Histogram of HSV quantized image

Hi i want to share with you the results of an HSV quantized image Histogram

i used an image 256X384 converted it into HSV and quantized it into (8X3X3) for H, S and V respectively and after that i made a weighted sum G= 9*H + 3*S + 3*V for this matrix i used this function:

histG=imhist(G,72)

but the output is like that:

histG =

        2820
           0
           0
           0
           0
           0
           0
           0
           0
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       13500

is this ok or i have something wrong ? if it is right please explain to me why i got this output.

Thank you

[Harsh Patel]

can you please provide full code from reading image to until quntization?

Answer 1

[Walter Roberson]

G= 9*H + 3*S + 3*V

is the wrong weighted sum. You want to add V, not 3*V .

It appears that only the low bin and the high bin are being used in the histogram, as if you had an error in your quantization that led to the results always being either (0,0,0) or the maximum value (8,3,3). The V vs 3*V would not account for this, but a quantization mistake would. Or, alternately, you might have good quantization but a black and white image is being quantized.

[yasmine]

sorry yes i wrote it wrongly

i made it G= 9*H + 3*S + V but it gives me the same output

[yasmine]

no am using RGB images.

here is the quantization am using:

%Quantizing the HUE

for row=1:BRows

for col=1:BCols

if (Hueblock11(row,col) <= 316 && Hueblock11(row,col) <= 20)

Hueblock11(row,col) = 0;

elseif (Hueblock11(row,col) >= 20 && Hueblock11(row,col) <= 40)

Hueblock11(row,col) = 1;

elseif (Hueblock11(row,col) >= 40 && Hueblock11(row,col) <= 75)

Hueblock11(row,col) = 2;

elseif (Hueblock11(row,col) >= 75 && Hueblock11(row,col) <= 155)

Hueblock11(row,col) = 3;

elseif (Hueblock11(row,col) >= 155 && Hueblock11(row,col) <= 190)

Hueblock11(row,col) = 4;

elseif (Hueblock11(row,col) >= 190 && Hueblock11(row,col) <= 270)

Hueblock11(row,col) = 5;

elseif (Hueblock11(row,col) >= 270 && Hueblock11(row,col) <= 295)

Hueblock11(row,col) = 6;

elseif (Hueblock11(row,col) >= 295 && Hueblock11(row,col) <= 316)

Hueblock11(row,col) = 7;

end

end %for col

end %for row

%figure(k+1), subplot(3,2,1), image(HistH1);

%figure(k+2), subplot(3,2,i), imshow(h);

% Quantizing of Saturation

for row=1:BRows

for col=1:BCols

if (Satblock1(row,col) >= 0 && Satblock1(row,col) <= 0.2)

Satblock1(row,col) = 0;

elseif (Satblock1(row,col) >= 0.2 && Satblock1(row,col) <= 0.7)

Satblock1(row,col) = 1;

elseif (Satblock1(row,col) >= 0.7 && Satblock1(row,col) <= 1)

Satblock1(row,col) = 2;

end % end of if

end % for col

end %for row

%Quantizing of Value

for row=1:BRows

for col=1:BCols

if (Valblock1(row,col) >= 0 && Valblock1(row,col) <= 0.2)

Valblock1(row,col) = 0;

elseif (Valblock1(row,col) >= 0.2 && Valblock1(row,col) <= 0.7)

Valblock1(row,col) = 1;

elseif (Valblock1(row,col) >= 0.7 && Valblock1(row,col) <= 1)

Valblock1(row,col) = 2;

end % end of if

end % for col

end %for row

[Walter Roberson]

Your first condition,

if (Hueblock11(row,col) <= 316 && Hueblock11(row,col) <= 20)

is incorrect, as it is equivalent to Hueblock11(row,col) <= 20

I do not know what your maximum value for Hueblock11() is, but I speculate that you are intending hue angles, in which case you likely want to be checking the range 316 to 360, and 0 to 20.

Your "if" chain is also faulty in that you include the boundary conditions on both the upper end of one range and the lower end of the next range.

You could achieve much greater efficiency by using histc() calls with no loops.

[yasmine]

i got your first point but would you explain using example the second point?

Thank you

[yasmine]

if (Hueblock11(row,col) >= 316 && Hueblock11(row,col) <= 360 && Hueblock11(row,col) >= 0 && Hueblock11(row,col) <= 20)

Hueblock11(row,col) = 0;

is that true for the first point ?(i was trying to cover the range from(316 to 20))

[Walter Roberson]

Do not modify HueBlock11 "in place": create a new output matrix.

Hbins = [0 20 40 75 ... 316 inf]);

Ht = histc(HueBlock11, Hbins)

H = Ht(1:end-2);

H(1) = H(1) + Ht(end-1);

Those last two lines are to put 316 upward in to the same bin as less than 20

[Walter Roberson]

Look at your code:

if (Hueblock11(row,col) <= 316 && Hueblock11(row,col) <= 20)

Hueblock11(row,col) = 0;

elseif (Hueblock11(row,col) >= 20 && Hueblock11(row,col) <= 40)

Hueblock11(row,col) = 1;

Now what if Hueblock11(row,col) is 20 _exactly_ ? That matches the first condition, but it also matches the second condition. The bin chosen then becomes a matter of the order you coded the tests, which is not robust. If you want the first bin to include 20 exactly, then do not have the second bin include 20 exactly in its range.

Answer 2

[Image Analyst]

yasmine: If you use imhist on floating point arrays, they need to be normalized in the range 0-1. I suggest you use hist() instead of imhist() - it doesn't have that requirement.

[pixelCounts binValues] = hist(G, numberOfBins);

[yasmine]

actually, the G matrix values are integers but greater than 1

min value= 0, max value=30

i tried to use the hist() but gave me same output also

[Image Analyst]

Can you upload your image to tinypic.com so we can try your code?

[Walter Roberson]

We would also need the code that constructs the original Hueblock11, Satblock1, and Valblock1.

[Image Analyst]

Well what did you do? Because if you start with a uint8 RGB image, and use code like this:

hsvImage = rgb2hsv(rgbImage);

H = hsvImage(:,:,1); % H is double in range 0-1

S = hsvImage(:,:,2); % S is double in range 0-1

V = hsvImage(:,:,3); % V is double in range 0-1

G = 9*H + 3*S + V; % G is double in range 0-13

[pixelCount, hueValues] = imhist(H);

then H, S, V, and G will all be double. H will be in the range 0-1, G will be in the range 0-13, and the histogram (pixelCount) will be continuous, not just in the first and last bin.

[Walter Roberson]

The 256X384 size and the (8,3,3) (72 bin) quantization scheme are fairly tied to a particular CBIR paper in which hue expressed in "degrees".

The documentation for rgb2hsv does not explicitly talk about return types when RGB images are being processed (only when a colormap is being processed.) Unfortunately I cannot test at the moment.

[Image Analyst]

From the histogram, both imhist and hist, it's clear that, for some reason G is binary. It has only two values even though it may be double instead of logical. I think we can only solve this if she gives the actual code.

[yasmine]

here is one pic:

http://i44.tinypic.com/sqm3o6.jpg

[yasmine]

i think my main problem is in the quantization of hue as the values of Hueblock11 array before quantization are: min value=0 and max value= 0.9167

so all the values enter in the first if condition

[yasmine]

Dear Walter Roberson

Do not modify HueBlock11 "in place": create a new output matrix.

Hbins = [0 20 40 75 ... 316 inf]);

Ht = histc(HueBlock11, Hbins)

H = Ht(1:end-2);

H(1) = H(1) + Ht(end-1);

Those last two lines are to put 316 upward in to the same bin as less than 20

please i need a real example.

[Walter Roberson]

After your line

hsvimage= rgb2hsv(images);

add

hsvimage(:,:,1) = hsvimage(:,:,1) * 360;

With regard to using histc(), the code I gave is a real example that only needs you to fill in the rest of the Hbins values

[yasmine]

Thank you so much, i didn know this tip before.

it worked well in quantization

but there is a point and i don know how to fix it:

after multiplying the Hue layer with 360 still the histogram of the Hue is only 2 values

this is before quantization

[yasmine]

knowing that in saturation and value the histogram is very good using imhist() although the values are double

the values of hue => min =0 , max= 359.4231

the values of saturation => min =0 , max= 1

the values of value => min=0.0392, max=0.9882

[Walter Roberson]

What code are you using at present to create the histogram ? If you are using imhist() then you cannot use that, as imhist() expects values to be in the range 0 to 1 (unless the values are uint8 or uint16 data type, which your values would not be.)

[yasmine]

ok so can i use this method for normalization:

y=(x-min(min(x)))/(max(max(x))-min(min(x)))

x and y are 2D arrays

[Walter Roberson]

More simply,

xmin = min(x(:));

y = (x - xmin) ./ (max(x(:)) - xmin);

On the other hand, if you use hist() instead of imhist() then you do not need to do this normalization at all.

[yasmine]

ok thank you so much

[Image Analyst]

Wait a minute. You accepted Walter's answer but never said what the problem was - why you had only two values in your histogram. I suggested that they were doubles and needed to be in the range 0-1 or else use hist() instead of imhist(). But you said NO, they are integers. Then later you admitted they were in fact doubles. So how did you finally get it to work so that you had values in more than 2 bins?

[Walter Roberson]

In the comment after she gave the link for the picture, she did say that before quantization the hues came out min value=0 and max value= 0.9167 -- so indeed not integers. But she probably figured she needed them to be integers as otherwise the 75's and 316's and so on would not be usable. And those values are taken right from the paper so they "had" to be right. Of course she could have divided all those boundary values by 360 and worked in double, but the Inertia of Authority is pretty powerful.

[Image Analyst]

OK fine, but why only two bins?

[Naushad Varish]

Please provide the code. It is still not working properly.

[Image Analyst]

naushad, I'm going to ask you the same thing. Because we gave code and the original poster accepted an answer. So there's no problem here, only with your code.