[issue5669] Extra heapq nlargest/nsmallest option for including ties

[George Sakkis]

George Sakkis <george.sakkis at gmail.com> added the comment:

> That should have been:  last = result[-1]; [last]*s.count(last).

Nice, though that's not equivalent if the objects' identity is
significant for what happens next (which typically is the case when ties
are preserved). The sorted/bisect solution doesn't have this problem.

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<http://bugs.python.org/issue5669>
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