如果你经常使用pandas，并且美学优先于性能，那么考虑内置函数pandas. series .drop_duplicate:

    import pandas as pd
    import numpy as np

    uniquifier = lambda alist: pd.Series(alist).drop_duplicates().tolist()

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

    alist = np.random.randint(low=0, high=1000, size=10000).tolist()

    print uniquifier(alist) == f7(alist)  # True

时间:

    In [104]: %timeit f7(alist)
    1000 loops, best of 3: 1.3 ms per loop
    In [110]: %timeit uniquifier(alist)
    100 loops, best of 3: 4.39 ms per loop

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 ) )]

sequence = ['1', '2', '3', '3', '6', '4', '5', '6']
unique = []
[unique.append(item) for item in sequence if item not in unique]

unique→[1、(2)、(3)、(6)、(4)、(5)]

在Python 3.7及以上版本中，字典可以保证记住它们的键插入顺序。这个问题的答案概括了当前的状况。

OrderedDict解决方案因此变得过时，没有任何import语句，我们可以简单地发出:

>>> lst = [1, 2, 1, 3, 3, 2, 4]
>>> list(dict.fromkeys(lst))
[1, 2, 3, 4]

Pandas用户应该查看Pandas .unique。

>>> import pandas as pd
>>> lst = [1, 2, 1, 3, 3, 2, 4]
>>> pd.unique(lst)
array([1, 2, 3, 4])

该函数返回一个NumPy数组。如果需要，可以使用tolist方法将其转换为列表。

如果你经常使用pandas，并且美学优先于性能，那么考虑内置函数pandas. series .drop_duplicate:

    import pandas as pd
    import numpy as np

    uniquifier = lambda alist: pd.Series(alist).drop_duplicates().tolist()

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

    alist = np.random.randint(low=0, high=1000, size=10000).tolist()

    print uniquifier(alist) == f7(alist)  # True

时间:

    In [104]: %timeit f7(alist)
    1000 loops, best of 3: 1.3 ms per loop
    In [110]: %timeit uniquifier(alist)
    100 loops, best of 3: 4.39 ms per loop