我们可以使用集合。计数器:

# tested on python 3.7
>>> from collections import Counter
>>> lst = ["1", "2", "20", "6", "210"]

>>> for i in Counter(lst):
>>>     print(i, end=" ")
1 2 20 6 210 

>>> for i in set(lst):
>>>     print(i, end=" ")
20 6 2 1 210

我们可以使用集合。计数器:

# tested on python 3.7
>>> from collections import Counter
>>> lst = ["1", "2", "20", "6", "210"]

>>> for i in Counter(lst):
>>>     print(i, end=" ")
1 2 20 6 210 

>>> for i in set(lst):
>>>     print(i, end=" ")
20 6 2 1 210

在数学中，有集和有序集(oset)。

set:唯一元素的无序容器(已实现) oset:唯一元素的有序容器

在Python中，只有集合是直接实现的。我们可以用常规字典键(3.7+)来模拟偏移量。

鉴于

a = [1, 2, 20, 6, 210, 2, 1]
b = {2, 6}

Code

oset = dict.fromkeys(a).keys()
# dict_keys([1, 2, 20, 6, 210])

Demo

删除复制，保留插入顺序。

list(oset)
# [1, 2, 20, 6, 210]

字典键上类似set的操作。

oset - b
# {1, 20, 210}

oset | b
# {1, 2, 5, 6, 20, 210}

oset & b
# {2, 6}

oset ^ b
# {1, 5, 20, 210}

细节

注意:无序结构并不排除有序元素。相反，维持秩序并不能得到保证。例子:

assert {1, 2, 3} == {2, 3, 1}                    # sets (order is ignored)

assert [1, 2, 3] != [2, 3, 1]                    # lists (order is guaranteed)

人们可能会很高兴地发现，列表和多集(mset)是两种更迷人的数学数据结构:

list:允许复制的有序元素容器(已实现) mset:允许复制的无序元素容器(NotImplemented)*

总结

Container | Ordered | Unique | Implemented
----------|---------|--------|------------
set       |    n    |    y   |     y
oset      |    y    |    y   |     n
list      |    y    |    n   |     y
mset      |    n    |    n   |     n*  

多重集可以通过collections.Counter()间接模拟，这是一个类似字典的多重度(计数)映射。

A set is an unordered data structure, so it does not preserve the insertion order. This depends on your requirements. If you have an normal list, and want to remove some set of elements while preserving the order of the list, you can do this with a list comprehension: >>> a = [1, 2, 20, 6, 210] >>> b = set([6, 20, 1]) >>> [x for x in a if x not in b] [2, 210] If you need a data structure that supports both fast membership tests and preservation of insertion order, you can use the keys of a Python dictionary, which starting from Python 3.7 is guaranteed to preserve the insertion order: >>> a = dict.fromkeys([1, 2, 20, 6, 210]) >>> b = dict.fromkeys([6, 20, 1]) >>> dict.fromkeys(x for x in a if x not in b) {2: None, 210: None} b doesn't really need to be ordered here – you could use a set as well. Note that a.keys() - b.keys() returns the set difference as a set, so it won't preserve the insertion order. In older versions of Python, you can use collections.OrderedDict instead: >>> a = collections.OrderedDict.fromkeys([1, 2, 20, 6, 210]) >>> b = collections.OrderedDict.fromkeys([6, 20, 1]) >>> collections.OrderedDict.fromkeys(x for x in a if x not in b) OrderedDict([(2, None), (210, None)])

有趣的是，人们总是用“现实问题”来开理论科学定义的玩笑。

如果设置有顺序，首先需要解决以下问题。 如果你的列表有重复的元素，当你把它变成一个集合时，顺序应该是什么?如果我们合并两个集合，顺序是什么?如果我们在相同的元素上相交两个不同顺序的集合是什么顺序?

另外，set在搜索特定键时要快得多，这在set操作中非常好(这就是为什么你需要set，而不是list)。

如果您真的关心索引，只需将其保存为列表即可。如果您仍然想对许多列表中的元素执行set操作，最简单的方法是为每个具有相同键的列表创建一个字典，并创建一个list值，其中包含原始列表中键的所有索引。

def indx_dic(l):
    dic = {}
    for i in range(len(l)):
        if l[i] in dic:
            dic.get(l[i]).append(i)
        else:
            dic[l[i]] = [i]
    return(dic)

a = [1,2,3,4,5,1,3,2]
set_a  = set(a)
dic_a = indx_dic(a)

print(dic_a)
# {1: [0, 5], 2: [1, 7], 3: [2, 6], 4: [3], 5: [4]}
print(set_a)
# {1, 2, 3, 4, 5}

这里有一个简单的方法:

x=[1,2,20,6,210]
print sorted(set(x))