1集合是一个无序的，不重复的数据组合，它的主要作用如下(set和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):
去重，把一个列表变成集合，就自动去重了
关系测试，测试两组数据之前的交集、差集、并集等关系
2（去重）
1 # -*- coding:utf-8 -*- 2 s={1,2,3,4,5,1,2,12,"tin",("tj",120),"tin",("tj",120)} 3 print(type(s)) 4 print(s)
输出结果：
1 <class 'set'> 2 {1, 2, 3, 4, 5, 'tin', 12, ('tj', 120)}
（关系测试）
交集:
1 # -*- coding:utf-8 -*- 2 s={1,2,3,4,5,1,2,12,"tin",("tj",120),"tin",("tj",120)} 3 s2={123,23,4,5,6,7,8,} 4 #交集，把相同的元素取出来 5 print(s2.intersecti(s)) #或 print(s2 & s)
输出结果：
{4, 5}
并集
1 s={1,2,3,4,5,1,2,12,"tin",("tj",120),"tin",("tj",120)} 2 s2={123,23,4,5,6,7,8,} #并集 3 print(s2.union(s)) #或 print(s2 | s)
输出结果：
{1, 2, 3, 4, 5, 6, 7, 8, ('tj', 120), 12, 'tin', 23, 123}
差集
1 s={1,2,3,4,5,1,2,12,"tin",("tj",120),"tin",("tj",120)} 2 s2={123,23,4,5,6,7,8,} #差集 3 print(s2.difference(s)) # 或 print(s2-s)
输出结果：
1 {6, 7, 8, 23, 123}
对称差集
1 s={1,2,3,4,5,1,2,12,"tin",("tj",120),"tin",("tj",120)} 2 s2={123,23,4,5,6,7,8,} 3 print(s2.symmetric_difference(s)) #或 print(s2 ^ s)
输入结果：
{1, 2, 3, 6, 7, 8, 12, 'tin', 23, ('tj', 120), 123}
随机删除元素：
1 s={1,2,3,4,5,1,2,12,"tin",("tj",120),"tin",("tj",120)} 2 s2={123,23,4,5,6,7,8,} 3 print(s2.pop()) #pop 随机删除元素
输出结果：4
删除元素（可以任意删除一项）：
1 s2={123,23,4,5,6,7,8,} 2 s.remove(2) 3 print(s)
输出结果：
{1, 3, 4, 5, 'tin', 12, ('tj', 120)}
父集：
1 s={1,2,3,4,5,1,2,12,"tin",("tj",120),"tin",("tj",120)} 2 s2={123,23,4,5,6,7,8,} 3 s.remove(2) 4 print(s) 5 print(s2.issuperset(s))
输出结果：
1 {1, 3, 4, 5, 12, ('tj', 120), 'tin'} 2 False
子集：
1 s={1,2,3,4,5,1,2,12,"tin",("tj",120),"tin",("tj",120)} 2 s2={123,23,4,5,6,7,8,} 3 s.remove(2) 4 print(s) 5 print(s2.issubset(s))
输出结果：
1 {1, 3, 4, 5, 'tin', 12, ('tj', 120)} 2 False
对称并集：
1 s={1,2,3,4,5,1,2,12,"tin",("tj",120),"tin",("tj",120)} 2 s2={123,23,4,5,6,7,8,} 3 s.remove(2) 4 print(s) 5 s2.difference_update(s) 6 print(s2)
输出结果：
1 {'tin', 1, 3, 4, 5, ('tj', 120), 12} 2 {6, 7, 8, 23, 123}
添加元素：
1 s={1,2,3,4,5,1,2,12,"tin",("tj",120),"tin",("tj",120)} 2 s2={123,23,4,5,6,7,8,} 3 s2.add(18) 4 print(s2) 5 s.add(21) 6 print(s)
输出结果：
1 {4, 5, 6, 7, 8, 18, 23, 123} 2 {('tj', 120), 1, 2, 3, 4, 5, 12, 'tin', 21}
合并多项元素：
1 s={1,2,3,4,5,1,2,12,"tin",("tj",120),"tin",("tj",120)} 2 s2={123,23,4,5,6,7,8,} 3 s.update(s2) 4 print(s)
输出结果：
{1, 2, 3, 4, 5, 6, 7, 8, 12, 23, 'tin', 123, ('tj', 120)}