No. if you have a square matrix of 3 rows, you'll need to pad 3 columns and
3 rows.


On Tue, Aug 20, 2013 at 11:11 AM, nikitsaraf <notifications@github.com>wrote:

> In test/test_dedupe.py:
>
> > @@ -172,6 +189,22 @@ def test_hierarchical(self):
> >      assert hierarchical(self.str_dupes,'S1', 0.5) == [set(['1', '2', '3']), set(['4','5'])]
> >      assert hierarchical(self.str_dupes,'S1', 0) == [set(['1', '2', '3', '4', '5'])]
> >
> > +  def test_hungarian(self):
> > +    hungarian = dedupe.clustering.clusterConstrained
> > +    assert hungarian(self.bipartite_dupes, 0.5) == [set([1, 6]),
>
> No, I understood that.
> But in the list of rows, there are two x and in the same way, the list of
> columns have two x.
>
> So, I was asking, that only one x would be included in the list of rows
> and one x in the list of columns right ?
>
> —
> Reply to this email directly or view it on GitHub<https://github.com/open-city/dedupe/pull/148/files#r5872783>
> .
>



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