如果您确实需要从heap中取出一个项目，但又想保留heap，您可以延迟执行，并在项目自然出现时丢弃它，而不是在列表中搜索它。
如果您将要删除的项目存储在黑名单set中，则每次heapq.heappop检查该项目是否在set中。如果存在，则丢弃它，然后再次heappop，直到您获得未列入黑名单的项目，或者heap为空

如果多个被删除的元素具有相同的值，那么黑名单集将是有问题的。相反，使用tombstone-counting-dict实现heap_remove：

def heap_remove(heap, value):
  tombstones[value] = tombstones.get(value, 0) + 1
  while len(heap) and heap[0] in tombstones and tombstones[heap[0]]:
      heappop(heap)

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正如预期的那样，你已经摊销了O（1）的移除时间，并且只要你不在其他地方的堆上popping，堆的top总是准确的。
考虑这个数字列表，它们首先被全部推入堆，然后以相同的顺序“移除”（而不是弹出）：
[三、三、二、七、一、四、二]
按预期工作：

After inserting 3 into heap, top = 3
After inserting 3 into heap, top = 3
After inserting 2 into heap, top = 2
After inserting 7 into heap, top = 2
After inserting 1 into heap, top = 1
After inserting 4 into heap, top = 1
After inserting 2 into heap, top = 1

型
但是移除是通过增加对象的tombstone来完成的。如果堆的顶部设置了tombstone，那么移除对象并递减tombstone计数器。

Called remove( 3 )
  Marking 3 for deletion
Called remove( 3 )
  Marking 3 for deletion
Called remove( 2 )
  Marking 2 for deletion
Called remove( 7 )
  Marking 7 for deletion
Called remove( 1 )
  Marking 1 for deletion
  Deleting top 1
    Updated heap is: [2, 2, 3, 7, 3, 4]
  Deleting top 1
    Updated heap is: [2, 3, 3, 7, 4]
Called remove( 4 )
  Marking 4 for deletion
Called remove( 2 )
  Marking 2 for deletion
  Deleting top 2
    Updated heap is: [3, 3, 4, 7]
  Deleting top 3
    Updated heap is: [3, 7, 4]
  Deleting top 3
    Updated heap is: [4, 7]
  Deleting top 4
    Updated heap is: [7]
  Deleting top 7
    Updated heap is: []

型
请注意，当第二个heap_remove(3)被调用时，@GP89的解决方案会崩溃，因为3已经在tombstone集中。
您可以探索此示例here。

有了以上两个想法，这里是一个完整的演示：我会尽快使它干净简洁。

from heapq import heappush, heappop
class Solution:
  def demo():
    deleted = {}
    h = [0]
    heappush(h, 789)
    heappush(h, 101)
    heappush(h, 101)
    self.remove(h, 101, deleted)
    max_val = self.peek(h, deleted)
  def remove(self, h, y, deleted):
    deleted[y] = deleted.get(y, 0) + 1
    while len(h) > 0 and h[0] == y and deleted[y] > 0:
        heappop(h)
        deleted[y] -= 1
  def peek(self, h, deleted):
    while len(h) > 0 and deleted.get(h[0],0) > 0:
        deleted[h[0]] -= 1
        heappop(h)
    return h[0]

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在这种方法中，我基本上是在字典中跟踪元素。所以，无论何时，我删除它，搜索过程都变成了O（1）。

import heapq
import collections
import itertools
class RemoveHeap:
    def __init__(self):
        self.h = []
        self.track = collections.defaultdict(collections.deque)
        self.counter = itertools.count()
    def insert_item(self, val):
        count = next(self.counter)
        item = [val, count, 'active']
        self.track[val].append(item)
        heapq.heappush(self.h, item)
    def delete_item(self, val):
        if val in self.track:
            items = self.track[val]
            for item in items:
                if item[2] == 'active':
                    item[2] = 'deleted'
                    break
    def pop_item(self):
        while len(self.h) > 0:
            item = heapq.heappop(self.h)
            item_track = self.track[item[0]]
            item_track.popleft()
            if len(item_track) == 0:
                del self.track[item[0]]
            else:
                self.track[item[0]] = item_track
            if item[2] == 'active':
                return item[0]
    def peek_item(self):
        item = self.h[0]
        if item[2] == 'deleted':
            x = self.pop_item()
            self.insert_item(x)
            return x
        return item[0]

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恐怕只有heapq没有这样的方法。因为从堆中搜索元素需要O(n)。
但是，您可以将它与dict之类的东西一起使用，这将为O(1)提供搜索条目的时间。

已删除：

我尝试使用字典记账，但我如何才能得到“创建测试”的索引时，它被插入？- Prakhar 3小时前
一种简单的方法是：

# remember to update this hdict when updating the heap.
hdict = { h[i][1]: i for i in range(len(h)) }

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然后你可以通过访问这个hdict而不是你的O(n)线性搜索来获得给定字符串的索引。