Count equal values in multiple matrices

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?

[Image Analyst]

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?

Answer 1

[Michael Haderlein]

I guess you don't want to overwrite X2. Introducing X3, the easiest way is:

 X3 = 1+(X1==X2);

[Christopher]

What if I have many matrices? Should I write

Y = 1+(X1==X2)+(X1==X3)+(X1==X4)+(X1==X5)+(X1==X6)+(X1==X7)+(X1==X8)+(X1==X9)+(X1==X10)+(X1==X11)+(X1==X12)+(X1==X13)+(X1==X14)+(X1==X15)+(X1==X16)+(X1==X17)+(X1==X18)

Or is there a faster way to do this? All the matrices are the same size. Thanks.

[Sean de Wolski]

You should avoid creating matrices like that:

faq

Answer 2

[Iain]

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);