Python zip()函数的用法和技巧(解锁并行迭代的魔力)

zip(iterable1, iterable2, ..., iterableN)

# 基础使用
names = ['Alice', 'Bob', 'Charlie']
ages = [25, 30, 35]
scores = [85, 92, 88]
# 打包三个列表
zipped = zip(names, ages, scores)
print(list(zipped))
# 输出: [('Alice', 25, 85), ('Bob', 30, 92), ('Charlie', 35, 88)]

# 基础使用
names = ['Alice', 'Bob', 'Charlie']
ages = [25, 30, 35]
scores = [85, 92, 88]
# 打包三个列表
zipped = zip(names, ages, scores)
print(list(zipped))
# 输出: [('Alice', 25, 85), ('Bob', 30, 92), ('Charlie', 35, 88)]

list1 = [1, 2, 3, 4]
list2 = ['a', 'b', 'c']
result = list(zip(list1, list2))
print(result)  # 输出: [(1, 'a'), (2, 'b'), (3, 'c')]

zipped = zip([1, 2], ['a', 'b'])
print(zipped)  # 输出: <zip object at 0x...>
print(list(zipped))  # 转换为列表查看内容

# 传统方式
names = ['Alice', 'Bob', 'Charlie']
ages = [25, 30, 35]
for i in range(len(names)):
    print(f"{names[i]} is {ages[i]} years old")
# 使用zip更优雅的方式
for name, age in zip(names, ages):
    print(f"{name} is {age} years old")

keys = ['name', 'age', 'city']
values = ['Alice', 25, 'New York']
# 快速创建字典
person = dict(zip(keys, values))
print(person)  # 输出: {'name': 'Alice', 'age': 25, 'city': 'New York'}

matrix = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
]
# 使用zip进行转置
transposed = list(zip(*matrix))
print(transposed)
# 输出: [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
# 如果需要列表而不是元组
transposed_list = [list(row) for row in zip(*matrix)]

# 将数据分为键值对
headers = ['id', 'name', 'score']
data = [
    [1, 'Alice', 95],
    [2, 'Bob', 88],
    [3, 'Charlie', 92]
]
for row in data:
    paired = dict(zip(headers, row))
    print(paired)

# 结合enumerate使用
items = ['apple', 'banana', 'cherry']
prices = [1.2, 0.8, 2.5]
for i, (item, price) in enumerate(zip(items, prices)):
    print(f"{i}: {item} - ${price}")

# 打包
pairs = [('a', 1), ('b', 2), ('c', 3)]
letters, numbers = zip(*pairs)
print(letters)  # 输出: ('a', 'b', 'c')
print(numbers)  # 输出: (1, 2, 3)

from itertools import zip_longest
list1 = [1, 2, 3, 4]
list2 = ['a', 'b', 'c']
# 使用zip_longest填充缺失值
result = list(zip_longest(list1, list2, fillvalue='N/A'))
print(result)
# 输出: [(1, 'a'), (2, 'b'), (3, 'c'), (4, 'N/A')]

# 同时处理多个列表的元素
list1 = [1, 2, 3]
list2 = [10, 20, 30]
# 计算对应元素的和
result = list(map(lambda x: x[0] + x[1], zip(list1, list2)))
print(result)  # 输出: [11, 22, 33]
# 更简洁的方式
result = [x + y for x, y in zip(list1, list2)]

# zip返回迭代器，内存友好
large_list1 = range(1000000)
large_list2 = range(1000000)
# 不会一次性创建所有元组
zipped = zip(large_list1, large_list2)

# 陷阱1: zip对象只能遍历一次
zipped = zip([1, 2], ['a', 'b'])
list1 = list(zipped)  # 第一次使用
list2 = list(zipped)  # 空列表！因为迭代器已耗尽
print(list2)  # 输出: []
# 解决方法
zipped = list(zip([1, 2], ['a', 'b']))  # 转换为列表
# 或重新创建迭代器
# 陷阱2: 处理不等长数据时要小心
list1 = [1, 2, 3, 4]
list2 = ['a', 'b', 'c']
for x, y in zip(list1, list2):
    # 注意: 元素4不会被处理
    pass

# 假设有两个CSV文件的数据列
names = ['Alice', 'Bob', '', 'Charlie', None]
ages = ['25', '30', '28', 'thirty-five', '40']
# 清理数据并配对
clean_data = []
for name, age in zip(names, ages):
    if name and name.strip() and age and age.isdigit():
        clean_data.append((name.strip(), int(age)))
print(clean_data)

students = ['Alice', 'Bob', 'Charlie']
scores = [85, 92, 85]
ages = [25, 23, 27]
# 按分数降序，年龄升序排序
sorted_students = [name for name, _, _ in 
                   sorted(zip(students, scores, ages), 
                          key=lambda x: (-x[1], x[2]))]
print(sorted_students)  # 输出: ['Bob', 'Alice', 'Charlie']

import os
# 批量重命名文件
original_names = ['file1.txt', 'file2.txt', 'file3.txt']
new_names = ['document1.txt', 'document2.txt', 'document3.txt']
for old, new in zip(original_names, new_names):
    # 实际应用中会调用 os.rename(old, new)
    print(f"Would rename {old} to {new}")