Python中如何按每次取5个元素的规则迭代遍历列表

方案1：列表切片实现（全Python版本兼容） #

这是兼容性最好的实现方式，不需要依赖任何第三方库或高版本语法：

L = ["a","b","c","d","e","f","g","h","i","j"]
# 按步长5遍历列表索引
for idx in range(0, len(L), 5):
    # 每次取长度为5的切片
    current_chunk = L[idx:idx+5]
    # 若需要过滤最后不足5个元素的轮次，可加判断 if len(current_chunk) ==5:
    first, second, third, fourth, fifth = current_chunk
    # 基础打印
    print(first, second, third, fourth, fifth)
    # 若需要和示例完全一致的带引号、逗号分隔格式，替换上面的print为下一行
    # print(', '.join(f'"{item}"' for item in current_chunk))

方案2：itertools.batched 实现（Python 3.12+ 适用） #

Python 3.12及以上版本内置了专门的分块迭代工具，语法更简洁：

from itertools import batched

L = ["a","b","c","d","e","f","g","h","i","j"]
for first, second, third, fourth, fifth in batched(L, 5):
    print(first, second, third, fourth, fifth)

内容的提问来源于stack exchange，提问作者jer_card