Python 中内置的最大堆 API

Question

默认 heapq 是最小队列实现，想知道是否有最大队列选项？谢谢。

我尝试使用 _heapify_max 作为最大堆的解决方案，但是如何动态处理 push/pop 元素？似乎 _heapify_max 只能在初始化期间使用。

import heapq

def heapsort(iterable):
    h = []
    for value in iterable:
        heapq.heappush(h, value)
    return [heapq.heappop(h) for i in range(len(h))]

if __name__ == "__main__":

    print heapsort([1, 3, 5, 7, 9, 2, 4, 6, 8, 0])

编辑，尝试 _heapify_max 似乎不适用于动态推送/弹出元素。我试过两种方法输出一样，输出都是，[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]。

def heapsort(iterable):
    h = []
    for value in iterable:
        heapq.heappush(h, value)
    return [heapq.heappop(h) for i in range(len(h))]

def heapsort2(iterable):
    h = []
    heapq._heapify_max(h)
    for value in iterable:
        heapq.heappush(h, value)
    return [heapq.heappop(h) for i in range(len(h))]

if __name__ == "__main__":

    print heapsort([1, 3, 5, 7, 9, 2, 4, 6, 8, 0])
    print heapsort2([1, 3, 5, 7, 9, 2, 4, 6, 8, 0])

提前致谢， 林

Answer 1

在最新的 cpython 源代码中有一个 _heappop_max 函数，您可能会发现它很有用：

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

如果您使用heapq._siftdown_max 更改heappush 逻辑，您应该会得到所需的输出：

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))]

输出：

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']

Answer 2

过去我只是简单地使用了sortedcontainers的SortedList，如：

> a = SortedList()
> a.add(3)
> a.add(2)
> a.add(1)
> a.pop()
3

它不是堆，但速度很快，可以根据需要直接工作。

如果你绝对需要它成为一个堆，你可以创建一个通用的否定类来保存你的项目。

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

但这会占用更多内存。