集合 （set）

set 是一个无序不重复的元素集，集合跟字典是无序的，不支持索引

创建集合：

第一种方式：通过{ }创建

>>> num={1,2,3,4,3,3,1}
>>> num
{1, 2, 3, 4}
>>> type(num)
<class 'set'>

第二种方式：通过set()

>>> li=set(['aaa','bbb','aaa'])
>>> li
{'aaa', 'bbb'}

可以去除列表里的重复项但不能保证顺序：

>>> num=[4,3,2,1,2,2]
>>> num=list(set(num))
>>> num
[1, 2, 3, 4]

class set(object):
    def add(self, *args, **kwargs): #添加,把传入的元素作为一个整个添加进去
        >>> st.add('python')
        >>> st
        {'d', 'f', 'python', 'a', 'b', 'c'}

    def update(self, *args, **kwargs): # 更新，把传入的元素拆分然后传入集合
        """
        >>> s1=set('boy')
        >>> s1
        {'b', 'y', 'o'}
        >>> s1.update('python')
        >>> s1
        {'t', 'y', 'p', 'n', 'h', 'b', 'o'}
        """
        pass

    def clear(self, *args, **kwargs): # 清空
        >>> st.clear()
        >>> st
        set()
        pass

    def copy(self, *args, **kwargs): # 拷贝

    def difference(self, *args, **kwargs): #差集。在s1集合,不在s2集合的,操作后可以输出返回值
        >>> s1=set(['aaa','bbb'])
        >>> s2=set(['bbb','ccc'])
        >>> s1.difference(s2)
        {'aaa'}

    def difference_update(self, *args, **kwargs): #将s1更新为s1-s2的差集。没有返回值
        >>> s1
         {'aaa', 'bbb'}
        >>> s2
         {'bbb', 'ccc'}
        >>> s1.difference_update(s2)
        >>> s1
         {'aaa'}
        >>> s2
         {'bbb', 'ccc'}

    def discard(self, *args, **kwargs):  # 删除 

    def remove(self, *args, **kwargs): # 删除
     #set.remove(obj)和set.discard(obj)的区别在于，当obj存在于set中时，都将其删除；但当obj不存在于set中时，remove()会报错，discard()不会。
    def pop(self, *args, **kwargs): # 随机删除
        >>> a=set('abcd')
        >>> a.pop()
        'b'
        >>> a.pop()
        'd'

    def intersection(self, *args, **kwargs): # 交集 s1 & s2
        >>> s1=set(['aaa','bbb'])
        >>> s2=set(['bbb','ccc'])
        >>> d=s1.intersection(s2)
        >>> print(d)
         {'bbb'}

    def intersection_update(self, *args, **kwargs): # 交集 s1更新为这个值

    def isdisjoint(self, *args, **kwargs): # 没有交集
       >>> s1.isdisjoint(s2)
       False

    def issubset(self, *args, **kwargs): #  s1的每一个元素都在s2中
       >>> s1=set('abc')
       >>> s2=set('abcdef')
       >>> s1.issubset(s2)
       True
    def issuperset(self, *args, **kwargs):  # s2的每一个元素都在s1中
       >>> s1.issuperset(s2)
       False

    def symmetric_difference(self, *args, **kwargs): # s1,s2的对称差，有返回值
        >>> s1=set(['aaa','bbb'])
        >>> s2=set(['bbb','ccc'])
        >>> d=s1.symmetric_difference(s2)
        >>> print(d)
         {'ccc', 'aaa'}

    def symmetric_difference_update(self, *args, **kwargs): # s1更新为s1,s2的对称差，没有返回值
        >>> s1=set(['aaa','bbb'])
        >>> s2=set(['bbb','ccc'])
        >>> s1.symmetric_difference_update(s2)
        >>> s1
         {'aaa', 'ccc'}

    def union(self, *args, **kwargs): # 并集  s1 | s2
        

还有一种不可变的集合：frozenset

>>> s = frozenset([1,2,3,4])
>>> s.add(0)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'frozenset' object has no attribute 'add'