Count equal values in multiple matrices
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I have a number of M by N matrices with random integers between 1 and 5, for example if I have:
X1 = [1 2 1 3 5 4 2 4 3; 5 3 3 2 5 4 3 2 3; 3 3 2 4 5 3 2 4 1; 4 5 3 4 1 4 2 3 1;]
X2 = [5 4 4 3 5 1 2 3 1; 5 3 2 4 2 3 1 4 3; 4 2 3 3 4 1 3 5 2; 1 3 1 3 4 5 3 2 4;]
The answer is
X2 = [1 1 1 2 2 1 2 1 1; 2 2 1 1 1 1 1 1 2; 1 1 1 1 1 1 1 1 1; 1 1 1 1 1 1 1 1 1;]
However, my real matrix is much larger and I have >>2 matrices to compare. What is a fast vectorized way to perform this operation?
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Image Analyst
2014-9-30
What if you have
X1 = [1 2 1 3 5 4 2 4 3; 5 3 3 2 5 4 3 2 3; 3 3 2 4 5 3 2 4 1; 4 5 3 4 1 4 2 3 1;]
X2 = [5 4 4 3 5 1 2 3 1; 5 3 2 4 2 3 1 4 3; 4 2 3 3 4 1 3 5 2; 1 3 1 3 4 5 3 2 4;]
X3 = [1 2 1 9 1 4 2 4 3; 5 3 3 2 5 4 3 2 3; 3 3 2 4 5 3 2 4 1; 4 5 3 4 1 4 2 3 1;]
X4 = [5 4 4 9 1 1 2 3 1; 5 3 2 4 2 3 1 4 3; 4 2 3 3 4 1 3 5 2; 1 3 1 3 4 5 3 2 4;]
Look at the 4th column - we have a pair of 3's and a pair of 9's. And in the 5th column, a pair of 1's. So in the output would it now be 4? In the extreme, you could just take a histogram of every element across the X arrays. The histogram will have 1's in non-repeated bins and 2, 3, 23, or whatever in bins that are repeated for some number. What do you want to do in that case?
回答(2 个)
Michael Haderlein
2014-9-30
I guess you don't want to overwrite X2. Introducing X3, the easiest way is:
X3 = 1+(X1==X2);
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Iain
2014-9-30
If you can concatenate your matrices in the 3rd dimension:
x_oth = X2;
x_oth(:,:,2) = X3;
...
Then you may be able to use
Y =1+ sum(bsxfun(@eq,X1,x_oth),3);
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