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去除 List 中重复元素

mylist

list

set

7 条回复 • 2015-01-14 23:45:19 +08:00

9hills

2015-01-11 17:32:56 +08:00

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

2015-01-11 21:33:41 +08:00 via iPhone

话说有没有 orderedset （

yiding

2015-01-11 21:52:49 +08:00

@dant 集合是三大特性之一，无序性

zergling

2015-01-11 22:04:39 +08:00

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

deepurple

2015-01-11 22:55:14 +08:00

刚看那个董伟明的《Python高级编程》的视频里讲到了这个

raquelken

2015-01-12 11:53:09 +08:00

@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

2015-01-14 23:45:19 +08:00

@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