集合

# a = [1,2,1,3,4,1,5,'qiancheng','qiancheng']
# b = set(a)
# for i in a:
# print(i)


# SET 不能有重复的值
# s = set(['qiangcheng','lixue'])
# s.add('uu')
# s.update('ooo') {'lixue', 'o', 'qiangchegng'}
# s.update('ops') {'p', 'lixue', 'o', 's', 'qiangchegng'}
# s.update([12,'eee']) {'eee', 12, 'qiangchegng', 'lixue'}
# s.remove('lixue')
# s.pop() 随机删除
# s.clear() 清空
#del s NameError: name 's' is not defined
# print(s)

# print(set('qqq') == set('qqqqqqqqq')) True
# print(set('qqq') < set('qqqwww')) True
# print(set('qqq') < set('qqq')) False
# print(set('qqqaa') or set('qqqwww')) {'q', 'a'}
# print(set('qqqaa') and set('qqqwww')) #{'q', 'w'}
# print(set('qqqaa') and set('bbbwww'))

a = set([1,2,3,4,5])
b = set([4,5,6,7,8])
# 交集
# print(a.intersection(b)) a & b {4, 5}
# 并集
# print(a.union(b)) #a | b {1, 2, 3, 4, 5, 6, 7, 8}
# 差集
# print(a.difference(b)) a - b {1, 2, 3} A里面有 B里面没有
#对称差集,又叫反向交集
# print(a.symmetric_difference(b)) a ^ b {1, 2, 3, 6, 7, 8}
# 子集
# print(a.issubset(b)) < False
# 超集
# print(a.issuperset(b)) > False a是否完全包含b








posted @ 2019-12-10 23:21  feichengwurao  阅读(76)  评论(0编辑  收藏  举报