在python的列表中找到前两个最大值的更多pythonic方法

Question

这些天我在python中设计了一些算法，但发现python中前两个最大的价值太丑陋且效率低下。

如何以高效或 Pythonic 的方式实现它？

Answer 1

大多数 Pythonic 方法是使用nlargest ：

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

Answer 2

我发现这始终比heapq.nlargest更快（对于 1,000,000 个项目的列表大约是 2 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

（在 MatthieuW 的建议下修改了功能）

这里是我的测试结果（ timeit正在采取永远，所以我用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

Answer 3

mylist = [100 , 2000 , 1 , 5]
mylist.sort()
biggest = mylist[-2:]

Answer 4

 a=int(input('Enter the first number:')) b=int(input('Enter the second Number:')) c=int(input('Ente the Third Number:')) if a>b and a>c: print('the value of A is',a,'highest velue') elif b>a and b>c: print('the value of B is',b,'highest velue') elif c>a and c>b: print('the value of C is',c,'highest velue') else: print('the value is equls')