集合操作

##################集合########################
# l=['a','b',1,'a','a']
# print(list(set(l))) #集合转化列表

# stus_linux={'alex','egon','张全蛋','李铁蛋','oldboy'}
# stus_python={'李二丫','wxx','liudehua','alex','egon'}

#即报名Linux又报名Python的学生:交集
# print(stus_linux & stus_python)
# print(stus_linux.intersection(stus_python))

#所有的学生:并集
# print(stus_linux | stus_python) # | 或
# print(stus_linux.union(stus_python))

#只报名Linux 没报Python 差集
# print(stus_linux - stus_python)
# print(stus_linux.difference(stus_python))

#没有同时报名两门课程的学生姓名:交差补集
#print(stus_linux ^ stus_python)
#print(stus_linux.symmetric_difference(stus_python))

# s1.isdisjoint() #如果两个集合没有交集 返回True
# s1={1,2,3}
# s2={4,5,6}
# print(s1.isdisjoint(s2))

#循环取值
# s1={1,'a','b','c','d'}
# for item in s1:
# print(item)

  

原文地址:https://www.cnblogs.com/zhangcaiwang1/p/9534355.html