Python -- set集合 类

标签：difference   symmetric_difference   update   intersection_update   

class set(object):

    def add(self, *args, **kwargs): # 添加一项

                >>> a

                {‘3‘, ‘1‘, ‘2‘}

                >>> a.add(4)

                >>> a

                {‘3‘, 4, ‘1‘, ‘2‘}

    def clear(self, *args, **kwargs): # 清除所有

                 >>> a

                {‘3‘, 4, ‘1‘, ‘2‘}

                >>> a.clear()

                >>> a

                set()

    def copy(self, *args, **kwargs): # 返回一个集合的浅拷贝

    def difference(self, *args, **kwargs): # 返回一个a中有但b中没有的元素新集合

                                            c = a-b；（注意与difference_update的区别）

                        >>> b

                        {‘4‘, ‘6‘, ‘5‘}

                        >>> a

                        {‘4‘, ‘3‘, ‘5‘, ‘1‘, ‘2‘}

                        >>> a.difference(b)

                        {‘3‘, ‘1‘, ‘2‘}

    def difference_update(self, *args, **kwargs): #  返回删除了b中所有元素后的a集合，

                                                    a = a-b；

                        >>> a

                        {‘4‘, ‘3‘, ‘1‘, ‘2‘}

                        >>> b

                        {‘4‘, ‘6‘, ‘5‘}

                        >>> a.difference_update(b)

                        >>> a

                        {‘3‘, ‘1‘, ‘2‘}

    def discard(self, *args, **kwargs): # 如果集合中存在某一元素则删除，没有也不会报错；

                         >>> a

                        {‘4‘, ‘3‘, ‘5‘, ‘1‘, ‘2‘}

                        >>> a.discard(x)

                        >>> a

                        {‘4‘, ‘3‘, ‘5‘, ‘1‘, ‘2‘}

    def intersection(self, *args, **kwargs): #  取交集，新创建一个set

                         >>> a.intersection(b)

                         {‘5‘}

                         >>> a

                         {‘5‘}

    def intersection_update(self, *args, **kwargs): #a中保留有交集元素，不创建一个新set；

                        >>> a

                        {‘4‘, ‘3‘, ‘5‘, ‘1‘, ‘2‘}

                        >>> b

                        {‘4‘, ‘6‘, ‘5‘}

                        >>> a.intersection(b)

                        {‘4‘, ‘5‘}

                        >>> a

                        {‘4‘, ‘3‘, ‘5‘, ‘1‘, ‘2‘}

    def isdisjoint(self, *args, **kwargs): # 如果没有交集返回true，否则返回false；

                        >>> a

                        {‘4‘, ‘3‘, ‘5‘, ‘1‘, ‘2‘}

                        >>> b

                        {‘4‘, ‘6‘}

                        >>> a.isdisjoint(b)

                        False

                        >>> b.remove(‘4‘)

                        >>> b

                        {‘6‘}

                        >>> a.isdisjoint(b)

                        True

                        >>> 

    def issubset(self, *args, **kwargs): # 是否为父级

                         >>> a

                        {‘4‘, ‘3‘, ‘5‘, ‘1‘, ‘2‘}

                        >>> b

                        {‘4‘, ‘5‘}

                        >>> a.issuperset(b)

                        True

    def issuperset(self, *args, **kwargs): # 判断b是否为a子集，如果是为True 否则false；

                        >>> a

                        {‘4‘, ‘3‘, ‘5‘, ‘1‘, ‘2‘}

                        >>> b

                        {‘4‘, ‘5‘}

                        >>> a.issuperset(b)

                        True

    def pop(self, *args, **kwargs): #  移除不需要加参数取最后一个元素，并在集合中删除该值

                                        如果为空，则返回keyerror

                         >>> a.update([6,7])

                         >>> a

                         {6, 7, ‘2‘, ‘5‘, ‘1‘}

                         >>> a.pop()

                          6

                         >>> a.pop()

                          7

    def remove(self, *args, **kwargs): #  需要加参数并且没有返回值，在集合中删除该元素；

                         >>> a

                        {‘4‘, ‘6‘, 5, 6, ‘3‘, ‘5‘, ‘1‘, ‘2‘}

                        >>> a.remove(5)

                        >>> a

                        {‘4‘, ‘6‘, 6, ‘3‘, ‘5‘, ‘1‘, ‘2‘}

    def symmetric_difference(self, *args, **kwargs): # 返回一个两者之一有的元素集合

                                                      并创建一个新的集合 c=a^b ；

                    >>> a

                    {‘4‘, ‘3‘, ‘5‘, ‘1‘, ‘2‘}

                    >>> b

                    {‘4‘, ‘6‘, ‘5‘}

                    >>> a.symmetric_difference(b)

                    {‘2‘, ‘6‘, ‘3‘, ‘1‘}

    def symmetric_difference_update(self, *args, **kwargs): # 返回一个两者之一有的元素集合

                     >>> a

                    {‘4‘, ‘3‘, ‘5‘, ‘1‘, ‘2‘}

                    >>> b

                    {‘4‘, ‘6‘, ‘5‘}

                    >>> a.symmetric_difference_update(b)

                    >>> a

                    {‘2‘, ‘6‘, ‘3‘, ‘1‘}

    def union(self, *args, **kwargs): # 返回一个新的集合，包含a，b中每一个元素；c=a+b 

                    >>> a

                    {‘4‘, ‘3‘, ‘5‘, ‘1‘, ‘2‘}

                    >>> b

                    {‘4‘, ‘6‘, ‘5‘}

                    >>> a.union(b)

                    {‘2‘, ‘4‘, ‘6‘, ‘3‘, ‘5‘, ‘1‘}

    def update(self, *args, **kwargs): # 可同时添加多个元素，而add只可同时添加一个；

                    >>> a

                    {‘2‘, ‘5‘, ‘1‘}

                    >>> a.update([3,4])

                    >>> a

                    {3, 4, ‘2‘, ‘5‘, ‘1‘}

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Python -- set集合 类

标签：difference   symmetric_difference   update   intersection_update   

原文地址：http://51enjoy.blog.51cto.com/8393791/1737736