定义集合

num = {1, 2, 3, 4, 5}

print(f"num：{num}\nnum数据类型为：{type(num)}")

求集合中元素个数

print(f"num中元素个数为：{len(num)}")

增加集合中的元素

num.add(6)

print(num) # {1,2,3,4,5,6}

删除集合中的元素

num.remove(4)

print(num) # {1,2,3,4,6}

从集合中移除并返回任意一共元素,会影响原来的集合

a = num.pop()

print(f"a:{a},num:{num}")

复制代码

# union (|)两个集求并集
name1 = {"bob", "jack", "tonnny", "amy"}
name2 = {"bob", "tom", "rose", "jack"}
name_a = name1.union(name2)
name_b = name2 | name1
print(name_a, "\n", name_b)

# intersection（&）求两个集合的交集
name = name1.intersection(name2)
nam = name1 & name2
print(f"name：{name},nam：{nam}")

# 求一个集合在另一个集合中的补集
n=name1.difference(name2) #{"tonnny","amy"}
m=name2.difference(name1) #{"tom", "rose"}
print(f"n：{n}\nm：{m}")

# {集合的表达式 for 自变量 可迭代对象}
set={i for i in range(1,5)}
print(set)

复制代码

Chinese={"小张","小王","小李","小刘","jack"}
Math={"小张","bob","rose","小刘","jack"}
English={"小张","tom","tony","amy"}

# 选课学生总人数-集合的并集All_num
All_num=Chinese|Math|English
print(f"选课学生总人数：{len(All_num)}人，分别为{All_num}")

#只选了Chinses的学生人数和姓名
Only_name=Chinese-Math-English
print(f"只选了Chinses的学生人数：{len(Only_name)}/学生姓名{Only_name}")

#只选择一门课的学生人数
# 只选语文
One_name1=Chinese-Math-English
# 只选数学
One_name2=Math-English-Chinese
#只选英语
One_name3=English-Chinese-Math
# 求并集
One_name=One_name1|One_name2|One_name3
print(f"只选择一门课的学生人：{One_name},总人数为{len(One_name)}")

#三门课都选的学生，求交集
ALL=Chinese&Math&English
print(f"三门课都选的学生：{ALL}，总人数为{len(ALL)}")