rescale histogram

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mohammed ABDEEN
mohammed ABDEEN am 30 Apr. 2012
hello all !
i need to rescale the x-axis(gray levels) of a histogram from 0-1 to 0-255 ! can anyone help me ?! ,i need it coz i've to calculate pdf then.. thank you
  2 Kommentare
Geoff
Geoff am 30 Apr. 2012
Post the code you are having issues with. If you are using the 'hist' command, you can solve a lot of issues by using 'histc' to generate an array, and then draw it into a bar graph.
mohammed ABDEEN
mohammed ABDEEN am 1 Mai 2012
histc doesnt satisfy my needs, coz i've a histogram of an image where the grey levels are in this interval [0,1] and there are like 4 pulses in a different positions,and i want to normalize it between [0 255] so i would have these 4 pulses or whatever in an integer number and not fractional number ....

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Antworten (3)

Image Analyst
Image Analyst am 30 Apr. 2012
If your image has value in the range 0-255 then how did you get bins in the 0-1 range? You must be looking at an image that has been transformed, perhaps by im2double. Maybe you should just take the histogram of the original image, or else multiply the image you took the histogram of by 255 and cast to uint8 and take the histogram of that instead.
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Image Analyst
Image Analyst am 2 Mai 2012
You still haven't convinced me, and I doubt you will. The probability values in the probability density function do not depend on the bin value. The probability values will of course depend on the width of the bins, and the location of the bins with respect to the true distribution, but simply scaling the bins (what you call i or gray level) will not change the PDF, and thus not change p in sum(p*log p), and not change entropy. Try it and see. If you don't see, ask me and I'll make a demo for you to prove it.
mohammed ABDEEN
mohammed ABDEEN am 3 Mai 2012
ok thnaks tomorrow i'll let u see what i'm doing i'll send u my code !

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Geoff
Geoff am 2 Mai 2012
Still not sure I understand your motive completely...
But it seems like you have this situation:
data = rand(100,1);
bins = (1:4)/4 - 1/8;
hist(data, bins);
And you want this:
hist(data*256, bins*256);
set(gca, 'XLim', [0 256]);
set(gca, 'XTick', linspace(0,256,5));
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mohammed ABDEEN
mohammed ABDEEN am 3 Mai 2012
ok i've missed the (.) in the formula . i did it!
mohammed ABDEEN
mohammed ABDEEN am 3 Mai 2012
but could u tell me ,which of these 2 codes is much better!
img1=img./255 % or img1=img.*255 ???
[counts x]=imhist(img1)
nz=find(counts)
p=count(nz)./(m*n) %m*n= number of pixels in img
sum(sum(p))
entropy=-sum(p.*log10p) % or log2 ???

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Junaid
Junaid am 2 Mai 2012
I guess if you normalize your histogram, you will your desired output .. May be ?
As after normalizing your histogram, the sum of all values will be 1. That is actually you want (all values in Histogram will be in between 0-1).
Easiest way to normalize is to divide histogram by sum of histogram.
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Junaid
Junaid am 3 Mai 2012
Oh... thanks for letting me know.
mohammed ABDEEN
mohammed ABDEEN am 3 Mai 2012
sorry junaid if i ddnt reply yet !! ... i'll write my code here after i've finished it.

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