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
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
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
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
Image Analyst
2022 年 2 月 8 日
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
Walter Roberson
2012 年 1 月 19 日
You should be able to extrapolate.
y = uint8(255 .* ((double(m)-double(min1))) ./ double(max1-min1));
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