how can i perform gray scale image normalization???
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i want to implement normalization to gray scale image to reduce the effect of illumination's differences.
the eq. of the grayscale normalization is :
y=((x-min)*255/(max-min))
x : gray scale value of original image.
y : gray scale value of op image(after normalization).
min : minimum gray scale for the original image.
max : maximum gray scale for the original image.
i tried to perform this by :
m=imread();
min1=min(min(m));
max1=max(max(m));
y=((m-min1).*255)./(max1-min1);
imshow(m);figure,imshow(y);
but it is wrong code .
i dont know why ?
is there any help?
regards
5 件のコメント
Image Analyst
2012 年 1 月 19 日
Perhaps because you didn't pass any filename into imread(). Perhaps because you left off the [] in imshow(y, []). It's floating so unless it's in the range 0-1 you need to give [] as the second arg to imshow().
I think you've been around here long enough to know that you need to tell us what the error message is (by copying and pasting the red text from the command window) because just saying "it is wrong code" is not sufficient for us to debug your code or figure out what is wrong or unexpected behavior.
mmm ssss
2012 年 1 月 19 日
Image Analyst
2012 年 1 月 19 日
Like I said, you probably just didn't display it correctly. See my answer below.
Walter Roberson
2012 年 1 月 19 日
We would need to see the current code.
Xylo
2014 年 3 月 11 日
you can use double() before main function....
y=double((m-min1).*255./(max1-min1)); and as m is a 2D variable, y should be 2D variable. i.e u have to write as for i=1:m for j=1:n y(i,j)=double((m(i,j)-min1).*255./(max1-min1)); end end
回答 (3 件)
Image Analyst
2012 年 1 月 19 日
Or you can simply do this:
normalizedImage = uint8(255*mat2gray(grayImage));
imshow(normalizedImage);
and not worry about the normalization because mat2gray will do it for you.
4 件のコメント
mmm ssss
2012 年 1 月 19 日
Image Analyst
2012 年 1 月 19 日
It does exactly what you asked for and exactly what all these other formulas here are doing. It maps the min of your array to 0 and the max of your array to 255. You can subtract the array from what you got with the other formulas and you'll see that everything is zero, meaning they're the same. If that doesn't have a "good effect" then none of the formulas here will either.
mmm ssss
2012 年 1 月 19 日
Image Analyst
2012 年 1 月 19 日
Then maybe their algorithm uses image normalization as just one step in the process and maybe you're not doing all the steps. Or else maybe their algorithm is not appropriate for the kind of video or images you have.
Syed Ahson Ali Shah
2022 年 2 月 8 日
編集済み: Syed Ahson Ali Shah
2022 年 2 月 10 日
1 投票
This is the Formula:
Normalized Image = (Original image - min of image) * ((newMax-newMin) / (ImageMax - ImageMin)) + newMin
where newMax and newMin is 255 and 0 respectively for the case when normalization is between 0 to 255.
2 件のコメント
No it's not:
Originalimage = [100, 200]
minofimage = min(Originalimage(:));
ImageMin = min(Originalimage(:));
ImageMax = max(Originalimage(:));
newMax = 255;
newMin = 0;
% Do the formula he gave.
NormalizedImage = (Originalimage - minofimage) * ((newMax-newMin) / (ImageMax - ImageMin)) + newMax
% Do my formula:
normalizedImage = uint8(255*mat2gray(Originalimage))
Syed Ahson Ali Shah
2022 年 2 月 10 日
There was typo mistake. I corrected now.
My answer is 100% correct. I guarantee
Walter Roberson
2012 年 1 月 18 日
I would suggest you use
y = uint8(255 .* ((double(m)-min1)) ./ (max1-min1));
With your existing code, the (x-min) would be okay, but multiplying by 255 would get saturation to 255 whenever the difference was not 0, and then you would get integer division of that 0 or 255 by the range interval.
5 件のコメント
mmm ssss
2012 年 1 月 18 日
Walter Roberson
2012 年 1 月 19 日
y = uint8(255 .* ((double(m)-double(min1))) ./ (max1-min1));
mmm ssss
2012 年 1 月 19 日
Walter Roberson
2012 年 1 月 19 日
You should be able to extrapolate.
y = uint8(255 .* ((double(m)-double(min1))) ./ double(max1-min1));
mmm ssss
2012 年 1 月 19 日
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