更多pythonic方法在python中查找列表中的前两个最大值

Question

这些天我在python中设计了一些算法,但是在python中找到前两个最大的值是太丑陋和低效.

如何以高效或pythonic的方式实现它？

Answer 1

大多数Pythonic方式是使用nlargest:

import heapq
values = heapq.nlargest(2, my_list)

Answer 2

我发现这一点比一般更快(约1,000件物品清单的2倍)heapq.nlargest:

def two_largest(sequence):
    first = second = 0
    for item in sequence:
        if item > second:
            if item > first:
                first, second = item, first
            else:
                second = item
    return first, second

(根据MatthieuW的建议修改的功能)

以下是我的测试结果(timeit永远是这样,所以我用过time.time()):

>>> from random import shuffle
>>> from time import time
>>> seq = range(1000000)
>>> shuffle(seq)
>>> def time_it(func, *args, **kwargs):
...     t0 = time()
...     func(*args, **kwargs)
...     return time() - t0
...

>>> #here I define the above function, two_largest().
>>> from heapq import nlargest
>>> time_it(nlargest, 2, seq)
0.258958101273
>>> time_it(two_largest, seq)
0.145977973938