# 堆队列
import heapq
# 怎样从一个集合中获得最大或者最小的 N 个元素列表?
nums = [1, 8, 2, 23, 7, -4, 18, 23, 42, 37, 2]
# 最大的3个
print(heapq.nlargest(3, nums))
# 最小的3个
print(heapq.nsmallest(3, nums))
portfolio = [
{'name': 'IBM', 'shares': 100, 'price': 91.1},
{'name': 'AAPL', 'shares': 50, 'price': 543.22},
{'name': 'FB', 'shares': 200, 'price': 21.09},
{'name': 'HPQ', 'shares': 35, 'price': 31.75},
{'name': 'YHOO', 'shares': 45, 'price': 16.35},
{'name': 'ACME', 'shares': 75, 'price': 115.65}
]
# key为比较值 key=func，这里用lambda
# 底层调用max(iterable, *[, default=obj, key=func])
print(heapq.nlargest(3, portfolio, key=lambda i: i['shares']))
print(heapq.nsmallest(3, portfolio, lambda i: i['price']))
# 堆化
heapq.heapify(nums)
print(nums) # [-4, 2, 1, 23, 7, 2, 18, 23, 42, 37, 8]
print(heapq.heappop(nums)) # -4
print(heapq.heappop(nums)) # 1
print(heapq.heappop(nums)) # 2
# 简单的优先级队列
class PriorityQueue:
def __init__(self):
self._queue = []
self._index = 0
def push(self, item, priority):
heapq.heappush(self._queue, (-priority, self._index, item)) # 最大的会成为最小的排到最前面，push是按照小->大，相同的比较_index
self._index += 1
def pop(self):
return heapq.heappop(self._queue)[-1]
class Item:
def __init__(self, name):
self.name = name
def __repr__(self):
return 'Item({!r})'.format(self.name)
print('='*20)
q = PriorityQueue()
q.push(Item('foo'), 1)
q.push(Item('bar'), 5)
q.push(Item('spam'), 4)
q.push(Item('grok'), 1)
q.push(Item('sam'), -1)
print(q.pop())
print(q.pop())
print(q.pop())
print(q.pop())
print(q.pop())
