python集合

>>> basket = {'apple', 'orange', 'apple', 'pear', 'orange', 'banana'}
>>> print(basket)                      # 这里演示的是去重功能
{'orange', 'banana', 'pear', 'apple'}
>>> 'orange' in basket                 # 快速判断元素是否在集合内
True
>>> 'crabgrass' in basket
False
>>> # 下面展示两个集合间的运算.
...
>>> a = set('abracadabra')
>>> b = set('alacazam')
>>> a                                  
{'a', 'r', 'b', 'c', 'd'}
>>> a - b                              # 集合a中包含而集合b中不包含的元素
{'r', 'd', 'b'}
>>> a | b                              # 集合a或b中包含的所有元素
{'a', 'c', 'r', 'd', 'b', 'm', 'z', 'l'}
>>> a & b                              # 集合a和b中都包含了的元素
{'a', 'c'}
>>> a ^ b                              # 不同时包含于a和b的元素
{'r', 'd', 'b', 'm', 'z', 'l'}

s.add(x) 或 s.update( x )
>>> thisset = set(("Google", "Runoob", "Taobao"))
>>> thisset.add("Facebook")
>>> print(thisset)
{'Taobao', 'Facebook', 'Google', 'Runoob'}

s.remove( x ) 或s.discard( x )
>>> thisset = set(("Google", "Runoob", "Taobao"))
>>> thisset.remove("Taobao")
>>> print(thisset)
{'Google', 'Runoob'}
>>> thisset.remove("Facebook")   # 不存在会发生错误
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
KeyError: 'Facebook'
>>>

len(s)
>>> thisset = set(("Google", "Runoob", "Taobao"))
>>> len(thisset)

s.clear()
>>> thisset = set(("Google", "Runoob", "Taobao"))
>>> thisset.clear()
>>> print(thisset)
set()

x in s
>>> thisset = set(("Google", "Runoob", "Taobao"))
>>> "Runoob" in thisset
True
>>> "Facebook" in thisset
False
>>>