imwrite, what it does with fractions and negative values

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James Latshaw
James Latshaw le 13 Juil 2021
Commenté : James Latshaw le 13 Juil 2021
Hello,
This is an odd question, but how does imwrite handle fractions or negative inputs?
I am interested in reproducing this effect.
For example If I randomly generate some data and save it as an image and then read the saved image, I will see that matlab (somehow) assigned a uint8 value for each pixel. What modiciations do I need to perform to 'a' in order to get it to be equal to 'b' ?
Thank you for taking time to read this.
-James
clc
a = rand(2,10)-.5;
imwrite(a,'cake.jpg');
b=imread('cake.jpg');
a
b

Réponse acceptée

Image Analyst
Image Analyst le 13 Juil 2021
Modifié(e) : Image Analyst le 13 Juil 2021
Try this illustrative demo:
a = rand(2,5) - 0.5 % Range is between -0.5 and +0.5
fileName = 'Delete_me.jpg';
imwrite(a, fileName);
b=imread(fileName)
delete(fileName); % Delete temporary file.
fileName = 'Delete_me.png';
imwrite(a, fileName);
b=imread(fileName)
delete(fileName); % Delete temporary file.
a2 = a*255
imwrite(a2, fileName);
% If image is floating point, it expects it to be in the range 0-1.
% Then it will scale 0-1 to 0-255 and clip negative to 0 and more than 1 to 255.
b=imread(fileName)
delete(fileName); % Delete temporary file.
a3 = uint8(a*255) % Make uint8. This may clip values <0 or >255 to 0 and 255.
imwrite(a3, fileName);
% If image is floating point, it expects it to be in the range 0-1.
% Then it will scale 0-1 to 0-255 and clip negative to 0 and more than 1 to 255.
b=imread(fileName) % Comes back unchanged as it should with uint8 values.
delete(fileName); % Delete temporary file.
fileName = 'Delete_me.mat';
save(fileName, 'a');
s = load(fileName);
b = s.a
delete(fileName); % Delete temporary file.
As you can see if it's a floating point number, imwrite will expect the number to be in the range 0-1 and it will multiply it by 255 and then cast to uint8. Numbers below 0 get clipped to 0, and numbers above 1 get clipped to 1 (so they'll be recalled as 255).
Moreover, run it a few times and you will probably see that the JPG recalled numbers are slightly different than the PNG numbers because of the lossy compression. So that's why you should use PNG, which is pretty much now a de facto standard format that most people use, and not JPG (which image analysts hate because it changes the values).
If you want to save the fractional part of the numbers then you should use MATLAB's proprietary format .MAT which will recall the floating point image with all its original, unchanged values.
I'm not sure exactly what you mean when you say "I am interested in reproducing this effect." What effect? What does "reproduce" mean to you.
  5 commentaires
Image Analyst
Image Analyst le 13 Juil 2021
Actually that doesn't agree with me. Note the last few words of that sentence "but I doubt it".
JPG convert the image to the spectral domain and then does a kind of low pass filter. Not exactly like a Fourier filter though because it does it block-by-block. That's what gives JPG images such lousy appearance with the notorious "JPG block artifacts" where you have huge rectangular blocks of uniform color. It's expecially bad in slowly varying areas, like the sky.
If your ground truth has positive and negative values, then you'd want to scale your images to 0-255 before saving to an image file. Use the same scaling factors for all images, or, to simulate possible changes in brightness (like you'd get if you did augmentation) you can scale each independently.
James Latshaw
James Latshaw le 13 Juil 2021
Thanks for all of your help and for your clarifiation. I selected yours as the answer.
Best Regards,
James

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Plus de réponses (2)

Mohammad Sami
Mohammad Sami le 13 Juil 2021
You can change to using random integer function instead of rand function.
Also change the image format to png instead of jpg.
a = randi([0,255],2,10,'uint8');
imwrite(a,'cake.png');
b=imread('cake.png');
isequal(a,b)
ans = logical
1
  1 commentaire
James Latshaw
James Latshaw le 13 Juil 2021
Thank you for you reply. This is interesting but I am trying to reproduce the transformation that imwrite seems to be applying to matrix 'a'. See my above reply Image Analyst.
:)

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Walter Roberson
Walter Roberson le 13 Juil 2021
%approximately
function imwrite(data, filename)
if isfloat(data)
data = uint8(data * 255);
end
and so on
end
This is a simplification, as the 255 depends upon the bit depth options.
Notice that the range of the floating point data is not tested first. imwrite() does not check to see whether the data happens to be in the range 0.0 to 255.0 and if so then simply uint8() it: imwrite() multiplies all floating point data by 255. If the floating point data happened to have values greater than 1.0 then the result of the multiplication will be greater than 255.0, and no matter what range that turns out to be, it is then processed by uint8() [unless 16 bit output was requested.]
uint8() has its usual properties:
  • any negative value "underflows" to 0.
  • any value greater than 255 "saturates" to 255.
  • any non-integer value is rounded to the nearest integer before being changed to integer data type.
  1 commentaire
James Latshaw
James Latshaw le 13 Juil 2021
Thank you for your reply.
I agree that when using imwrite() that values <0 should be 0. But if that is the case then how do you explain the below?
a = rand(2,10)-.5;
imwrite(a,'cake.jpg');
b=imread('cake.jpg');
a =
-0.3219 -0.4433 -0.1642 -0.2911 0.1754 0.4121 0.2455 0.0619 0.0972 -0.3659
-0.1404 0.0219 -0.3243 0.4052 -0.0315 -0.3960 0.2363 -0.3158 -0.2001 -0.2874
>> b
b =
2×10 uint8 matrix
0 0 0 19 30 99 74 13 15 9
2 4 11 80 12 12 53 10 8 4
look at a(1,4) = negative value but b(1,4) = positive
Also, b(2,2)=4 and b(2,10)=4 but a(2,2) and a(2,10) are opposite sign. Opposite sign values are produce the same output value of 4.
I think something to do with the compression of the JPG images (using the eignevalues) is causing this effect. I am interested in exploring why this is happening becuse this 'transformation' is having a positive imact of my trianing data for a neural network. Just trying to undertand what is happening inside imwrite.
Thanks,
James

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