Histogram of HSV quantized image

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yasmine 2011-12-26

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https://ww2.mathworks.cn/matlabcentral/answers/24749-histogram-of-hsv-quantized-image

评论： Image Analyst 2018-5-11

采纳的回答： Walter Roberson

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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 =

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

Thank you

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Harsh Patel 2017-4-4

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

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采纳的回答

Walter Roberson 2011-12-26

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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.

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yasmine 2011-12-26

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 2011-12-26

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 2011-12-26

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.

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