使用 heapq 降序

Question

我正在使用 Python 的 heapq 模块按升序和降序获取数据。

对于升序，我使用的是最小堆，效果很好，如下所示：

>>> from heapq import heapify, heappop
>>> heap = [9, 3, 1, 5, 6, 2, 7]
>>> heapify(heap)
>>> heappop(heap)
1
>>> heappop(heap)
2
>>> heappop(heap)
3

对于降序，我尝试了不同的方法，但它们都有一些缺点：

使用负值作为优先级进行反向排序。我必须使用单独的列表来使数据可重用。如果原始列表很大，则复制列表的成本很高。

>>> from heapq import heapify, heappop
>>> heap = [9, 3, 1, 5, 6, 2, 7]
>>> heap_neg = [-x for x in heap]
>>> heapify(heap_neg)
>>> -heappop(heap_neg)
9
>>> -heappop(heap_neg)
7
>>> -heappop(heap_neg)
6

优先使用负值元组，这也是浪费space。我不想将整数列表存储为元组列表。

>>> from heapq import heapify, heappop
>>> heap = [(-9, 9), (-3, 3), (-1, 1), (-5, 5), (-6, 6), (-2,2), (-7,7)]
>>> heapify(heap)
>>> heappop(heap)[1]
9
>>> heappop(heap)[1]
7
>>> heappop(heap)[1]
6

在 heapify 中缺少使用键排序。类似于：

>>> from heapq import heapify, heappop
>>> heap = [9, 3, 1, 5, 6, 2, 7]
>>> heapify(heap, key=lambda x:-x) # This doesn't work as heapify don't have key parameter

如果我使用heapq._heapify_max（堆），我将不得不在每个元素弹出后_heapify_max。喜欢：

>>> from heapq import _heapify_max, heappop
>>> heap = [9, 3, 1, 5, 6, 2, 7]
>>> _heapify_max(heap)
>>> heappop(heap)
9
>>> heappop(heap)  # popping without _heapify_max gives wrong result
1
>>> _heapify_max(heap)
>>> heappop(heap) # popping after _heapify_max gives correct result
7

有什么方法可以让我获得降序顺序，类似于我获得升序顺序的方式？ :)

Answer 1

我认为在这种情况下你正在寻找一个排序的链表，我修改了我找到的一个 here 所以它会按升序插入（我添加了 pop 函数，由于某种原因它不是' t 在代码中，但我认为您可能需要它):

# Python program to insert in sorted list

# Node class 
class Node:

    # Constructor to initialize the node object
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:

    # Function to initialize head
    def __init__(self):
        self.head = None

    def sortedInsert(self, new_node):

        # Special case for the empty linked list 
        if self.head is None:
            new_node.next = self.head
            self.head = new_node

        # Special case for head at end
        elif self.head.data <= new_node.data:
            new_node.next = self.head
            self.head = new_node

        else :

            # Locate the node before the point of insertion
            current = self.head
            while(current.next is not None and
                 current.next.data > new_node.data):
                current = current.next

            new_node.next = current.next
            current.next = new_node

    # Function to insert a new node at the beginning
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

    # Utility function to prit the linked LinkedList
    def printList(self):
        temp = self.head
        while(temp):
            print(temp.data),
            temp = temp.next

    def pop(self):
        val = self.head.data
        self.head = self.head.next
        return val


# Driver program
llist = LinkedList()
new_node = Node(5)
llist.sortedInsert(new_node)
new_node = Node(10)
llist.sortedInsert(new_node)
new_node = Node(7)
llist.sortedInsert(new_node)
new_node = Node(3)
llist.sortedInsert(new_node)
new_node = Node(1)
llist.sortedInsert(new_node)
new_node = Node(9)
llist.sortedInsert(new_node)
print("Create Linked List")
llist.printList()

如您所见，只是将 >= 更改为 <=，它完美地完成了工作

Answer 2

正如我们在评论中讨论的那样，当您从空堆开始并将值添加为你去。由于这是找到值流的运行中位数的用例，因此在添加值时取反应该就可以了。

这是我编写的运行中值生成器，只是为了仔细检查它是否按我预期的方式工作：

def running_median(iterable):
    left_q = [] # heap of smaller-than-median elements, stored negated
    right_q = [] # heap of larger-than-median elements

    for value in iterable:
        if len(left_q) == len(right_q): # push to left_q when they're equal size
            if len(right_q) > 0 and value > right_q[0]:
                value = heapq.heapreplace(right_q, value)
            heapq.heappush(left_q, -value)
        else: # push to right_q only when it's (strictly) smaller
            if value < -left_q[0]:
                value = -heapq.heapreplace(left_q, -value)
            heapq.heappush(right_q, value)

        # len(left_q) is always >= len(right_q) so we never yield right_q[0]
        if len(left_q) > len(right_q):
            yield -left_q[0]
        else:
            yield (-left_q[0] + right_q[0]) / 2

left_q 堆存储小于或等于中值的值。每个值在被压入时都会被取反，因此对它使用正常的最小堆操作可以使它像最大堆一样工作。我们只需要记住重新否定我们从中取出的任何值，回到原来的符号。

Answer 3

有私有方法（在 python 3.8 测试）

import heapq


if __name__ == '__main__':
    a = [1, 3, 2, 5]

    heapq._heapify_max(a)

    for item in range(0, len(a)):
        print(heapq._heappop_max(a)

结果是

sorted heap  5
sorted heap  3
sorted heap  2
sorted heap  1

但是对于某些人来说，使用私有方法可能看起来不够正确。出于这个原因，我们可以通过将您的对象放在修改后的包装器中来更改顺序

class DescOrder:
    def __init__(self, entity):
        self.entity = entity

    def __lt__(self, o):
        return self.entity.__gt__(o.entity)

    def __repr__(self):
        return str(self.entity)

def check_sorting(a, b):
    new_heap = []

    for element in a:
        heapq.heappush(new_heap, DescOrder(element))

    for index in range(0, len(b)):
        assert heapq.heappop(new_heap).entity == b[index]


if __name__ == '__main__':
    check_sorting([5, 1, -1, 3, 2], [5, 3, 2, 1, -1])
    check_sorting([5, 2, -1, 3, 1], [5, 3, 2, 1, -1])
    check_sorting([-1, 2, 5, 3, 1], [5, 3, 2, 1, -1])

使用 heapq 降序

Descending order using heapq

python

heap

python-2.7

python-3.x