Computing the integral of a binary image
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Hello fellows:
I am trying to figure out how to compute the integral of the square of the difference between 2 images. So far I know that I have to convert the image to a binary one, but in order to compute the integral I am thinking of plotting the profile of the intensity values of each image than subtract one from another and integrate after I square the result. What do you think about that methodology? If it is wrong or you have a better one can you share keywords for the approach to do that?
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Walter Roberson
2012-7-2
The integral of the square of the difference between two binary images is
sum(BinaryImage1(:) ~= BinaryImage2(:))
That is:
0 - 0 = 0; 0^2 = 0; and (0 ~= 0) = 0
0 - 1 = -1; (-1)^2 = 1; and (0 ~= 1) = 1
1 - 0 = 1; 1^2 = 1; and (1 ~= 0) = 1
1 - 1 = 0; 0^2 = 0; and (1 ~= 1) = 0
and thus the square of the difference is the same as ~= of the values.
2 个评论
Star Strider
2012-7-2
Isn't that the same as 'xor'?
sum(xor(A,B))
Sean de Wolski
2012-7-2
@Star Strider: yup:
xor([1 1 0 0],[0 1 1 0])
ne([1 1 0 0],[0 1 1 0])
更多回答（1 个）
Image Analyst
2012-7-2
Sounds like you just want the RMS difference between the two images, so I guess I don't understand why you think you must convert the images to binary images, or why " plotting the profile" is necessary. For gray scale images, why not just do
rmsDifference = sqrt(sum((double(grayImage1(:)) - double(grayImage2(:))) .^2));
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2012-7-2
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