from collections.abc import Iterable, Iterator, Generator

# 检查
print(isinstance([1, 2, 3], Iterable))   # True
print(isinstance([1, 2, 3], Iterator))   # False
print(isinstance(iter([1, 2, 3]), Iterable))  # True
print(isinstance(iter([1, 2, 3]), Iterator))  # True

class IteratorProtocol:
    """迭代器协议的完整契约"""

    def __iter__(self):
        """返回迭代器对象本身"""
        return self

    def __next__(self):
        """返回下一个元素，没有元素时抛出 StopIteration"""
        if self._exhausted:
            raise StopIteration
        # ... 返回下一个值

# for x in iterable:
#     print(x)

# 等价于：
_iterator = iter(iterable)     # 1. 获取迭代器
while True:
    try:
        x = next(_iterator)     # 2. 逐个取值
    except StopIteration:       # 3. 捕捉结束信号
        break
    print(x)                    # 4. 执行循环体
del _iterator                   # 5. 清理迭代器

class OldStyleIterable:
    """旧式可迭代对象——只有 __getitem__"""
    def __init__(self, data):
        self.data = data

    def __getitem__(self, index):
        if index >= len(self.data):
            raise IndexError     # Python 捕获 IndexError 并转为 StopIteration
        return self.data[index]

# 仍然可以 for 循环！
for x in OldStyleIterable([1, 2, 3]):
    print(x)  # 1, 2, 3

# ❌ 列表：一次性加载所有数据到内存
all_lines = file.readlines()  # 大文件时内存爆炸

# ✅ 迭代器：逐行读取，每次只占一行内存
for line in file:             # file 本身是迭代器
    process(line)

# ❌ 列表推导式：立即生成全部结果
squares = [x**2 for x in range(10**7)]  # 内存爆炸！

# ✅ 迭代器：按需生成
squares = (x**2 for x in range(10**7))  # 几乎不占内存

class Counter:
    """无限递增的迭代器——演示惰性求值"""
    def __init__(self, start=0):
        self.current = start

    def __iter__(self):
        return self

    def __next__(self):
        result = self.current
        self.current += 1
        return result

# 永远不会构造完 10 亿个元素的列表
# 只会在 for 循环中逐个生成
counter = Counter()
for i, val in enumerate(counter):
    if i >= 10:
        break
    print(val)  # 0, 1, 2, ..., 9

import itertools

# 1. count + islice = 惰性切片
evens = itertools.count(start=0, step=2)
first_10 = list(itertools.islice(evens, 10))  # [0, 2, 4, ... 18]

# 2. cycle 实现轮询
servers = ['S1', 'S2', 'S3']
pool = itertools.cycle(servers)
assignments = [next(pool) for _ in range(7)]  # ['S1','S2','S3','S1','S2','S3','S1']

# 3. chain 合并多个迭代器
result = list(itertools.chain([1, 2], [3, 4], 'AB'))  # [1, 2, 3, 4, 'A', 'B']

# 4. zip_longest 处理不等长
a = [1, 2, 3]
b = ['a', 'b']
padded = list(itertools.zip_longest(a, b, fillvalue='N/A'))
# [(1, 'a'), (2, 'b'), (3, 'N/A')]

# 5. groupby 分组（必须已排序）
data = [('A', 1), ('A', 2), ('B', 3), ('B', 4)]
for key, group in itertools.groupby(data, key=lambda x: x[0]):
    print(key, list(group))  # A [(A,1),(A,2)]  B [(B,3),(B,4)]

nums = iter([1, 2, 3])

# 第一次遍历：正常
for x in nums:
    print(x)  # 1, 2, 3

# 第二次遍历：什么都没有！
for x in nums:
    print(x)  # (无输出) — 迭代器已经耗尽

# ✅ 解决：需要重新创建迭代器
nums = iter([1, 2, 3])

# ❌ RuntimeError: Set changed size during iteration
items = {1, 2, 3, 4, 5}
for item in items:
    if item % 2 == 0:
        items.remove(item)

# ✅ 正确做法：先复制再遍历
for item in list(items):  # list() 创建副本
    if item % 2 == 0:
        items.remove(item)

# 小心：传给多个函数可能导致意外耗尽
def check_positive(nums):
    return all(x > 0 for x in nums)

def sum_squares(nums):
    return sum(x**2 for x in nums)

data = (x for x in [1, 2, -3, 4])  # 生成器
print(check_positive(data))  # False（遍历到 -3 时停止）
print(sum_squares(data))     # 只得到 4²=16，因为生成器已部分耗尽！

# ✅ 改用列表
data = [1, 2, -3, 4]

# chain 嵌套 chain：没问题，惰性的
deep = itertools.chain.from_iterable(
    itertools.chain.from_iterable(...)
)

# 但打印调试时会立即求值
list(deep)  # 小心内存！