如何在 Python 中只获取路径的最后一部分？

Question

In python , suppose I have a path like this:

/folderA/folderB/folderC/folderD/

How can I get just the folderD part?

Answer 1

Use os.path.normpath , then os.path.basename :

>>> os.path.basename(os.path.normpath('/folderA/folderB/folderC/folderD/'))
'folderD'

The first strips off any trailing slashes, the second gives you the last part of the path. Using only basename gives everything after the last slash, which in this case is '' .

Answer 2

With python 3 you can use the pathlib module (pathlib.PurePath for example):

>>> import pathlib

>>> path = pathlib.PurePath('/folderA/folderB/folderC/folderD/')
>>> path.name
'folderD'

If you want the last folder name where a file is located:

>>> path = pathlib.PurePath('/folderA/folderB/folderC/folderD/file.py')
>>> path.parent.name
'folderD'

Answer 3

You could do

>>> import os
>>> os.path.basename('/folderA/folderB/folderC/folderD')

UPDATE1: This approach works in case you give it /folderA/folderB/folderC/folderD/xx.py. This gives xx.py as the basename. Which is not what you want I guess. So you could do this -

>>> import os
>>> path = "/folderA/folderB/folderC/folderD"
>>> if os.path.isdir(path):
        dirname = os.path.basename(path)

UPDATE2: As lars pointed out, making changes so as to accomodate trailing '/'.

>>> from os.path import normpath, basename
>>> basename(normpath('/folderA/folderB/folderC/folderD/'))
'folderD'

Answer 4

Here is my approach:

>>> import os
>>> print os.path.basename(
        os.path.dirname('/folderA/folderB/folderC/folderD/test.py'))
folderD
>>> print os.path.basename(
        os.path.dirname('/folderA/folderB/folderC/folderD/'))
folderD
>>> print os.path.basename(
        os.path.dirname('/folderA/folderB/folderC/folderD'))
folderC

Answer 5

I was searching for a solution to get the last foldername where the file is located, I just used split two times, to get the right part. It's not the question but google transfered me here.

pathname = "/folderA/folderB/folderC/folderD/filename.py"
head, tail = os.path.split(os.path.split(pathname)[0])
print(head + "   "  + tail)

Answer 6

I like the parts method of Path for this:

grandparent_directory, parent_directory, filename = Path(export_filename).parts[-3:]
log.info(f'{t: <30}: {num_rows: >7} Rows exported to {grandparent_directory}/{parent_directory}/{filename}')

Answer 7

If you use the native python package pathlib it's really simple.

>>> from pathlib import Path
>>> your_path = Path("/folderA/folderB/folderC/folderD/")
>>> your_path.stem
'folderD'

Suppose you have the path to a file in folderD.

>>> from pathlib import Path
>>> your_path = Path("/folderA/folderB/folderC/folderD/file.txt")
>>> your_path.name
'file.txt'
>>> your_path.parent
'folderD'

Answer 8

A naive solution(Python 2.5.2+):

s="/path/to/any/folder/orfile"
desired_dir_or_file = s[s.rindex('/',0,-1)+1:-1] if s.endswith('/') else s[s.rindex('/')+1:]

Answer 9

During my current projects, I'm often passing rear parts of a path to a function and therefore use the Path module. To get the n-th part in reverse order, I'm using:

from typing import Union
from pathlib import Path

def get_single_subpath_part(base_dir: Union[Path, str], n:int) -> str:
    if n ==0:
        return Path(base_dir).name
    for _ in range(n):
        base_dir = Path(base_dir).parent
    return getattr(base_dir, "name")

path= "/folderA/folderB/folderC/folderD/"

# for getting the last part:
print(get_single_subpath_part(path, 0))
# yields "folderD"

# for the second last
print(get_single_subpath_part(path, 1))
#yields "folderC"

Furthermore, to pass the n-th part in reverse order of a path containing the remaining path, I use:

from typing import Union
from pathlib import Path

def get_n_last_subparts_path(base_dir: Union[Path, str], n:int) -> Path:
    return Path(*Path(base_dir).parts[-n-1:])

path= "/folderA/folderB/folderC/folderD/"

# for getting the last part:
print(get_n_last_subparts_path(path, 0))
# yields a `Path` object of "folderD"

# for second last and last part together 
print(get_n_last_subparts_path(path, 1))
# yields a `Path` object of "folderc/folderD"

Note that this function returns a Path object which can easily be converted to a string (eg str(path) )

Answer 10

path = "/folderA/folderB/folderC/folderD/"
last = path.split('/').pop()

Answer 11

str = "/folderA/folderB/folderC/folderD/"
print str.split("/")[-2]