Python中路径的第一个元素

5 投票
4 回答
9065 浏览
提问于 2025-04-16 04:20

我有一堆路径,我想以一种通用的方式提取每个路径的第一个元素,我该怎么做呢?

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

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

提前谢谢你们!

4 个回答

2

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)

输出结果:

['abs', 'path', 'foo']
['rel', 'path']
['just-a-file']
2

有一个库可以处理路径的分割,这个方法在不同的平台上都能用,但它只把路径分成两部分:

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]
7
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']

当然可以!请把你想要翻译的内容发给我,我会帮你用简单易懂的语言解释清楚。

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