For every matrix row find rows within the same matrix that have more than one common element in same index below the row

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Pseudoscientist
Pseudoscientist 2019-4-27
编辑： Pseudoscientist 2019-4-27
Let's say I have matrix A, every row contains three values and for every row of the matrix A need to find other rows that have more than one common element in same index below the present row. For example for row i, we are interested in rows i+1:end. Matrix A is sorted. If for a row there's no rows found that fulfil this condition the result cell should have value 0 in that row number.
A = [
1 2 3
1 4 5
3 4 5
1 2 4
1 2 5
2 4 5]
Result =
4,5
3,5,6
6
5
0
I'm looknig for a fast way to do that, the fastest I've done takes 90 minutes with 230k rows
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Walter Roberson
Walter Roberson 2019-4-27
Why is the output for the second row not 3,5,6, since [1 4 5] matches [1 2 5] at both 1 and 5 ? Why is the output for the 4th row not 5, since [1 2 4] matches [1 2 5] at both 1 and 2 ?
It looks to me as if you have a secret additional condition, that once a row below has matched, it is removed from further consideration.

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采纳的回答

the cyclist
the cyclist 2019-4-27
Here is a start at a solution for you.
A = [
1 2 3
1 4 5
3 4 5
1 2 4
1 2 5
2 4 5];
for i = 1:size(A,1)-1
r{i} = find(sum(A(i,:)==A(i+1:end,:),2)>=2)+i;
end
Some notes:
  • It stores the results in a cell array, r, to avoid the ragged-edge issue that Walter points out.
  • It stores an empty set, rather than a zero, if no matching rows are found.
  • It finds more rows than you've pointed out, for the reason that Walter mentioned in his second comment above.
All of these issues are easily resolved, but you need to make a better specification of your rules and output formatting.

更多回答（1 个）

Walter Roberson
Walter Roberson 2019-4-27
t1 = tril(A(:,1) == A(:,1).', -1);
t2 = tril(A(:,2) == A(:,2).', -1);
t3 = tril(A(:,3) == A(:,3).', -1);
mask = t1+t2+t3 > 1;
Then extract the positions from this. For example,
sort( mask .* (1:size(t1,1).', 'descend')
would give an array in which the columns contain the indices per row, zero padded at the end. Or you could
arrayfun(@(IDX) find(mask(:,IDX)), 1:size(mask,2), 'uniform', 0)

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