最好的方法:

from heapq import *
h = [5, 7, 9, 1, 3]
h_neg = [-i for i in h]
heapify(h_neg)            # heapify
heappush(h_neg, -2)       # push
print(-heappop(h_neg))    # pop
# 9

扩展int类并重写__lt__是一种方法。

import queue
class MyInt(int):
    def __lt__(self, other):
        return self > other

def main():
    q = queue.PriorityQueue()
    q.put(MyInt(10))
    q.put(MyInt(5))
    q.put(MyInt(1))
    while not q.empty():
        print (q.get())


if __name__ == "__main__":
    main()

你可以使用

import heapq
listForTree = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]    
heapq.heapify(listForTree)             # for a min heap
heapq._heapify_max(listForTree)        # for a maxheap!!

如果你想要弹出元素，使用:

heapq.heappop(minheap)      # pop from minheap
heapq._heappop_max(maxheap) # pop from maxheap

我还需要使用max-heap，我处理的是整数，所以我只是包装了我需要的heap中的两个方法，如下所示:

import heapq


def heappush(heap, item):
    return heapq.heappush(heap, -item)


def heappop(heap):
    return -heapq.heappop(heap)

然后，我只是分别用heappush()和heappop()替换了我的heapq.heappush()和heappop()调用。

我创建了一个名为heap_class的包，它实现了最大堆，还将各种堆函数包装到一个与列表兼容的环境中。

>>> from heap_class import Heap
>>> h = Heap([3, 1, 9, 20], max=True)
>>> h.pop()
20
>>> h.peek()  # same as h[0]
9
>>> h.push(17)  # or h.append(17)
>>> h[0]  # same as h.peek()
17
>>> h[1]  # inefficient, but works
9

从最大堆中获得最小堆。

>>> y = reversed(h)
>>> y.peek()
1
>>> y  # repr is inefficient, but correct
Heap([1, 3, 9, 17], max=False)
>>> 9 in y
True
>>> y.raw()  # underlying heap structure
[1, 3, 17, 9]

正如其他人所提到的，在max堆中处理字符串和复杂对象在heapq中是相当困难的，因为它们不同 否定的形式。heap_class实现简单:

>>> h = Heap(('aa', 4), ('aa', 5), ('zz', 2), ('zz', 1), max=True)
>>> h.pop()
('zz', 2)

支持自定义键，并与后续的推/追加和弹出一起工作:

>>> vals = [('Adam', 'Smith'), ('Zeta', 'Jones')]
>>> h = Heap(vals, key=lambda name: name[1])
>>> h.peek()  # Jones comes before Smith
('Zeta', 'Jones')
>>> h.push(('Aaron', 'Allen'))
>>> h.peek()
('Aaron', 'Allen')

(实现是建立在heapq函数上的，所以它都是用C语言或C语言包装的，除了Python中max heap上的heappush和heapreplace)

最简单的方法 就是把每个元素都转换成负数，问题就解决了。

import heapq
heap = []
heapq.heappush(heap, 1*(-1))
heapq.heappush(heap, 10*(-1))
heapq.heappush(heap, 20*(-1))
print(heap)

输出如下所示:

[-20, -1, -10]