Python集合

集合（set）和字典（dict）类似，也是一组key的集合，但不存储value。由于key不能重复，所以，在set中，没有重复的key。

定义一个集合需要一个列表（list）作为输入集合：

list_1 = set([15,44,789,45,65,87])
print(list_1,type(list_1))#{65, 44, 45, 15, 789, 87} <class 'set'>

 自带重复元素过滤

list_1 = set([15,44,789,45,789,65,87,87])
print(list_1,type(list_1))#{65, 44, 45, 15, 789, 87} <class 'set'>

 基本操作

添加 add（key）

list_3 =set([1101,1102,1103])
list_3.add(899)
print(list_3)#{899, 1101, 1102, 1103}

 添加多项 update（list）

list_3 =set([1101,1102,1103])
list_3.update([899,810,827])
print(list_3)#{899, 810, 1101, 1102, 1103, 827}

 删除 remove(key)

list_3 =set([1101,1102,1103])
list_3.remove(1101)
print(list_3)#{1102, 1103}

如果要删除一个不存在的元素会报错

list_3 =set([1101,1102,1103])
list_3.remove(1201)
####################
Traceback (most recent call last):
  File "E:/pywww/day02/04.py", line 26, in <module>
    list_3.remove(1201)
KeyError: 1201

 可以使用另外一种方法 discard

list_3 =set([1101,1102,1103])
print(list_3.discard(1101))#None
print(list_3.discard(1201))#None
print(list_3)#{1102, 1103}

 discard 没有返回值

统计长度len(s)

list_3 =set([1101,1102,1103])
print(len(list_3))#3

 成员判断 in not in

list_3 =set([1101,1102,1103])
print(1101 in list_3)#True
print(1109 in list_3)#False
print(1101 not in list_3)#False
print(1109 not in list_3)#True

关系操作

集合交集 intersection

求两个集合中相同的key

list_1 = set([15,44,789,45,789,65,87,87])
list_2 = set([5,789,64,38,62,56,45])
print(list_1.intersection(list_2))#{789, 45}

 集合并集 union

求两个集合去重后的key的集合

list_1 = set([15,44,789,45,789,65,87,87])
list_2 = set([5,789,64,38,62,56,45])

print(list_1.union(list_2))#{64, 65, 5, 38, 44, 45, 15, 789, 87, 56, 62}

集合差集 difference

查找 list_1中 在 list_2 里面没有的key

list_1 = set([15,44,789,45,789,65,87,87])
list_2 = set([5,789,64,38,62,56,45])

print(list_1.difference(list_2))#{65, 44, 87, 15}

集合子集

list_1 = set([15,44,789,45,789,65,87,87])
list_2 = set([5,789,64,38,62,56,45])
list3 = set([15,789,65])
print(list3.issubset(list_1))#True
print(list_2.issubset(list_1))#False

集合父集

list_1 = set([15,44,789,45,789,65,87,87])
list_2 = set([5,789,64,38,62,56,45])
list3 = set([15,789,65])
print(list_1.issuperset(list3))#True
print(list_1.issuperset(list_2))#False

对称差集

两个集合中互相都没有的key的集合

list_1 = set([15,44,789,45,789,65,87,87])
list_2 = set([5,789,64,38,62,56,45])
print(list_2.symmetric_difference(list_1))#{64, 65, 5, 15, 87, 38, 44, 56, 62}

 判断两个集合是否有交集

list_1 = set([15,44,789,45,789,65,87,87,1101])
list_2 = set([5,789,64,38,62,56,45])
list_3 =set([1101,1102,1103])
print(list_2.isdisjoint(list_3))#True
print(list_1.isdisjoint(list_3))#False

 运算符

list_1 = set([15,44,789,45,789,65,87,87,1101])
list_2 = set([5,789,64,38,62,56,45])
list_3 =set([1101,1102,1103])

交集 &

print(list_1 & list_2)#{789, 45}

 并集 |

print(list_1| list_2)#{64, 65, 5, 38, 44, 1101, 45, 15, 789, 87, 56, 62}

 差集 -

print(list_1 - list_2)#{65, 44, 1101, 87, 15}

 在list_1 中不在list_2中的key

对称差集 ^

print(list_1 ^ list_2)#{64, 65, 5, 38, 87, 56, 44, 1101, 62, 15}

 项在 list_1 或 list_2 中,但不会同时出现在二者中