# 创建元组
fruits = ("apple", "banana", "cherry")
numbers = (1, 3, 5, 7, 9)
mixed = (1, "hello", True, 3.14)
nested = (("a", "b"), (1, 2, 3))

# 创建单元素元组（注意逗号）
single = (42,)  # 必须有逗号，否则会被视为普通括号

# 访问元组元素
print(fruits[0])  # 输出: apple
print(numbers[2])  # 输出: 5
print(mixed[-1])  # 输出: 3.14 (负数索引从末尾开始)
print(nested[1][0])  # 输出: 1 (访问嵌套元组)

# 元组不可变（尝试修改会报错）
# fruits[1] = "grape"  # 这会引发 TypeError

# 元组长度
fruits = ("apple", "banana", "cherry", "date")
print(len(fruits))  # 输出: 4

# 元组切片
print(fruits[1:3])  # 输出: ('banana', 'cherry') (索引1到2)
print(fruits[:2])   # 输出: ('apple', 'banana') (从开始到索引1)
print(fruits[2:])   # 输出: ('cherry', 'date') (从索引2到结束)
print(fruits[-2:])  # 输出: ('cherry', 'date') (最后两个元素)
print(fruits[::2])  # 输出: ('apple', 'cherry') (每隔一个元素)
print(fruits[::-1]) # 输出: ('date', 'cherry', 'banana', 'apple') (反转元组)

# 连接元组
t1 = (1, 2, 3)
t2 = (4, 5, 6)
combined = t1 + t2
print(combined)  # 输出: (1, 2, 3, 4, 5, 6)

# 重复元组
repeated = t1 * 3
print(repeated)  # 输出: (1, 2, 3, 1, 2, 3, 1, 2, 3)

# 元组嵌套
nested = (t1, t2)
print(nested)  # 输出: ((1, 2, 3), (4, 5, 6))
print(nested[0][1])  # 输出: 2

# 基本解包
t = (1, 2, 3)
a, b, c = t
print(a)  # 输出: 1
print(b)  # 输出: 2
print(c)  # 输出: 3

# 交换变量
x = 10
y = 20
x, y = y, x  # 元组解包实现变量交换
print(x)  # 输出: 20
print(y)  # 输出: 10

# 解包部分元素（使用*）
t = (1, 2, 3, 4, 5)
first, *middle, last = t
print(first)  # 输出: 1
print(middle)  # 输出: [2, 3, 4]
print(last)  # 输出: 5

# 函数返回多个值
def get_name_and_age():
  return "John", 30

name, age = get_name_and_age()
print(name)  # 输出: John
print(age)  # 输出: 30

# count() - 统计元素出现次数
t = (1, 2, 3, 2, 4, 2)
print(t.count(2))  # 输出: 3

# index() - 返回元素第一次出现的索引
print(t.index(2))  # 输出: 1
print(t.index(2, 2))  # 输出: 3 (从索引2开始查找)

# 检查元素是否存在
print(3 in t)  # 输出: True
print(5 in t)  # 输出: False

# 元组转列表
t = (1, 2, 3)
lst = list(t)
print(lst)  # 输出: [1, 2, 3]

# 列表转元组
lst = [4, 5, 6]
t = tuple(lst)
print(t)  # 输出: (4, 5, 6)

# 字符串转元组
s = "hello"
t = tuple(s)
print(t)  # 输出: ('h', 'e', 'l', 'l', 'o')

# 元组转字符串（需要先转换元素）
t = ('a', 'b', 'c')
s = ''.join(t)
print(s)  # 输出: abc

# 元组转集合
t = (1, 2, 2, 3)
s = set(t)
print(s)  # 输出: {1, 2, 3} (去重)

# 使用 for 循环遍历元组
fruits = ("apple", "banana", "cherry")
for fruit in fruits:
  print(fruit)  # 依次输出: apple, banana, cherry

# 使用索引遍历元组
for i in range(len(fruits)):
  print(fruits[i])  # 依次输出: apple, banana, cherry

# 使用 enumerate() 获取索引和值
for index, fruit in enumerate(fruits):
  print(f"Index {index}: {fruit}")  # 依次输出: Index 0: apple, Index 1: banana, Index 2: cherry

# 使用 while 循环遍历元组
i = 0
while i < len(fruits):
  print(fruits[i])
  i += 1  # 依次输出: apple, banana, cherry

# 生成器表达式
squares_gen = (x**2 for x in range(5))
print(squares_gen)  # 输出: <generator object <genexpr> at 0x...>

# 转换为元组
squares_tuple = tuple(squares_gen)
print(squares_tuple)  # 输出: (0, 1, 4, 9, 16)

# 直接使用生成器表达式
sum_squares = sum(x**2 for x in range(5))
print(sum_squares)  # 输出: 30

# 元组作为字典键
coordinates = {(0, 0): "原点", (1, 1): "第一象限", (-1, 1): "第二象限"}
print(coordinates[(0, 0)])  # 输出: 原点

# 列表不能作为字典键（因为列表是可变的）
# invalid_dict = {[0, 0]: "原点"}  # 这会引发 TypeError

# 使用元组存储数据库记录
records = [
  (1, "John", 25),
  (2, "Mary", 30),
  (3, "Bob", 35)
]

# 遍历记录
for id, name, age in records:
  print(f"ID: {id}, 姓名: {name}, 年龄: {age}")

# 函数接受元组参数
def print_person_info(person):
  name, age, city = person
  print(f"{name} 今年 {age} 岁，住在 {city}")

person = ("John", 30, "New York")
print_person_info(person)  # 输出: John 今年 30 岁，住在 New York

# 函数返回元组
def get_min_max(numbers):
  return min(numbers), max(numbers)

numbers = [5, 2, 8, 1, 9]
min_val, max_val = get_min_max(numbers)
print(f"最小值: {min_val}, 最大值: {max_val}")  # 输出: 最小值: 1, 最大值: 9

# 使用 * 解包元组作为函数参数
def add(a, b, c):
  return a + b + c

t = (1, 2, 3)
result = add(*t)  # 等价于 add(1, 2, 3)
print(result)  # 输出: 6

# 适合使用元组的场景
# 1. 函数返回多值
def get_user_info():
  return "John", 30, "New York"  # 返回元组

# 2. 固定配置数据
COLORS = ("red", "green", "blue")  # 常量元组

# 3. 字典键
points = {(0, 0): "原点", (1, 1): "第一象限"}  # 元组作为键

# 适合使用列表的场景
# 1. 动态数据
shopping_list = ["apple", "banana"]
shopping_list.append("cherry")  # 可以修改

# 2. 需要排序的数据
numbers = [3, 1, 4, 1, 5]
numbers.sort()  # 可以排序

# 3. 需要频繁修改的数据
scores = [85, 90, 78]
scores[1] = 92  # 可以修改