mylist = list(set(mylist))

[jacklong]

去除 List 中重复元素

[9hills]

set下，list顺序就变了。你再list也没用

保持顺序：
a = set()
l2 = [i for i in l1 if not (i in a or a.add(i))]

[dant]

话说有没有 orderedset （

[4mrqn07k]

@dant 集合是三大特性之一，无序性

[zergling]

或者：
from collections import OrderedDict
l2 = OrderedDict(zip(l1, l1)).keys()

[deepurple]

刚看那个董伟明的《Python高级编程》的视频里讲到了这个

[raquelken]

@zergling 用dict的key去重必须是list元素是可以hashed的，而且还有这个方法 {}.fromkeys(mylist).keys()

[dongweiming]

@deepurple 我来填坑: 标准答案是这里 https://docs.python.org/2/faq/programming.html#id40

以上都是一些trick

In [1]: mylist = ['a', 'b', 'r', 'a', 'b', 'k', 'v', 'b']

In [2]: %timeit list(set(mylist))
The slowest run took 7.63 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 781 ns per loop

In [3]: %timeit {}.fromkeys(mylist).keys()
The slowest run took 4.30 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 887 ns per loop

In [4]: from collections import OrderedDict

In [5]: %timeit OrderedDict.fromkeys(mylist).keys()
The slowest run took 11.18 times longer than the fastest. This could mean that an intermediate result is being cached
100000 loops, best of 3: 18.5 µs per loop

In [6]: %timeit list({}.fromkeys(mylist))
The slowest run took 5.99 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 995 ns per loop