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Image Analyst
Image Analyst il 11 Ott 2012
Wouldn't it just go like this
difference = single(image1) - single(image2);
squaredError = difference .^ 2;
meanSquaredError = sum(squaredError(:)) / numel(image1);
rmsError = sqrt(meanSquaredError);
Of course you could compact that all into one line if you want.

5 Commenti

Mahua Nandy(Pal)
Mahua Nandy(Pal) il 14 Ott 2012
Thanks a lot. Please let me know why single() has to be used?
you need to use single so that you can have negative numbers. If your image is uint8 (0-255 gray levels like most images), then it is UNSIGNED, which means anything that should be negative will be clipped at zero:
smallNumber = uint8(5);
bigNumber = uint8(200);
u8 = smallNumber - bigNumber
s8 = single(smallNumber) - single(bigNumber)
In the command window:
u8 =
0
s8 =
-195
Mahua Nandy(Pal)
Mahua Nandy(Pal) il 16 Ott 2012
Thank you for your lucid clarification.
MAT-Magic
MAT-Magic il 18 Gen 2020
Modificato: Image Analyst il 18 Gen 2020
@Image Analyst, can I use this formula for two vectors having same length?
difference = single(image1) - single(image2);
squaredError = difference .^ 2;
meanSquaredError = sum(squaredError(:)) / numel(image1);
rmsError = sqrt(meanSquaredError);
?
Yes, you can, but I'd use mean() instead of sum to simplify it:
differenceImage = single(image1) - single(image2);
squaredErrorImage = differenceImage .^ 2;
meanSquaredError = mean(squaredErrorImage(:)) % A scalar
rmsError = sqrt(meanSquaredError)
And if they're vectors of the same shape (row or column) then you don't even need the (:).

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Daniel Shub
Daniel Shub il 11 Ott 2012
Modificato: Daniel Shub il 11 Ott 2012
Just to be a little bit difference. If you have the DSP system toolbox you can do
step(dsp.RMS('Dimension', 'all'), x)
where x is your error signal. So in the case of two imagines (image1 and image2)
image1 = randn(128);
image2 = randn(128);
x = image1-image2;

6 Commenti

Image Analyst
Image Analyst il 11 Ott 2012
Would x = cat(3, image1, image2)?
Daniel Shub
Daniel Shub il 11 Ott 2012
IA what?
Image Analyst
Image Analyst il 11 Ott 2012
He has two images, say image1 and image2 where image1 is the reference image and another image that is the misaligned one. He has attempted to "fix" (register) by aligning it with image1 and that "registered" image is image2. Where would those two images go into your formula?
Daniel Shub
Daniel Shub il 11 Ott 2012
I got it, dsp.RMS calculates the RMS of an n-d signal. So to get the RMS error, x needs to be the error signal. In the case of two images x is the difference between the images.
Ishara Nipuni
Ishara Nipuni il 25 Gen 2019
I calculated the RMS value of my image registration algorithm by using your code. But I can't understand how to do a validation for my registration algorithm with th RMS value. Can you tell me please how can I com to conclusions about the accuracy of my registration algorithm with the use of RMS values?
Ishara Nipuni
Ishara Nipuni il 25 Gen 2019
I calculated the RMS value with getting the same image as image1 and image2. But the value was not zero. But I think that it should be zero. Can you please explain me about it?

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