python - 我将如何基于 Python 中的匹配路径字符串创建列表？

Question

我基本上是在尝试将文件路径字符串分隔到同一路径中的文件列表中。假设我有这个列表:

files = ['user/hey.jpg','user/folder1/1.txt','user/folder1/folder2/random.txt',
         'user/folder1/blah.txt','user/folder3/folder4/folder5/1.txt',
         'user/folder3/folder4/folder5/3.txt','user/folder3/folder4/folder5/2.txt',
         'user/1.jpg']

我希望得到这个输出(不关心顺序):

[
 ['user/folder1/1.txt','user/folder1/blah.txt']
 ['user/folder1/folder2/random.txt']
 ['user/folder3/folder4/folder5/1.txt',
  'user/folder3/folder4/folder5/2.txt','user/folder3/folder4/folder5/3.txt']
 ['user/1.jpg','user/hey.jpg']
]

抱歉，如果我的示例草率。在手机上打字并不好玩。感谢您的帮助!

Answer 1

使用 os.path.dirname 作为 sorted 和 groupby 函数的键应该会引导您得到结果:

import os
from itertools import groupby
[list(g) for _, g in groupby(sorted(files, key=os.path.dirname), key=os.path.dirname)]

#[['user/hey.jpg', 'user/1.jpg'],
# ['user/folder1/1.txt', 'user/folder1/blah.txt'],
# ['user/folder1/folder2/random.txt'],
# ['user/folder3/folder4/folder5/1.txt',
#  'user/folder3/folder4/folder5/3.txt',
#  'user/folder3/folder4/folder5/2.txt']]