深入理解Python中解包错误的解决方案

# 基本解包示例
coordinates = (3, 5, 7)
x, y, z = coordinates
print(f"x={x}, y={y}, z={z}")  # 输出: x=3, y=5, z=7

# 错误示例：期望5个值，实际只有4个
try:
    a, b, c, d, e = [1, 2, 3, 4]
    print("解包成功")
except ValueError as e:
    print(f"错误: {e}")  # 输出: not enough values to unpack (expected 5, got 4)

# 错误示例：错误的字典解包方式
user_data = {"name": "Alice", "age": 25, "city": "Beijing"}

try:
    k1, v1, k2, v2, k3, v3 = user_data.items()
except ValueError as e:
    print(f"字典解包错误: {e}")

# 错误示例：函数返回值数量不匹配
def calculate_stats(data):
    """计算数据的基本统计量"""
    return len(data), sum(data), min(data), max(data)

data_list = [10, 20, 30, 40]

try:
    count, total, minimum, maximum, average = calculate_stats(data_list)
except ValueError as e:
    print(f"函数返回值解包错误: {e}")

# 正确解包：变量数量精确匹配
data = [1, 2, 3, 4, 5]
first, second, third, fourth, fifth = data
print(f"解包结果: {first}, {second}, {third}, {fourth}, {fifth}")

# 使用 * 操作符处理不定数量解包
mixed_data = [1, 2, 3, 4, 5, 6, 7, 8, 9]

# 提取前两个元素，其余放入列表
first, second, *remaining = mixed_data
print(f"前两个: {first}, {second}")
print(f"剩余元素: {remaining}")

# 提取首尾元素，中间部分收集
start, *middle, end = mixed_data
print(f"开头: {start}, 结尾: {end}")
print(f"中间部分: {middle}")

def safe_unpack(iterable, expected_count, default_value=None):
    """
    安全解包函数
    :param iterable: 可迭代对象
    :param expected_count: 期望解包数量
    :param default_value: 默认值
    :return: 解包后的元组
    """
    actual_count = len(iterable)
    
    if actual_count < expected_count:
        # 补充默认值
        extended_list = list(iterable) + [default_value] * (expected_count - actual_count)
        return tuple(extended_list)
    elif actual_count > expected_count:
        # 截断多余元素
        return tuple(iterable[:expected_count])
    else:
        return tuple(iterable)

# 使用安全解包函数
short_list = [1, 2, 3]
result = safe_unpack(short_list, 5, default_value=0)
a, b, c, d, e = result
print(f"安全解包结果: {a}, {b}, {c}, {d}, {e}")

# 处理多维数据的解包
matrix = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
]

# 正确的行列解包方式
for row in matrix:
    if len(row) == 3:  # 检查每行元素数量
        a, b, c = row
        print(f"行数据: {a}, {b}, {c}")

# 使用 zip 进行转置解包
transposed = list(zip(*matrix))
print(f"转置矩阵: {transposed}")

# 嵌套数据结构解包
nested_data = [(1, 'a', True), (2, 'b', False), (3, 'c', True)]

for number, letter, flag in nested_data:
    print(f"数字: {number}, 字母: {letter}, 标志: {flag}")

# 复杂嵌套解包
complex_data = [('Alice', (25, 'Engineer')), ('Bob', (30, 'Designer'))]

for name, (age, profession) in complex_data:
    print(f"{name}: {age}岁, {profession}")

# 正确的字典解包方式
user_profile = {
    "name": "Alice",
    "age": 25,
    "email": "alice@example.com",
    "city": "Beijing"
}

# 方法1: 直接解包键值对
for key, value in user_profile.items():
    print(f"{key}: {value}")

# 方法2: 选择性解包
name, age = user_profile['name'], user_profile['age']
print(f"姓名: {name}, 年龄: {age}")

# 方法3: 使用字典get方法安全解包
phone = user_profile.get('phone', '未提供')
print(f"电话: {phone}")

class SafeUnpacker:
    """安全的解包工具类"""
    
    @staticmethod
    def unpack_with_validation(iterable, expected_count, context=""):
        """
        带验证的解包方法
        """
        if len(iterable) != expected_count:
            raise ValueError(
                f"解包数量不匹配 {context}: "
                f"期望 {expected_count}, 实际 {len(iterable)}"
            )
        return iterable
    
    @staticmethod
    def unpack_with_defaults(iterable, expected_count, defaults):
        """
        带默认值的解包方法
        """
        result = list(iterable)
        while len(result) < expected_count:
            result.append(defaults[len(result)])
        return tuple(result[:expected_count])

# 使用示例
data = [1, 2, 3]

try:
    validated_data = SafeUnpacker.unpack_with_validation(data, 4, "测试数据")
    a, b, c, d = validated_data
except ValueError as e:
    print(f"验证错误: {e}")

# 使用默认值解包
default_data = SafeUnpacker.unpack_with_defaults(data, 5, [0, 0, 0, 100, 200])
x, y, z, w, v = default_data
print(f"带默认值的解包: {x}, {y}, {z}, {w}, {v}")

import time
from collections import namedtuple

# 使用命名元组提高代码可读性和安全性
Person = namedtuple('Person', ['name', 'age', 'email'])

def process_data_performance():
    """性能优化的解包方式"""
    large_data = [('user' + str(i), i, f'user{i}@example.com') for i in range(10000)]
    
    # 传统解包方式
    start_time = time.time()
    for item in large_data:
        name, age, email = item
    traditional_time = time.time() - start_time
    
    # 命名元组方式
    start_time = time.time()
    for item in large_data:
        person = Person(*item)
    namedtuple_time = time.time() - start_time
    
    print(f"传统解包时间: {traditional_time:.4f}s")
    print(f"命名元组时间: {namedtuple_time:.4f}s")
    print(f"性能提升: {((traditional_time - namedtuple_time)/traditional_time)*100:.1f}%")

process_data_performance()