Counting number of same neighbourhood pixels between two matrices.

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Stewart Tan
Stewart Tan 2019-9-10
评论： Shunichi Kusano 2019-9-10
So i have two made up matrices, which represents an image block.
p1 = [4 7 1 2;
4 5 3 1;
8 9 1 10;
8 19 2 1];
p2 = [4 7 1 2;
4 5 3 2;
5 1 0 11;
8 19 2 0]
and what i want to do is to count how many neighbourhood pixels are similar between both matrices above. To do so, i first locate the centre pixel in each matrices:
centre_pix = floor(([4 4]+1)/2)
p1_centre = p1(2,2) %2,2 is the output of centre pix
p2_centre = p2(2,2)
Hence the number 5 in row 2, column 2 in each matrix is the centre pixel.
What I'm trying to do now is to count the number of neighbourhood pixel of the centre pixel that are similar between the two matrices.
Which would be 4,7,1,4,3 therefore the count is 5.

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采纳的回答

Shunichi Kusano
Shunichi Kusano 2019-9-10
编辑：Shunichi Kusano 2019-9-10
Hi Tan,
If "similar" means "exactly the same" or "the difference is zero", it is not difficult to implement. What the following codes do is to take the difference between the two matrices, then to count the number of neighbourhood pixels whose value is zero.
dif_p = double(p1) - double(p2);
indSame = dif_p == 0; %
fun = @(x) nnz(x) - uint8(x(2,2));
result = nlfilter(indSame, [3 3], fun)
hope this helps.

更多回答（1 个）

Bruno Luong
Bruno Luong 2019-9-10
编辑：Bruno Luong 2019-9-10
This command counts the number of 3x3 neigbouring pixels that exactly match
conv2(p1==p2,ones(3),'same')

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