允许您选择任意数量的最大或最小的项目

import heapq
heap = [23, 7, -4, 18, 23, 42, 37, 2, 8, 2, 23, 7, -4, 18, 23, 42, 37, 2]
heapq.heapify(heap)
print(heapq.nlargest(3, heap))  # [42, 42, 37]
print(heapq.nsmallest(3, heap)) # [-4, -4, 2]

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()

python中有内置堆，但我只是想分享一下，如果有人像我一样想自己构建它。 我是python的新手，不要判断我是否犯了错误。 算法是有效的，但效率我不知道

class Heap :

    def __init__(self):
        self.heap = []
        self.size = 0


    def add(self, heap):
        self.heap = heap
        self.size = len(self.heap)

    def heappush(self, value):
        self.heap.append(value)
        self.size += 1


    def heapify(self, heap ,index=0):

        mid = int(self.size /2)
        """
            if you want to travel great value from bottom to the top you need to repeat swaping by the hight of the tree
            I  don't how how can i get the  height of the tree that's why i use sezi/2
            you can find height by this formula
            2^(x) = size+1  why 2^x because tree is growing exponentially 
            xln(2) = ln(size+1)
            x = ln(size+1)/ln(2)
        """

        for i in range(mid):
            self.createTee(heap ,index)

        return heap

    def createTee(self,  heap ,shiftindex):

        """
        """
        """

            this pos reffer to the index of the parent only parent with children
                    (1)
                (2)      (3)           here the size of list is 7/2 = 3
            (4)   (5)  (6)  (7)        the number of parent is 3 but we use {2,1,0} in while loop
                                       that why a put pos -1

        """
        pos = int(self.size /2 ) -1
        """
            this if you wanna sort this heap list we should swap max value in the root of the tree with the last
            value in the list and if you wanna repeat this until sort all list you will need to prevent the func from
            change what we already sorted I should decrease the size of the list that will heapify on it

        """

        newsize = self.size - shiftindex
        while pos >= 0 :
            left_child = pos * 2 + 1
            right_child = pos * 2 + 2
            # this mean that left child is exist
            if left_child < newsize:
                if right_child < newsize:
                    # if the right child exit we wanna check if left child > rightchild
                    # if right child doesn't exist we can check that we will get error out of range
                    if heap[pos] < heap[left_child] and heap[left_child]  > heap[right_child] :
                        heap[left_child] , heap[pos] = heap[pos], heap[left_child]
                # here if the righ child doesn't exist
                else:
                    if heap[pos] < heap[left_child] :
                        heap[left_child] , heap[pos] = heap[pos], heap[left_child]
            # if the right child exist
            if right_child < newsize :
                if heap[pos] < heap[right_child] :
                    heap[right_child], heap[pos] = heap[pos], heap[right_child]
            pos -= 1

        return heap

    def sort(self ):
        k = 1
        for i in range(self.size -1 ,0 ,-1):
            """
            because this is max heap we swap root with last element in the list

            """
            self.heap [0] , self.heap[i] = self.heap[i], self.heap[0]
            self.heapify(self.heap ,k)
            k+=1

        return self.heap


h = Heap()
h.add([5,7,0,8,9,10,20,30,50,-1] )
h.heappush(-2)
print(" before heapify ")
print(h.heap)
print(" after heapify ")
print(h.heapify(h.heap,0))
print(" after sort ")
print(h.sort())

输出:

之前heapify [5,7,0,8,9,10,20,30,50， -1， -2]

heapify后 [50, 30, 20, 8, 9, 10, 0, 7, 5， -1， -2]

排序后 [-2， -1, 0,5,7,8,9,10,20,30,50]

希望您能理解我的代码。如果有什么你不明白的地方，请发表评论，我会尽力帮助你

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

允许您选择任意数量的最大或最小的项目

import heapq
heap = [23, 7, -4, 18, 23, 42, 37, 2, 8, 2, 23, 7, -4, 18, 23, 42, 37, 2]
heapq.heapify(heap)
print(heapq.nlargest(3, heap))  # [42, 42, 37]
print(heapq.nsmallest(3, heap)) # [-4, -4, 2]

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。