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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 Comments
Xylo
Xylo on 11 Mar 2014
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

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Answers (3)

Image Analyst
Image Analyst on 19 Jan 2012
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 Comments
mmm ssss
mmm ssss on 19 Jan 2012
image analyst
i think that your opininon is correct but, in many paper they use the grayscale normalization to reduce the differences in illumination.
Image Analyst
Image Analyst on 19 Jan 2012
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.

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Syed Ahson Ali Shah
Syed Ahson Ali Shah on 8 Feb 2022
Edited: Syed Ahson Ali Shah on 10 Feb 2022
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 Comments
Image Analyst
Image Analyst on 8 Feb 2022
No it's not:
Originalimage = [100, 200]
Originalimage = 1×2
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
NormalizedImage = 1×2
255 510
% Do my formula:
normalizedImage = uint8(255*mat2gray(Originalimage))
normalizedImage = 1×2
0 255
Syed Ahson Ali Shah
Syed Ahson Ali Shah on 10 Feb 2022
There was typo mistake. I corrected now.
My answer is 100% correct. I guarantee

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Walter Roberson
Walter Roberson on 18 Jan 2012
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 Comments
Walter Roberson
Walter Roberson on 19 Jan 2012
You should be able to extrapolate.
y = uint8(255 .* ((double(m)-double(min1))) ./ double(max1-min1));
mmm ssss
mmm ssss on 19 Jan 2012
can you see the modified code:
clear all
>> m=imread('E:\master_matlab\HandVein_DataSet\0010hv3.bmp');
min1=min(min(m));
max1=max(max(m));
y = uint8(255 .* ((double(m)-double(min1))) ./ double(max1-min1));
>> imshow(m);
>> figure,imshow(y);title(,normalization,);
i implemented also image analyst'method:
J = filter2(fspecial('sobel'), m);
K = mat2gray(m);
figure,imshow(K);
can you give me your opinion in the resultant images (Y&K) , is they good or the image before normalization is more suitable.

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