元组与集合

元组的创建

元组是不可变序列，不可以增删改

# 使用()创建
tuple1 = ("t1", "t2", 10)
print(tuple1)
tuple1 = "t1", "t2", 20  # 省略小括号
print(tuple1)

# 使用tuple函数创建
tuple2 = tuple(("t1", "t2", 30))
print(tuple2)
print(type(tuple2))

# 只包含一个元素的元组
tuple3 = ("python",)
print(tuple3)
print(type(tuple3))

# 空元组
tuple4 = ()
tuple5 = tuple()
print(tuple4)
print(type(tuple4))
print(tuple5)
print(type(tuple5))

元组元素不可变

t = (10, [20, 30], 40)
print(t, type(t))
print(t[0], type(t[0]))  # 10 <class 'int'>
print(t[1], type(t[1]))  # [20, 30] <class 'list'>
print(t[2], type(t[2]))  # 40 <class 'int'>

# 元组的元素不可变；若元素本身是可变对象，其元素中的值可变，但对象本身引用不变
list1 = t[1]
print(list1, id(list1))
list1[0] = 30
print(list1, id(list1))

元组的遍历

t = (10, 20, 30)
for i in t:
    print(i)

集合的创建

set 集合中元素不重复，且无序

set1 = {1, "Python", 2, 3, 4, 3, 2}
print(set1)  # {1, 2, 3, 4, 'Python'};set集合中元素不重复，且无序

set2 = set(range(6))
print(set2)
set2 = set([1, 3, 5, 7])
print(set2)
set2 = set((7, 5, 3, 1))
print(set2)
set2 = set("python")
print(set2)  # {'y', 'o', 't', 'h', 'p', 'n'};set集合会把字符串拆分
set2 = set({7, 5, 3, 10})
print(set2)
set2 = set()
print(set2)  # set()；空集合

集合的操作

# 集合元素的判断
set1 = {10, 20, 30, 40, 50}
print(10 in set1)
print(10 not in set1)

# 集合元素的新增
# 一次新增一个元素
set1 = {10, 20, 30, 40, 50}
set1.add(60)
print(set1)
# 一次新增多个元素
set1 = {10, 20, 30, 40, 50}
set1.update([70, 80, 100])
print(set1)  # {80, 50, 20, 100, 70, 40, 10, 30}
set1 = {10, 20, 30, 40, 50}
set1.update({70, 80, 101})
print(set1)
set1 = {10, 20, 30, 40, 50}
set1.update((70, 80, 102))
print(set1)

# 集合元素的删除
set1 = {10, 20, 30, 40, 50}
set1.remove(10)
print(set1)  # {50, 20, 40, 30}
set1 = {10, 20, 30, 40, 50}
# set1.remove(100)  # KeyError: 100;元素在集合中不存在
set1.discard(100)  # 删除不存在元素，不会抛异常
set1 = {10, 20, 30, 40, 50}
set1.pop()
print(set1)  # {20, 40, 10, 30};删除其中一个元素
set1 = {10, 20, 30, 40, 50}
set1.clear()
print(set1)  # set()：清空集合元素

# 集合的遍历
set1 = {10, 20, 30, 40, 50}
for i in set1:
    print(i)

集合的关系

# 相等判断，元素相同则相等
s1 = {10, 20, 30, 40}
s2 = {40, 20, 30, 10}
print(s1 == s2)  # True

# 子集判断
s1 = {10, 20, 30, 40}
s2 = {10, 20, 30}
s3 = {10, 20, 60}
print(s2.issubset(s1))  # True
print(s3.issubset(s1))  # False

# 超集判断
s1 = {10, 20, 30, 40}
s2 = {10, 20, 30}
s3 = {10, 20, 60}
print(s1.issuperset(s2))  # True
print(s1.issuperset(s3))  # False

# 交集判断
s1 = {10, 20, 30, 40}
s2 = {100, 200, 300}
s3 = {10, 20, 60}
print(s1.isdisjoint(s2))  # True;没有交集返回True
print(s1.isdisjoint(s3))  # False；

集合的数学操作

s1 = {10, 20, 30, 40}
s2 = {10, 20, 50, 60}
# 取交集
print(s1.intersection(s2))  # {10, 20}
print(s1 & s2)
# 取并集
print(s1.union(s2))  # {40, 10, 50, 20, 60, 30}
print(s1 | s2)
# 取差集
print(s1.difference(s2))  # {40, 30}
print(s1 - s2)
# 取对称差集
print(s1.symmetric_difference(s2))  # {40, 50, 60, 30}
print(s1 ^ s2)

集合生成式

s1 = {i * 2 for i in range(6)}
print(s1)  # {0, 2, 4, 6, 8, 10}