Question

我有一个路径列表，我需要以便携方式提取每个路径的第一个元素，我该怎么做？

['/abs/path/foo',
 'rel/path',
 'just-a-file']

到

['abs', 'rel', 'just-a-file']

提前致谢奥利

Answer 1

In [69]: import os

In [70]: paths
Out[70]: ['/abs/path/foo', 'rel/path', 'just-a-file']

In [71]: [next(part for part in path.split(os.path.sep) if part) for path in paths]
Out[71]: ['abs', 'rel', 'just-a-file']

Answer 2

有一个库调用以独立于平台的方式处理拆分路径，但它只分为两部分：

import os.path

def paths(p) :
  head,tail = os.path.split(p)
  components = []
  while len(tail)>0:
    components.insert(0,tail)
    head,tail = os.path.split(head)
  return components

for p in ['/abs/path/foo','rel/path','just-a-file'] :
  print paths(p)[0]

Answer 3

使用更新的pathlib方法...

import PurePath from pathlib
import os

# Separates the paths into parts and prints to the console...
def print_path_parts(path: str):

    path = PurePath(path)
    parts = list(path.parts)

    # From your description, looks like you don't want the root.
    # Pop it off.
    if parts[0] == os.sep:
        parts.pop(0)

    print(parts)

# Array of path strings...
paths = ['/abs/path/foo',
         'rel/path',
         'just-a-file']

# For each path, print parts to the console
for path in paths:
    print_path_parts(path)

输出：

['abs', 'path', 'foo']
['rel', 'path']
['just-a-file']

Answer 4

为什么不使用正则表达式？

>>> import re
>>> paths = ['/abs/path/foo',
...          'rel/path',
...          'just-a-file']
>>> 
>>> [re.match(r'\/?([^\/]+)', p).groups()[0] for p in paths]
['abs', 'rel', 'just-a-file']

和Windows：

>>> paths = [r'\abs\path\foo',
...          r'rel\path',
...          r'just-a-file',
...          r'C:\abs\path\foo',
...          r'C:rel\path',
...          r'C:just-a-file']
>>> 
>>> [re.match(r'(?:[A-Za-z]:)?\\?([^\\]+)', p).groups()[0] for p in paths]
['abs', 'rel', 'just-a-file', 'abs', 'rel', 'just-a-file']

python中路径的第一个元素

4 个答案: