Subtracting two numbers gives 0 as answer

12 vues (au cours des 30 derniers jours)
Kostantinos Thanasis
Kostantinos Thanasis le 30 Oct 2020
Commenté : Ameer Hamza le 30 Oct 2020
I have two matrices that represent a picture in the gray scale. Values 0-255.
I want to calculate a metric called MSE which uses the formula
where p1(w,h) coresponds to the value of the pixel (w,h) in the first picture. Same for p2. Then all of that squared. For simplicity i used to for loops to iterate the matrices. Each iteration a find the difference,then to the power of 2 then add it to the total.
The problem is that,let's say for the first iteration the p1(w,h) = 156 and the p2(w,h) = 169. When i subtract them i do not get -13. I get 0.
May be this is a silly question but im new to this and have been trying to solve this for hours.
ΝΟΤΕ: I was always getting a 0 in return so i did one operation per line to find the issue.

Réponse acceptée

Ameer Hamza
Ameer Hamza le 30 Oct 2020
Modifié(e) : Ameer Hamza le 30 Oct 2020
You are getting zero because the images are probably in uint8 format and negative results are given a value of zero in that case
>> uint8(10)-uint8(1)
ans =
uint8
9
>> uint8(1)-uint8(10)
ans =
uint8
0
The solution is to convert the matrices to double() before subtraction. At beginning of your function, add the lines
startingImage = double(startingImage);
compressedImage = double(compressedImage);
mse = 0; ...
...
Also note that you can do these things without loop.
startingImage = double(startingImage);
compressedImage = double(compressedImage);
mse = mean((startingImage-compressedImage).^2, 'all')
or if you are using an older version of MATLAB
mse = mean((startingImage(:)-compressedImage(:)).^2)
  2 commentaires
Kostantinos Thanasis
Kostantinos Thanasis le 30 Oct 2020
Do not know the policy for commenting here(in respect to the stack overflow one) but thank you very much. You explained the issue i was facing very well so i could have solved it on my own but also also provided me with the solution.
I am gratefull.
Ameer Hamza
Ameer Hamza le 30 Oct 2020
I am glad to be of help! :)

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