如何按父文件夹对路径列表进行动态分组？

给定如下路径列表，需要根据路径中的目标文件夹（如tent1、tent2）将路径自动分组到对应的子列表中，无需硬编码文件夹名称：

paths = [
    '/var/lib/cons/states/tent1/tops-ok_2022_11_28', 
    '/var/lib/cons/states/tent1/tops-ok_2022_11_27',  
    '/var/lib/cons/states/tent2/tops-ok_2022_11_28', 
    '/var/lib/cons/states/tent2/tops-ok_2022_11_27', 
    '/var/lib/cons/states/tent3/tops-ok_2022_11_28', 
    '/var/lib/cons/states/tent3/tops-ok_2022_11_27', 
    '/var/lib/cons/states/tent4/tops-ok_2022_11_28', 
    '/var/lib/cons/states/tent4/tops-ok_2022_11_27',
]

解决方案 #

可以通过字典动态分组：提取每个路径中的目标文件夹名称作为键，将对应路径存入值列表中，最后将字典的值转换为列表即可得到结果。

方法一：使用defaultdict简化代码

from collections import defaultdict

grouped = defaultdict(list)
for path in paths:
    # 拆分路径，取倒数第二个部分作为目标文件夹名
    folder = path.split('/')[-2]
    grouped[folder].append(path)

result = list(grouped.values())

方法二：普通字典实现（无需额外模块）

grouped = {}
for path in paths:
    folder = path.split('/')[-2]
    if folder not in grouped:
        grouped[folder] = []
    grouped[folder].append(path)

result = list(grouped.values())

最终结果 #

运行上述代码后，result的值与期望一致：

[['/var/lib/cons/states/tent1/tops-ok_2022_11_28', 
  '/var/lib/cons/states/tent1/tops-ok_2022_11_27'], 
 ['/var/lib/cons/states/tent2/tops-ok_2022_11_28', 
  '/var/lib/cons/states/tent2/tops-ok_2022_11_27'],
 ['/var/lib/cons/states/tent3/tops-ok_2022_11_28', 
  '/var/lib/cons/states/tent3/tops-ok_2022_11_27'],
 ['/var/lib/cons/states/tent4/tops-ok_2022_11_28', 
  '/var/lib/cons/states/tent4/tops-ok_2022_11_27']]

内容的提问来源于stack exchange，提问作者dean89