MizardX的答案提供了多种方法的良好集合。

这是我自言自语时想到的:

mylist = [x for i,x in enumerate(mylist) if x not in mylist[i+1:]]

只是从外部module1中添加这样一个功能的另一个(非常高性能的)实现:

>>> from iteration_utilities import unique_everseen
>>> lst = [1,1,1,2,3,2,2,2,1,3,4]

>>> list(unique_everseen(lst))
[1, 2, 3, 4]

计时

我做了一些计时(Python 3.6)，这些表明它比我测试的所有其他替代方案都快，包括OrderedDict.fromkeys, f7和more_itertools.unique_everseen:

%matplotlib notebook

from iteration_utilities import unique_everseen
from collections import OrderedDict
from more_itertools import unique_everseen as mi_unique_everseen

def f7(seq):
    seen = set()
    seen_add = seen.add
    return [x for x in seq if not (x in seen or seen_add(x))]

def iteration_utilities_unique_everseen(seq):
    return list(unique_everseen(seq))

def more_itertools_unique_everseen(seq):
    return list(mi_unique_everseen(seq))

def odict(seq):
    return list(OrderedDict.fromkeys(seq))

from simple_benchmark import benchmark

b = benchmark([f7, iteration_utilities_unique_everseen, more_itertools_unique_everseen, odict],
              {2**i: list(range(2**i)) for i in range(1, 20)},
              'list size (no duplicates)')
b.plot()

为了确保这一点，我还做了一个重复的测试，看看是否有区别:

import random

b = benchmark([f7, iteration_utilities_unique_everseen, more_itertools_unique_everseen, odict],
              {2**i: [random.randint(0, 2**(i-1)) for _ in range(2**i)] for i in range(1, 20)},
              'list size (lots of duplicates)')
b.plot()

一个只包含一个值:

b = benchmark([f7, iteration_utilities_unique_everseen, more_itertools_unique_everseen, odict],
              {2**i: [1]*(2**i) for i in range(1, 20)},
              'list size (only duplicates)')
b.plot()

在所有这些情况下，iteration_utilities。Unique_everseen函数是最快的(在我的电脑上)。

这iteration_utilities。unique_everseen函数也可以处理输入中的不可哈希值(但是当值是可哈希值时，性能是O(n*n)而不是O(n))。

>>> lst = [{1}, {1}, {2}, {1}, {3}]

>>> list(unique_everseen(lst))
[{1}, {2}, {3}]

1免责声明:我是该软件包的作者。

对于不可哈希类型(例如列表的列表)，基于MizardX的:

def f7_noHash(seq)
    seen = set()
    return [ x for x in seq if str( x ) not in seen and not seen.add( str( x ) )]

这里有一个简单的方法:

list1 = ["hello", " ", "w", "o", "r", "l", "d"]
sorted(set(list1 ), key=list1.index)

输出如下:

["hello", " ", "w", "o", "r", "l", "d"]

1. 这些解决方案很好…… 为了在保留秩序的同时删除重复项，本页其他地方提出了优秀的解决方案:

seen = set()
[x for x in seq if not (x in seen or seen.add(x))]

以及变化，例如:

seen = set()
[x for x in seq if x not in seen and not seen.add(x)]

确实很受欢迎，因为它们简单、极简，并部署了正确的哈希以获得最佳效率。关于这些方法的主要抱怨似乎是，将方法see .add(x)“返回”的不变量None用作逻辑表达式中的常量(因此是多余的/不必要的)值(只是为了它的副作用)是笨拙和/或令人困惑的。

2. …but they waste one hash lookup per iteration. Surprisingly, given the amount of discussion and debate on this topic, there is actually a significant improvement to the code that seems to have been overlooked. As shown, each "test-and-set" iteration requires two hash lookups: the first to test membership x not in seen and then again to actually add the value seen.add(x). Since the first operation guarantees that the second will always be successful, there is a wasteful duplication of effort here. And because the overall technique here is so efficient, the excess hash lookups will likely end up being the most expensive proportion of what little work remains.

3.相反，让布景完成它的工作吧! 注意，上面的例子只调用set。加上预见，这样做总是会导致集合成员的增加。集合本身永远没有机会拒绝副本;我们的代码片段实际上已经篡夺了这个角色。使用显式的两步测试和设置代码剥夺了set自身排除这些重复的核心能力。

4. 单哈希查找代码: 下面的版本将每次迭代的哈希查找次数减少了一半，从两次减少到只有一次。

seen = set()
[x for x in seq if len(seen) < len(seen.add(x) or seen)]

这里有一些替代选项:http://www.peterbe.com/plog/uniqifiers-benchmark

最快的一个:

def f7(seq):
    seen = set()
    seen_add = seen.add
    return [x for x in seq if not (x in seen or seen_add(x))]

为什么要赋值。添加到seen_add而不是只调用see . Add ?Python是一种动态语言，解析可见。每次迭代添加比解析一个局部变量代价更大。观察。Add可能会在迭代之间发生更改，而运行时还不够聪明，无法排除这种情况。为了安全起见，它必须每次检查对象。

如果您计划在同一个数据集上大量使用这个函数，那么使用一个有序集可能会更好:http://code.activestate.com/recipes/528878/

O(1)每次操作的插入、删除和成员检查。

(小额外注意:see .add()总是返回None，所以以上值只是作为一种尝试更新集合的方式，而不是逻辑测试的组成部分。)