Histogram of HSV quantized image
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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 =
2820
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
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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
1 个评论
采纳的回答
Walter Roberson
2011-12-26
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.
7 个评论
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.
更多回答(1 个)
Image Analyst
2011-12-26
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);
21 个评论
Naushad Varish
2018-5-11
编辑:Naushad Varish
2018-5-11
Please provide the code. It is still not working properly.
Image Analyst
2018-5-11
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.
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