How to extract colour descriptor (set of features) from the image?

Hello, I have the following task. I want to compute the color histogram of an Image in HSV color space. For that I want to uniformly quantize the HSV color space into 240 cubes in which the quantization of each channel is 15 hues (H), 4 saturations (S) and 4 intensities (V ) respectively. And after making this color histogram the values of each bin is my 240 dim vector of color features which I want to get finally. Could someone please help me with the m-code for this task or with the high level plan how to create it? I will really appreciate it. Thank you.

Answer 1

[Image Analyst]

Loop over every pixel in the image and determine the bin, then increment the count. Here's a start

[rows, columns, numberOfColorChannels] = size(rgbImage);
hsvImage = rgb2hsv(rgbImage); % Ranges from 0 to 1.
hsvHist = zeros(15,4,4);
for col = 1 : columns
    for row = 1 : rows
        hBin = floor(hsvImage(row, column, 1) * 15);
        sBin = floor(hsvImage(row, column, 2) * 4);
        vBin = floor(hsvImage(row, column, 3) * 4);
        hsvHist(hBin, sBin, vBin) = hsvHist(hBin, sBin, vBin) + 1;
    end
end

[Andrew Tim]

Thank you very much!

[Kondal Rao J]

@Image Analyst Sir, I'm getting the following error can you help me

Subscript indices must either be real positive integers or logicals.

Error in Binning_HSVimage (line 12) hsvHist(hBin, sBin, vBin) = hsvHist(hBin, sBin, vBin) + 1;

[Gopichandh Danala]

He mistyped col with colums above

for col = 1 : columns
    for row = 1 : rows
        hBin = floor(hsvImage(row, col, 1) * 15);
        sBin = floor(hsvImage(row, col, 2) * 4);
        vBin = floor(hsvImage(row, col, 3) * 4);
        hsvHist(hBin, sBin, vBin) = hsvHist(hBin, sBin, vBin) + 1;
    end
end

[Image Analyst]

Right, thanks for catching that. I think Kondal must have also fixed it or else he would not have reached the line that he says is throwing an error. The error indicates that hBin, sBin, or vBin would be negative, zero, or fractional. One of them must be zero so I'd add one to the bins.

[rows, columns, numberOfColorChannels] = size(rgbImage);
hsvImage = rgb2hsv(rgbImage); % Ranges from 0 to 1.
hsvHist = zeros(15,4,4);
for col = 1 : columns
    for row = 1 : rows
        hBin = floor(hsvImage(row, col, 1) * 15) + 1;
        sBin = floor(hsvImage(row, col, 2) * 4) + 1;
        vBin = floor(hsvImage(row, col, 3) * 4) + 1;
        hsvHist(hBin, sBin, vBin) = hsvHist(hBin, sBin, vBin) + 1;
    end
end

[Kondal Rao J]

@Image Analyst, Sir, I tried to find the HSV histogram of an Image. This histogram should be used as input feature vector for classifier. My object is color naming for real world applications. Can you give any suggestion. Is there any other way I can train my classifier apart from color Histogram. Here is my code

I = imread('peppers.png');
rgbImage=imresize(I,[256 256]);
figure,imshow(rgbImage);

[rows, columns, numberOfColorChannels] = size(rgbImage);
hsvImage = rgb2hsv(rgbImage); % Ranges from 0 to 1.
hsvHist = zeros(15,4,4);
for col = 1 : columns
    for row = 1 : rows
        hBin = floor(hsvImage(row, col, 1) * 15)+ 1;
        sBin = floor(hsvImage(row, col, 2) * 4)+ 1;
        vBin = floor(hsvImage(row, col, 3) * 4)+ 1;
        hsvHist(hBin, sBin, vBin) = hsvHist(hBin, sBin, vBin) + 1;
    end
end

the following is the error::

"Index exceeds matrix dimensions.

Error in Binning_HSVimage (line 14)
        hsvHist(hBin, sBin, vBin) = hsvHist(hBin, sBin, vBin) + 1;"

[Image Analyst]

The problem is that there is exactly 0 and exactly 1 somewhere in the s channel. So if we put 0 into bin 1, then 0.9999999 would go into bin 4, but if s=1 exactly, it wants to go into bin 5, which is bigger than we preallocated hsvHist to be. But we don't want the dimensions to be (16,5,5) because the only thing that would go into bin 5 would be those rare pixels whose value was exactly 1.0000. So to put them into bin 4 instead, we just have to multiply by something just a hair smaller than 4, like 3.9999. Here is code that works:

rgbImage = imread('peppers.png');
rgbImage = imresize(rgbImage, [256 256]);
figure
imshow(rgbImage);

[rows, columns, numberOfColorChannels] = size(rgbImage);
hsvImage = rgb2hsv(rgbImage); % Ranges from 0 to 1.
hsvHist = zeros(16,4,4);
for col = 1 : columns
    for row = 1 : rows
        hBin = floor(hsvImage(row, col, 1) * 15.9999)+ 1;
        sBin = floor(hsvImage(row, col, 2) * 3.9999)+ 1;
        vBin = floor(hsvImage(row, col, 3) * 3.9999)+ 1;
        hsvHist(hBin, sBin, vBin) = hsvHist(hBin, sBin, vBin) + 1;
    end
end

[Kondal Rao J]

@Image analyst Sir, Its working. Thank you very much. Thank you for your valuable suggestion and time.

[Kondal Rao J]

@Image Analyst, Sir, Is it possible do that in CIELAB color Space?. How to bin CIELAB color space?

[Image Analyst]

Sure, just use

labImage = rgb2lab(rgbImage);

Be aware that L, A, and B don't have the range 0-1 like HSV so you'll have to normalize them.

[Dharmayug Aiyavoo]

what about normal rgb color histogram feature extraction with 512 dimension?

[Image Analyst]

What about it?

Go ahead and do it if you want to, I won't get upset.

[Aswathy M A]

Subscript indices must either be real positive integers or logicals.

Error in Untitled (line 16)

hsvHist(hBin, sBin, vBin) = hsvHist(hBin, sBin, vBin) + 1;

i got this as error..

[Image Analyst]

One of them is probably zero. If it's zero, just add 1 to it.

[shalaw faraj]

Dear @Image Analyst

Could you help me please? do you have any email or any number for communication ? I have a question please. Thanks

[nissrine Neyy]

@Image Analyst when creating :

hsvHist = zeros(16,4,4);

the 16 represents the rows, 4 the columns and the second 4 represents the dimention, is that right ? if yes why do we need 4 dimentions while we have only 3 planes (or dimentions) of the HSV (H,V,S) ?

[Image Analyst]

nissrine, note that an array defined as zeros(n1, n2, n3) is always a 3-D array regardless of what the values of n1, n2, and n3 are.

Now h, s, and v are basically continuous variables in the range of 0-1 and the poster wanted to divide the range of h into 16 bins and s and v into 4 bins.

[nissrine Neyy]

Okey, got it now .. thank you so much :)