Shuffle matrix rows so that common elements are in completely different order than in a said row

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Pseudoscientist
Pseudoscientist 2019-4-26
编辑： Matt J 2019-4-26
Say we have a matrix row 'current_row'
1 2 3
and a matrix A
1 2 5
1 3 6
2 4 3
The task is to re-arrange matrix A so that the common elements with 'current:_row' in each row are in compelely different indices.
One solution would be:
5 1 2
6 3 1
2 3 4
So that
for i=1:size(A,1)
find(current_row == A(i,:))
end
gives always three []'s

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采纳的回答

Matt J
Matt J 2019-4-26
编辑：Matt J 2019-4-26
Here is a vectorized implementation of what John described.
[m,n]=size(A);
I=repmat((1:m).',1,n);
B=A; %copy for shuffling
pool=find(any(B==current_row,2));
while ~isempty(pool)
p=numel(pool);
[~,J]=sort(rand(p,n),2);
B(pool,:)=B( sub2ind([m,n], I(pool,:), J) ); %shuffle rows in 'pool'
pool=find(any(B==current_row,2)); %narrow the search
end
result=Anew;

更多回答（1 个）

John D'Errico
John D'Errico 2019-4-26
What you are looking for is a variation of derangement - thus a permutation such that no element is in the same location as it started.
A simple solution? Just continue to shuffle each other row until it meets the goal, that no common elements are in the same location as they are in current_row. Stop when it does, and then go on to the next row. WTP?
You can probably improve on that scheme a bit, by finding the elements in common between the two rows. Then first permute them into ANY locations, as long as they are not the same as the current_row. Having done that derangement operation, then permute the other elements into any random permutation.

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