怎么从一个集合中获取最大或最小的N个元素列表？
heapq模块有两个函数：nlargest() 和 nsmallest() 可以完美解决这个问题。
In [39]: import heapq In [40]: nums = [2, 5, 3, 4] In [41]: heapq.nlargest(2,nums) Out[41]: [5, 4] In [42]: heapq.nsmallest(2,nums) Out[42]: [2, 3]
两个函数都能接受一个关键字参数，用于更复杂的数据结构中：
以price值进行比较
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} ] cheap = heapq.nsmallest(3, portfolio, key=lambda s: s[‘price‘]) expensive = heapq.nlargest(3, portfolio, key=lambda s: s[‘price‘])
原文：http://www.cnblogs.com/xiaoming279/p/6230391.html