Array Correlation - compute array difference with its neighbor elements

2 ビュー (過去 30 日間)
ichi
ichi 2017 年 10 月 29 日
コメント済み: Walter Roberson 2017 年 10 月 30 日
Hi,
Do you know the fastest way to compute array mean difference with its neighbor elements in a N x N matrix? and is there a possibility to perform for second neighbors, third and so on.
Suppose you have
a=[1 4 5 3 6 ; 8 10 0 9 11 ; 2 5 20 15 9 ; 8 12 4 6 0 ; 7 9 18 2 5]
so I want a code to give this result for mean difference of first neighbors:
d(1,1)=[(1-4)+(1-8)]/2, d(1,2)=[(4-1)+(4-5)+(4-10)]/3, d(2,2)=[(10-8)+(10-0)+(10-4)+(10-5)]/4 and etc. For second neighbors: dd(3,3)=[(20-2)+(20-9)+(20-5)+(20-18)/4].
thank you

回答 (1 件)

Walter Roberson
Walter Roberson 2017 年 10 月 29 日
mask = [0 1 0; 1 0 1; 0 1 0];
Ncount = conv2(ones(size(a)), mask, 'same');
d = double(a) - conv2(double(a), mask, 'same')./Ncount;
You can leave out the double() if you are sure that a will not be integer data type (images are usually integer data type.)
For second difference, use an appropriate larger mask such as
[0 0 1 0 0; 0 0 0 0 0; 1 0 0 0 1; 0 0 0 0 0; 0 0 1 0 0]
  2 件のコメント
ichi
ichi 2017 年 10 月 30 日
Thank you for the answer, I knew the conv2(a,mask,'same'), gives you the sum of the neighbors of each element, but this method is so good.
Walter Roberson
Walter Roberson 2017 年 10 月 30 日
If the question is resolved, please Accept the answer.

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