Ordering rows based on value
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Suppose I have a matrix of values as follows:
matches = [1,2;1,4;2,5;3,4;3,6;5,6;];
Is there a way to "connect these rows together based on a common value in the rows column. The desired result is as follows:
connected = [1,2,5,6,3,4,1]
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채택된 답변
Andrei Bobrov
2012년 4월 13일
k1 = bsxfun(@minus,matches(:,1),matches(:,2)');
[ii,jj] = find(k1 == 0);
a = unique([ii;jj]);
out1 = unique(matches(a,:)');
f=matches(setdiff(1:6,a),:);
[ff,bb,cc] = unique(f','first');
[~,id] = sort(bb,'descend');
out2 = ff(id);
connected = [out1;out2(2:end)]
추가 답변 (2개)
Geoff
2012년 4월 13일
Try this:
matches = [1,2;1,4;2,5;3,4;3,6;5,6;];
connected = matches(1);
while size(matches,1) > 0
[r,c] = find(matches == connected(end), 1);
if isempty(r)
disp('End of chain');
break;
end
connected(end+1) = matches(r, 3-c);
matches(r,:) = [];
end
disp(connected);
It's a little destructive. It systematically removes rows from the matches matrix until there are no more links or there are no rows left. It copes with ambiguity (if there is any) by selecting the first available value (determined by whatever find deems as the first value). In some some cases, this might end the chain prematurely.
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Geoff
2012년 4월 13일
Well it depends on your data. For this data, it's fine. And maybe you'll always throw 'sane' data at it. I just imagined that some data sets won't produce a closed loop, will branch, or contain a side-loop that itself is closed.
Imagine the above data with [6 7; 7 8; 8 3] on it. If we happen to follow [6 7] instead of [3 6] then we'll touch all the values: [1 2 5 6 7 8 3 4 1]. But if we follow [3 6] first, we'll skip over the side-loop and produce the original output.
Anyway, I just think about that stuff, but I don't know what your data is representing or what its constraints are. If this solves your problem I'm happy. Don't forget to close off the question by accepting.
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