Default heapq is min queue implementation and wondering if there is an option for max queue? Thanks.
I tried the solution using _heapify_max for max heap, but how to han
In the past I have simply used sortedcontainers's SortedList
for this, as:
> a = SortedList()
> a.add(3)
> a.add(2)
> a.add(1)
> a.pop()
3
It's not a heap, but it's fast and works directly as required.
If you absolutely need it to be a heap, you could make a general negation class to hold your items.
class Neg():
def __init__(self, x):
self.x = x
def __cmp__(self, other):
return -cmp(self.x, other.x)
def maxheappush(heap, item):
heapq.heappush(heap, Neg(item))
def maxheappop(heap):
return heapq.heappop(heap).x
But that will be using a little more memory.
There is a _heappop_max function in the latest cpython source that you may find useful:
def _heappop_max(heap):
"""Maxheap version of a heappop."""
lastelt = heap.pop() # raises appropriate IndexError if heap is empty
if heap:
returnitem = heap[0]
heap[0] = lastelt
heapq._siftup_max(heap, 0)
return returnitem
return lastelt
If you change the heappush
logic using heapq._siftdown_max
you should get the desired output:
def _heappush_max(heap, item):
heap.append(item)
heapq._siftdown_max(heap, 0, len(heap)-1)
def _heappop_max(heap):
"""Maxheap version of a heappop."""
lastelt = heap.pop() # raises appropriate IndexError if heap is empty
if heap:
returnitem = heap[0]
heap[0] = lastelt
heapq._siftup_max(heap, 0)
return returnitem
return lastelt
def heapsort2(iterable):
h = []
heapq._heapify_max(h)
for value in iterable:
_heappush_max(h, value)
return [_heappop_max(h) for i in range(len(h))]
Output:
In [14]: heapsort2([1,3,6,2,7,9,0,4,5,8])
Out[14]: [9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
In [15]: heapsort2([7, 8, 9, 6, 4, 2, 3, 5, 1, 0])
Out[15]: [9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
In [16]: heapsort2([19,13,15,17,11,10,14,20,18])
Out[16]: [20, 19, 18, 17, 15, 14, 13, 11, 10]
In [17]: heapsort2(["foo","bar","foobar","baz"])
Out[17]: ['foobar', 'foo', 'baz', 'bar']