how to use "heapq" module as a max-heap?
Steven D'Aprano
steve at REMOVE-THIS-cybersource.com.au
Sun Mar 29 11:25:05 CEST 2009
(Top posting and incorrect quoting corrected.)
On Sun, 29 Mar 2009 16:48:53 +0800, apollo wrote:
> > as we all known, in the standard module 'heapq', we can easily get
> > the smallest item from the heap. i.e. it's an implementation of
> > min-heap.
> >
> > my question is how to use 'heapq' to extract the biggest item from the
> > heap? is it possible?
> >
> > thanks in advance.:)
>
> if the 'heapq' module supports user-defined comparasion, this problem
> will be much easier.
It certain does. Here's one way of turning the min-heap into a fake max-
heap:
>>> class Backwards(float):
... def __lt__(self, other):
... return not float.__le__(self, other)
... def __le__(self, other):
... return not float.__lt__(self, other)
... def __gt__(self, other):
... return not float.__ge__(self, other)
... def __ge__(self, other):
... return not float.__gt__(self, other)
...
>>> x, y = Backwards(27), Backwards(1270)
>>> x > y
True
>>> y < x
True
>>> heap = []
>>> for x in xrange(20):
... heapq.heappush(heap, Backwards(x))
...
>>> heapq.heappop(heap)
19.0
>>> heapq.heappop(heap)
18.0
>>> heap
[17.0, 16.0, 13.0, 15.0, 8.0, 10.0, 12.0, 9.0, 14.0, 2.0, 7.0, 1.0, 5.0,
4.0, 11.0, 0.0, 6.0, 3.0]
The problem with this approach is that when you pop the "largest" item,
and compare it to something else, it will compare *smaller* because of
the custom comparisons, so now you have to remove the custom comparisons
to revert to the standard behaviour. The book-keeping for this could be
awful.
However, another approach is to use the nlargest function:
>>> heap = []
>>> for i in (5, 6, 7, 0, 1, 10, 2, 4, 3, 9, 8):
... heapq.heappush(heap, i)
...
>>> heap
[0, 1, 2, 3, 5, 10, 7, 6, 4, 9, 8]
>>> heapq.nlargest(1, heap)
[10]
The problem with this is that the value is not popped off the heap. Also,
I'm not sure how efficient nlargest is.
--
Steven
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