Why my calculation of standard deviation of an image is different from the built-in function

1 visualización (últimos 30 días)
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;

Respuestas (4)

Jan
Jan el 28 de Jun. de 2013
Editada: Jan el 28 de Jun. de 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 comentarios
ZhG
ZhG el 28 de Jun. de 2013
Actually, it is the same after I had modified my program as your answer. I am confused. The result obtained by built-in function is only 9.6938. But a result of 72.0412 was obtained by my program.
Jan
Jan el 28 de Jun. de 2013
Editada: Jan el 28 de Jun. de 2013
I have converted the image to the type double to avoid saturation effects with integer types. So please try my code and not a program modified like my code.

Iniciar sesión para comentar.


ZhG
ZhG el 28 de Jun. de 2013
However, I found where the problem happens. If I calculate standard deviation of an image without converting it into double, it will generate a result of 9.6938. However, it will does return 72.0412 afte I transformed it to double.
Is there anyone know what the problem is?
  3 comentarios
Jan
Jan el 28 de Jun. de 2013
Editada: Jan el 28 de Jun. de 2013
Operations on integer types are affected of the saturaion:
uint8(250) + uint8(250) == uint8(255) !!
Iain
Iain el 28 de Jun. de 2013
Wind-up is a type of saturation, and I use the terms almost interchangably.

Iniciar sesión para comentar.


Image Analyst
Image Analyst el 28 de Jun. de 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()?

ZhG
ZhG el 28 de Jun. de 2013
thank you all, guys.

Categorías

Más información sobre Images en Help Center y File Exchange.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by