Fastest calculation method to: Count elements in a matrix, in the neighborhood of some element, having some value

For an element (i,j) in a matrix I want to calculate the number of neighboring elements which have the same value as that in (i,j). I currently have made the straightforward code:

for i=2:ynum+brd2-1
                    for j=2:xnum+brd2-1
                        if x(i,j)==x(i-1,j+1)
                            e(i,j)=e(i,j)+1;
                        end;
                        if x(i,j)==x(i,j+1)
                            e(i,j)=e(i,j)+1;
                        end;
                        if x(i,j)==x(i+1,j+1)
                            e(i,j)=e(i,j)+1;
                        end;
                        if x(i,j)==x(i-1,j)
                            e(i,j)=e(i,j)+1;
                        end;
                        if x(i,j)==x(i+1,j)
                            e(i,j)=e(i,j)+1;
                        end;
                        if x(i,j)==x(i-1,j-1)
                            e(i,j)=e(i,j)+1;
                        end;
                        if x(i,j)==x(i,j-1)
                            e(i,j)=e(i,j)+1;
                        end;
                        if x(i,j)==x(i+1,j-1)
                            e(i,j)=e(i,j)+1;
                        end;
                        e(i,j)=8-e(i,j);
                    end
                end

which works just fine and is designed for a 9 point stencil (I look at the 8 nearest neighbors).

The problem is that it is slow (or probably much slower than another method) and I want to do the same thing with a 37 point stencil that looks like this:

    000
   00000
  0000000
  000X000
  0000000
   00000
    000

where the x is (i,j), instead of

000
0X0
000

I assume I should use some kind of countif or sum(sum())methods, but I am new to matlab and do not know what the fastest operations are.

Is it fastest to count over a rectangle around the circle and then subtract the 3 points near the vertices?

Thanks

[Matt J]

Are there any special restrictions that on the matrix data that you're working with? I would guess, for example, that the x(i,j) values are all integers. Otherwise, you would be comparing x(i,j) with its neighbors using a tolerance for floating point differences.

[Image Analyst]

I guess I'm not understanding why you don' just use the other method which you say is faster. Care to explain?

Answer 1

[Matt J]

I would expect this to be faster. It's for a 3x3 stencil, but it can easily be generalized.

    stencil=zeros(3);
    stencil(5)=1;

    e=zeros(size(x));

     for i=[1:4,6:8]

        stencil(i)=-1;

         e(2:end-1,2:end-1)=e(2:end-1,2:end-1) + ~conv2(x,stencil,'valid');

        stencil(i)=0; 

     end