Why my calculation of standard deviation of an image is different from the built-in function
1 vue (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Hello,
My code is shown below, and I opened the built-in function. I found it just calculates the square root or variance. So, the method is the same. I don't know why.
function ret = yStdDeva(im)
tic;
[tR, tC] = size(im);
mean1 = mean(im(:));
% Calculate square of difference between pixels and mean
sdpm = (double(im) - mean1).^2;
sum_sdpm = sum(sdpm(:));
% Calculate variance
imvar = (1/(tR*tC ))*sum_sdpm;
% Calculate Standard deviation
ret = (imvar)^(1/2);
toc;
0 commentaires
Réponses (4)
Jan
le 28 Juin 2013
Modifié(e) : Jan
le 28 Juin 2013
You simply use a wrong formula: You have to normalize by the number of elements minus 1. In addition there can be a difference between SQRT and ^0.5 due to the different numerical implementations.
d = double(im);
C = d(:) - mean(d(:));
S = sqrt(sum(C .* C, 1) ./ (numel(d) - 1)):
4 commentaires
Image Analyst
le 28 Juin 2013
I don't see much difference - just a really slight difference (less than a thousandth of a gray level):
im=imread('cameraman.tif');
% Your way:
[tR, tC] = size(im);
mean1 = mean(im(:));
% Calculate square of difference between pixels and mean
sdpm = (double(im) - mean1).^2;
sum_sdpm = sum(sdpm(:));
% Calculate variance
imvar = (1/(tR*tC ))*sum_sdpm;
% Calculate Standard deviation
ret = (imvar)^(1/2)
% Standard, built-in way:
std(double(im(:)))
In the command window:
ret =
62.3412396872523
ans =
62.3417153186086
Why are you wanting to do it yourself anyway? Why not just use std()?
Voir également
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!