convert a binary image to RGB color space

Hello, I have RGB image that I converted to binary to do some processing and I want to reconvert it back to its original color RGB, is it possible? I did some researchs on the internet but unfortunately I couldn't find a good answer. thank you in advance.

Answer 1

[Walter Roberson]

No it is not possible. However, you can take

FilteredImage = repmat(YourBinaryImage, 1, 1, 3) .* YourColorImage;

if what you are doing is using YourBinaryImage to select which pixels should become black in your color image.

Answer 2

[Image Analyst]

You can't convert the binary image, but fortunately you should still have the original RGB image, don't you? If not, why not?

[Youssra]

hello, thank you all for your answer.

The main problem is that I'm working on color image watermarking, and basicaly I convert my watermak (logo) to a binary image then apply an embedding code, after that in extraction phase I got a binary extracted watermark that I need to convert to RGB to get it look like the original one. I'm working on blind watermarking wich means I can't use the original watermak for extraction

[Walter Roberson]

When you extract the bits, you get a vector of bit values, call it BitV.

ByteV = uint8( dec2bin(char(reshape(BitV, 8, []) .' + '0')) );

now that you have ByteV you can reshape() it into a 3D array of the appropriate size, which would be an image.

[Youssra]

Thank you @Walter, I will try your answer and let you know ^^

[Youssra]

Hello , I tried the answer that @ Walter gave me but I realised that the size of the result is completely different. for example I tried on 32*32 binary image (wich is RGB image originally) but at the end I got 1024*6 image and doesn't look like the original one :/

[Walter Roberson]

32 rows * 32 columns * 3 colors * 8 bits = 24576 bits embedded. Regroup the 8 bits and convert to bytes = 32 * 32 * 3 = 3072 bytes which happens to be 1024 * 3. I do not know where your factor of 2 came from: check to see how many bits your code embeds and how many bits it extracts. Once you get it down to 1024 * 3 then you rehape() that to 32 * 32 * 3.

1024 * 6 would make sense if your image is composed of uint16 instead of uint8, which is possible but not all that common.