How to reduce the number of colors in an image with OpenCV?

Question

I have a set of image files, and I want to reduce the number of colors of them to 64. How can I do this with OpenCV?

I need this so I can work with a 64-sized image histogram. I'm implementing CBIR techniques

What I want is color quantization to a 4-bit palette.

Answer 1

There are many ways to do it. The methods suggested by jeff7 are OK, but some drawbacks are:

• method 1 have parameters N and M, that you must choose, and you must also convert it to another colorspace.
• method 2 answered can be very slow, since you should compute a 16.7 Milion bins histogram and sort it by frequency (to obtain the 64 higher frequency values)

I like to use an algorithm based on the Most Significant Bits to use in a RGB color and convert it to a 64 color image. If you're using C/OpenCV, you can use something like the function below.

If you're working with gray-level images I recommed to use the LUT() function of the OpenCV 2.3, since it is faster. There is a tutorial on how to use LUT to reduce the number of colors. See: Tutorial: How to scan images, lookup tables... However I find it more complicated if you're working with RGB images.

void reduceTo64Colors(IplImage *img, IplImage *img_quant) {
    int i,j;
    int height   = img->height;   
    int width    = img->width;    
    int step     = img->widthStep;

    uchar *data = (uchar *)img->imageData;
    int step2 = img_quant->widthStep;
    uchar *data2 = (uchar *)img_quant->imageData;

    for (i = 0; i < height ; i++)  {
        for (j = 0; j < width; j++)  {

          // operator XXXXXXXX & 11000000 equivalent to  XXXXXXXX AND 11000000 (=192)
          // operator 01000000 >> 2 is a 2-bit shift to the right = 00010000 
          uchar C1 = (data[i*step+j*3+0] & 192)>>2;
          uchar C2 = (data[i*step+j*3+1] & 192)>>4;
          uchar C3 = (data[i*step+j*3+2] & 192)>>6;

          data2[i*step2+j] = C1 | C2 | C3; // merges the 2 MSB of each channel
        }     
    }
}