如何在lis中按路径对文件进行分组

Question

从这个字符串列表中可以得到：

['/User/someuser/file.py', '/tmp/download/file.zip', '/tmp/download/file2.zip', '/', '/usr/local/bin', '/tmp/file.txt']

列这个清单

[('/tmp/download', ('file.zip', 'file2.zip')), ('/tmp', ('file.txt',)),
 ('/User/someuser', ('file.py',)), ('/', ()), ('/usr/local/bin', ())]

Answer 1

使用^{}将路径名拆分为dirname和basename。然后使用itertools.groupby将文件分组在一起。你知道吗

import os
from itertools import groupby


l= ['/User/someuser/file.py', '/tmp/download/file.zip', '/tmp/download/file2.zip', '/', '/usr/local/bin', '/tmp/file.txt']

print([(k, tuple(i[1] for i in g)) for k, g in groupby(map(os.path.split, sorted(l)), key=lambda x: x[0])])

印刷品

[('/', ('',)), ('/User/someuser', ('file.py',)), ('/tmp/download', ('file.zip', 'file2.zip')), ('/tmp', ('file.txt',)), ('/usr/local', ('bin',))]