how can we do "8-coarse division of RGB color space" for dominant color extraction?

this is basically for quantized color.

Answer 1

[Image Analyst]

Try rgb2ind() or this kmeans example

[preet]

can u tell me how to divide RGB color space in 8 coarse in malab?

[Walter Roberson]

It is not possible to do "8-coarse division" (whatever that is, I don't know), without "using any method". Making mystical passes over it and saying "Abaracadabara, you are now 8-coarse divided" would still be using some method.

[preet]

Walter Roberson

tell me the function ( method )in matlab for coarse division

[Walter Roberson]

What is "coarse division" ?

[preet]

its divide the rgb color space into 8 blocks.like this

http://140.117.166.1/eehome/ISCOM2005/SubmitPaper/UploadPapers/ISCON05_00089.pdf

my problem in this pdf page 5.

[Image Analyst]

It looks like the FAQ applies here: http://matlab.wikia.com/wiki/FAQ#Can_you_program_up_the_algorithm_in_this_article_for_me_and_explain_it_to_me.3F, though it looks somewhat interesting so if I can find the time today or tomorrow maybe I can read the paper.

[preet]

i appreciate if u do for me... happy new year..

Answer 2

[Walter Roberson]

dImage = im2double(YourImage);
binned_image = round(dImage);
quadrant_num = binned_image(:,:,1) * 4 + binned_image(:,:,2) * 2 + binned_image(:,:,1);

quadrant 0 would be (low red, low blue, low green), quadrant 1 would be (low red, low blue, high green), quadrant 2 would be (low red, high blue, low green), and so on up to quadrant 7 as (high red, high blue, high green)

You might want to add 1 to the quadrant number, if you want to use it as an index.

[preet]

why did u multiply 4 and 2 with red and green component.... quadrant_num = binned_image(:,:,1) * 4 + binned_image(:,:,2) * 2 + binned_image(:,:,3);

[Walter Roberson]

Each plane, R, G, B, will have value 0 or 1 in binned_image. To convert that to figure out which of the 8 areas is being referred to, the easiest way is to interpret the 3 values together as a set of three bits in a binary number, and we might as well us the bit order R then G then B. You could, if you really wanted, write this as

(left-shift Red bit-contribution by one place, OR in Green bit-contribution), left shift the result by one place, OR in the Blue bit-contribution

but if you work that through you will find that that corresponds to multiplying the Red bit-contribution by 4, multiplying the Green bit-contribution by 2, multiplying the Blue bit-contribution by 1, and adding those values all together. Which is what I did.

[preet]

when i try to do this then my image data loss.it similarly like quantization of binary image

[Walter Roberson]

Are you using

imagesc(quadrant_num)

or alternately

imshow(quadrant_num, [])

to see the different bin numbers as different colors ?

[Image Analyst]

Yes, that's the nature of quantization. Your image will look posterized.

[preet]

Walter Roberson it should be like to use rgb2ind() function but its not?

[Image Analyst]

Correct. The rgb2ind() does not reveal the algorithm that it uses in the help, but it's most likely not what Walter did. You can get an overview of several common/popular methods in this Wikipedia article: http://en.wikipedia.org/wiki/Color_quantization

[Walter Roberson]

The paper says,

The color space is equally divided into 8 coarse partitions according to the quantity of each color component.

So you do not want to do something like rgb2ind() does, as rbg2ind() does not divide the color-space equally. The code I gave does divide the color-space equally, and the value I return for each point is an index like rgb2ind() would return, with the minor exception that for technical reasons, rgb2ind() would correspond to taking uint8() of quadrant_num

[preet]

i said i need the result like rgb2ind()..

[Walter Roberson]

Okay, here it is:

dImage = im2double(YourImage);
binned_image = round(dImage);
quadrant_num = uint8( binned_image(:,:,1) * 4 + binned_image(:,:,2) * 2 +   binned_image(:,:,1) );

quadrant_num will now be exactly like rgb2ind() in that it will return a 2D array of uint8() values, each of which is an "ind". Keep in mind that exactly like rgb2ind(), the value 0 represents the first color in the color map, and 1 represents the second color, and so on.