Build a relation between matrix components
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Hi, I wanted to ask whether there is an easier method to calculate the sum of relations for each component by considering the neghborhood of each point of a matrix. For example, by considering a 3*3 matrix:
a=[a11,a12,a13; a21,a22,a23; a31,a32,a33]
I want to calculate the sum of relations, in which the relation is:
r=(a-b)/(a+b)
So for example, for component a11, I would have:
r11=(a11-a12)/(a11+a12) + (a11-a22)/(a11+a22) + (a11-a21)/(a11+a21)
Essentially I want to calculate the sum of relations around each point. For example, for a11, there are 3 other points, so I have 3 relations summed up; for another point, for example, point a22, I have 8 other points in the neighborhood of a22, so I would have 8 relations that would be summed up. I've written a function for it but it's very long. A small portion of it:
function [c] = relation(y, i, j, m, n)
c=0;
if i~=1 && j~=1 && i~=m && j~=n
c = (y(i,j) - y(i,j+1))/ (y(i,j) + y(i,j+1));
c = c + (y(i,j) - y(i+1,j+1)) / (y(i,j) + y(i+1,j+1));
c = c + (y(i,j) - y(i+1,j))/ (y(i,j) + y(i+1,j));
c = c + (y(i,j) - y(i+1,j-1))/ (y(i,j) + y(i+1,j-1));
c = c + (y(i,j) - y(i-1,j-1))/ (y(i,j) + y(i-1,j-1));
c = c + (y(i,j) - y(i-1,j))/ (y(i,j) + y(i-1,j));
c = c + (y(i,j) - y(i-1,j+1))/ (y(i,j) + y(i-1,j+1));
c = c + (y(i,j) - y(i,j-1))/ (y(i,j) + y(i,j-1));
end
if i==1 && j==1
c = (y(i,j) - y(i,j+1))/ (y(i,j) + y(i,j+1));
c = c + (y(i,j) - y(i+1,j+1)) / (y(i,j) + y(i+1,j+1));
c = c + (y(i,j) - y(i+1,j))/ (y(i,j) + y(i+1,j));
end
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.
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I wanted to ask whether there would be an easier way to calculate the sum of relations.
Thank you.
2 件のコメント
採用された回答
Rik
2021 年 7 月 1 日
It is almost possible to use a convolution to solve your question, but I couldn't see how, so I did it with a replicated array.
% example inputs
A = reshape(1:16,4,4);
n = numel(A)-1;
%first we pad the array with NaNs
A_pad=NaN(size(A)+2);
A_pad(2:(end-1),2:(end-1))=A;
%now we can replicate this for the 8 different shifts we need to apply
n=0;
B=repmat(A,[1 1 8]);
for k1=[-1 0 1]
tmp1=circshift(A_pad,k1,1);
for k2=[-1 0 1]
if k1==0 && k2==0,continue,end % skip the null shift
tmp2=circshift(tmp1,k2,2);
n=n+1;
B(:,:,n)=tmp2(2:(end-1),2:(end-1));
end
end
%calculate relations
%NB: pre-R2016b you need to use A=repmat(A,[1 1 8]); first
R=(A-B)./(A+B);
%sum over the third dimension, using only the valid shifts
R=sum(R,3,'omitnan')
2 件のコメント
Rik
2021 年 8 月 3 日
As with all scripts in Matlab: yes. Can you guess how? Your first guess is probably correct.
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