按目录名与文件名多层分组排序Python路径的实现问题

问题描述 #

需要对给定的路径列表按以下规则进行嵌套分组排序：

1. 一级分组：按路径中的TENT1/TENT2目录划分
2. 二级分组：按文件名前缀（如metr-ok、weig-ok）划分，无论文件是直接位于TENT目录下还是其下的Job子目录中

原使用groupby结合dirname的代码无法实现预期结构，需求解。

输入路径列表：

paths = [
    '/var/lib/conc/states/TENT1/Job1/metr-ok_2022_11_28',
    '/var/lib/conc/states/TENT1/Job1/metr-ok_2022_11_29',
    '/var/lib/conc/states/TENT1/Job1/weig-ok_2022_11_28',
    '/var/lib/conc/states/TENT1/Job1/weig-ok_2022_11_29',
    '/var/lib/conc/states/TENT1/down-ok_2022_11_27',
    '/var/lib/conc/states/TENT1/down-ok_2022_11_28',
    '/var/lib/conc/states/TENT1/serv-ok_2022_11_28',
    '/var/lib/conc/states/TENT1/serv-ok_2022_11_29',
    '/var/lib/conc/states/TENT2/Job2/metr-ok_2022_11_28',
    '/var/lib/conc/states/TENT2/Job2/metr-ok_2022_11_29',
    '/var/lib/conc/states/TENT2/Job2/weig-ok_2022_11_28',
    '/var/lib/conc/states/TENT2/Job2/weig-ok_2022_11_29',
    '/var/lib/conc/states/TENT2/down-ok_2022_11_27',
    '/var/lib/conc/states/TENT2/down-ok_2022_11_28',
    '/var/lib/conc/states/TENT2/serv-ok_2022_11_28',
    '/var/lib/conc/states/TENT2/serv-ok_2022_11_29',
]

预期输出结构：

[
    [
        [
            '/var/lib/conc/states/TENT1/Job1/metr-ok_2022_11_28',
            '/var/lib/conc/states/TENT1/Job1/metr-ok_2022_11_29'
        ],
        [
            '/var/lib/conc/states/TENT1/Job1/weig-ok_2022_11_28',
            '/var/lib/conc/states/TENT1/Job1/weig-ok_2022_11_29'
        ],
        [
            '/var/lib/conc/states/TENT1/down-ok_2022_11_27',
            '/var/lib/conc/states/TENT1/down-ok_2022_11_28',
        ],
        [
            '/var/lib/conc/states/TENT1/serv-ok_2022_11_28',
            '/var/lib/conc/states/TENT1/serv-ok_2022_11_29'
        ],
    ],
    [
        [
            '/var/lib/conc/states/TENT2/Job2/metr-ok_2022_11_28',
            '/var/lib/conc/states/TENT2/Job2/metr-ok_2022_11_29'
        ],
        [
            '/var/lib/conc/states/TENT2/Job2/weig-ok_2022_11_28',
            '/var/lib/conc/states/TENT2/Job2/weig-ok_2022_11_29'
        ],
        [
            '/var/lib/conc/states/TENT2/down-ok_2022_11_27',
            '/var/lib/conc/states/TENT2/down-ok_2022_11_28',
        ],
        [
            '/var/lib/conc/states/TENT2/serv-ok_2022_11_28',
            '/var/lib/conc/states/TENT2/serv-ok_2022_11_29',
        ],
    ]
]

原代码问题分析 #

1. 分组键错误：使用dirname作为分组键只能按文件所在目录分组，无法实现按文件名前缀跨目录分组的需求。
2. 未提前排序：itertools.groupby要求输入序列必须先按分组键排序，否则会将同键但不连续的元素分到不同组。

解决方案代码 #

from itertools import groupby
from os.path import basename

def get_tent_group(path):
    # 提取路径中的TENT1/TENT2作为一级分组键
    return path.split('/')[4]

def get_file_prefix(path):
    # 提取文件名前缀（如metr-ok、weig-ok）作为二级分组键
    filename = basename(path)
    return '_'.join(filename.split('_')[:2])

# 先按一级分组键、再按二级分组键排序，确保同组元素连续
sorted_paths = sorted(paths, key=lambda x: (get_tent_group(x), get_file_prefix(x)))

# 构建嵌套分组结构
result = []
for _, tent_group in groupby(sorted_paths, key=get_tent_group):
    inner_groups = []
    for _, file_group in groupby(tent_group, key=get_file_prefix):
        inner_groups.append(list(file_group))
    result.append(inner_groups)

# 验证输出
print(result)

代码说明 #

1. 分组键定义：
   • get_tent_group：通过分割路径提取TENT1/TENT2部分作为一级分组依据。
   • get_file_prefix：提取文件名中_前两位（如metr-ok）作为二级分组依据，实现跨目录同前缀文件的分组。
2. 排序处理：先按一级键、再按二级键排序，保证groupby能正确识别连续的同组元素。
3. 嵌套分组：先按TENT分组，再在每个TENT组内按文件前缀分组，最终形成预期的三层嵌套列表。

内容的提问来源于stack exchange，提问作者dean89