Python · 2 分钟阅读

`zip()` 函数

zip() 把多个可迭代对象按位置配对，产出“每一列”的元组。配合
* 解包运算符，还能把配对好的结果再“拆回”原始形状。

1. 基本用法

a = [1, 2, 3]
b = ["a", "b", "c"]
print(list(zip(a, b)))
# [(1, 'a'), (2, 'b'), (3, 'c')]

要点：

strs = ["flower", "flow", "flight"]
print(list(zip(*strs)))
# [('f', 'f', 'f'), ('l', 'l', 'l'), ('o', 'o', 'i'), ('w', 'w', 'g')]

把 *strs 看成“把 strs 解包为独立参数”：

zip(*strs)   ≡   zip("flower", "flow", "flight")

zip 再“把每个串同位置的字符放成一组”：

zip 的结果 = [第 0 列, 第 1 列, 第 2 列, ...]
           = [(f,f,f), (l,l,l), (o,o,i), (w,w,g), ...]

这正是 最长公共前缀 这类题目的关键技巧：把列凑齐，按列做集合去重或
all / any 判定。

names = ["Alice", "Bob", "Carol"]
ages  = [30, 25, 28]

for name, age in zip(names, ages):
    print(name, age)

keys = ["name", "age", "city"]
vals = ["Alice", 30, "NY"]
print(dict(zip(keys, vals)))
# {"name": "Alice", "age": 30, "city": "NY"}

matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
transposed = list(zip(*matrix))
# [(1, 4, 7), (2, 5, 8), (3, 6, 9)]

list1 = [1, 2, 3]
list2 = [10, 20, 30]
list3 = [100, 200, 300]

for a, b, c in zip(list1, list2, list3):
    print(a + b + c)

import itertools
a = [1, 2, 3, 4]
b = ["x", "y"]
print(list(itertools.zip_longest(a, b, fillvalue=0)))
# [(1, 'x'), (2, 'y'), (3, 0), (4, 0)]

pairs = [(1, 'a'), (2, 'b'), (3, 'c')]
nums, letters = zip(*pairs)
print(nums)        # (1, 2, 3)
print(letters)     # ('a', 'b', 'c')

这正是“转置”的反向操作。