_rgb = permute(skin_rgb,[3 2 1]);
skin_rgb = skin_rgb(:);

[Muhammad Farhan Mughal]

I apologize for this, in the following code now i am working with only one image "i = 2000" . As you can see that aa is selected from image I so it must have occur once. "a2" variable has location of "aa" in "rgb_hist" array. This implies that skin_rgb(a2) should atleast be equal to 1. But it is not. Which in my understanding is due to the order of skin_rgb.

>> aa = 1 + [I(500,500,1),I(500,500,2),I(500,500,3)];
>> [a1,a2] = ismember(aa,rgb_hist,'rows');
>> skin_rgb(a2)

ans =

     0

For easiness I am providing the rgb_hist variable order again

rgb_hist = [1 1 1 ; 1 1 2 ; .... ; 1 1 256; 1 2 1 ; 1 2 2; ,.... ;1 256 256 ; 2 1 1;..... ; 256 256 256];

Thank you.

[Bruno Luong]

Original RGB value starts from 0 ends at 255. This is the reason we add 1 before doing accumarray.

So skin_rgb(10,20,30) actually stores the number of image pixels that have the RGB = [09,19,29]

[Muhammad Farhan Mughal]

Yes, I understad that, but in the folowing code I have already added 1 in variable "aa".

>> aa = 1 + [I(500,500,1),I(500,500,2),I(500,500,3)];
>> [a1,a2] = ismember(aa,rgb_hist,'rows');
>> skin_rgb(a2)
ans =
     0

[Bruno Luong]

Then your rgb_hist is wrong.

This test code returns 1 everytime

% Such simple generate thing that you never bother to post
[B,G,R] = ndgrid(1:256);
rgb_hist = [R(:),G(:),B(:)];
% Quick check for order
rgb_hist(1:10,:)
rgb_hist(end+(-9:0),:)

% Random data
I = floor(256*rand(100,100,3));
aa = 1 + [I(100,100,1),I(100,100,2),I(100,100,3)]

% Counting algo
skin_rgb = accumarray(aa,1,256+[0 0 0]);
skin_rgb = permute(skin_rgb,[3 2 1]);
skin_rgb = skin_rgb(:);

% Check
[a1,a2] = ismember(aa,rgb_hist,'rows');
skin_rgb(a2) % <- This returns correctly 1

[Muhammad Farhan Mughal]

Thanks a lot