Question

我基本上试图将文件路径字符串分隔到同一路径中的文件列表中。假设我有这个清单：

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']

我希望获得此输出（不关心订单）：

[
 ['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']
]

很抱歉，如果我的例子很草率。在手机上打字并不好玩。任何帮助表示赞赏！

Answer 1

使用os.path.dirname作为sorted中的密钥，groupby函数会引导您获得结果：

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']]

Answer 2

试试这个：

import collections
results = defaultdict(list)
for path in files:
    results[os.path.dirname(path)].append(path)

Answer 3

另一个选择，获取一组唯一目录，然后使用嵌套列表理解。

from os.path import dirname

unique_dirs = set(dirname(s) for s in files)
output = [[f for f in files if dirname(f) == directory] 
          for directory in unique_dirs]

Answer 4

你可以这样做

>>> sorted_files = [ x for x in sorted(files, key=lambda x: x.count("/"))]
>>> max_dash = max([x.count("/") for x in sorted_files])
>>> max_dash
4
>>> newlist = [[y for y in sorted_files if y.count("/")==x] for x in range(max_dash)]
>>> newlist
[[], ['user/hey.jpg', 'user/1.jpg'], ['user/folder1/1.txt', 'user/folder1/blah.txt'], ['user/folder1/folder2/random.txt']]

如何基于Python中匹配的路径字符串创建列表？

4 个答案: