Combine a (m x n x p) matrix (image) of 8 bit numbers to a (m x n) matrix of 24 bit numbers and vice versa

Say there is a matrix of (m x n x p), esp. a color image with R G and B channel. Each channel information is 8-bit integer.

But, for an analysis, the three 8-bit values have to be combined to get a 24-bit value and the analysis is done on the (m x n) matrix of 24-bit values.

After the analysis, the matrix has to be decomposed back to three 8-bit channels for displaying the results.

What I am doing right now:

• Iterating through all the values in the matrix
• Convert each decimal value to binary (using `dec2bin`)
• Combine the three binary values together to get a 24-bit number (using `strcat` and `bin2dec`)

Code:

    for i=1:m
        for j=1:n
            new_img(i,j) = bin2dec(strcat(...
                sprintf('%.8d',str2double(dec2bin(img(i,j,1)))), ...
                sprintf('%.8d',str2double(dec2bin(img(i,j,2)))), ...
                sprintf('%.8d',str2double(dec2bin(img(i,j,3))))));
        end
    end

For the decomposition back to three 8-bits after analysis, the exact reverse process is done, still iterating through (m x n) values.

The problem is huge computation time.

I know that this is the not the correct way of doing this. Is there any matrix operation that I can do to achieve this so that the computation is done quickly?

Answer 1

[Walter Roberson]

img32 = uint32(img);
new_img = img32(:,:,1) * 2^16 + img32(:,:,2) * 2^8 + img32(:,:,3);

and now new_img is a uint32() image as appropriate.

[Shahidh K]

Thanks for the answer, that was great! But, how do we get the 8-bit 3D matrix back from the 32-bit new_img?

[Shahidh K]

I hope this does the conversion back to 8-bits

r_img = zeros(size(img1));

r_img(:,:,1) = new_img ./ 2^16;
r_img(:,:,2) = new_img ./ 2^8;
r_img(:,:,3) = new_img ;

[Walter Roberson]

 r_img = zeros(size(img1), class(img1));
 r_img(:,:,3) = mod(new_img, 256);
 t = (new_img - r_img(:,:,3)) / 256;
 r_img(:,:,2) = mod(t, 256);
 t = (t - r_img(:,:,2)) / 256;
 r_img(:,:,1) = t;

[Nitesh Methani]

Can you please explain the logic? I am not sure I understood it right..Thanks.

[Walter Roberson]

Suppose we were working in decimal and we were decomposing the number 738 . Then, r_img(:,:,3) = mod(738, 10) = 8. t = (738-8)/10 = 730/10 = 73. r_img(:,:,2) = mod(73,10) = 3. t = (t-rimg(:,:,2))/10 = (73-3)/10 = 7 . rimg(:,:,1) = 7

Each mod() extracts the last "digit" in the base we are working with. Subtracting that last digit from the original number is always going to give a number that is exactly divisible by the base we were working with, because 'mod' is defined as the remainder after division and when you subtract off the remainder you have to get something that divides evenly. You just repeat that process a couple of times.