How to get only the last part of a path in Python?

Question

In python, suppose I have a path like this:

/folderA/folderB/folderC/folderD/

How can I get just the folderD part?

Hieronymus

path.split('/')[len(path.split('/'))-1:][0] – Sanatorium

Answer 1

Use os.path.normpath to strip off any trailing slashes, then os.path.basename gives you the last part of the path:

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

Using only basename gives everything after the last slash, which in this case is ''.

Biparty

I initially thought rstrip('/') would be simpler but then quickly realised I'd have to use rstrip(os.path.sep), so obviously the use of normpath is justified. – Ken

This doesn't seem to work on Windows long paths, e.g., '\\\\?\\D:\\A\\B\\C\\' and '\\\\?\\UNC\\svr\\B\\C\\' (returns an empty string) This solution works for all cases. – Imidazole

@jinnlao's answer is much much better nowadays. – Clarenceclarenceux

I'd add str.strip around it, too, because based on input, you might have whitespaces, but that might just be me – Airel

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'

Rigby

well, just to point it out PurePath vs Path, the first provides "purely computational operations" (no I/O), while the other ("concrete path") inherits from the first but also provides I/O operations. – Alberich

@Alberich To expand on this, this answer applies to non-pure paths as well – Dustup

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'

Woodchopper

In Python-think, os.path.basename(".../") correctly yields ''. Yes, I, too, find that sub-optimal. The ...basename(...normpath... solution below is canonical, though. – Snowinsummer

@lars yeah! just saw that in that case normalize the path first before feeding it to basename. os.path.basename(os.path.normpath('/folderA/folderB/folderC/folderD/')) – Woodchopper

UPDATE2 is the best approach I have found so far. – Lamonica

thumbs up for the UPDATE2 – Aretta

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

Renz

What does your approach solve different/better than the aforementioned? – Nurse

@user1767754: though terse, this does illustrate a solution that works with our without a file name, while also showing the pitfall if a final directory does not have a trailing / (and without needing purepath) – Marketa

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)

Nurse

You can also use os.path.basename(os.path.dirname(pathname)) which would give folderD in your example. See answer from @Mike Miterer – Marketa

Answer 6

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.name
'folderD'

Quarry

In your 2nd example stem will only return file not file.txt. If you want file.txt you need your_path.name. Pathlib Docs – Industrialize

Should be your_path.parent.name that gives "folderD" – Levin

Actually, it was the result which was wrong since it was meant as an example of what the output would look like, but for clarity's sake, I like your suggestion better. – Quarry

Answer 7

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}')

Corabella

My favorite, even cleaner: '/'.join(Path(export_filename).parts[-3:]) – Tadich

Answer 8

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 Pathobject which can easily be converted to a string (e.g. str(path))

Urbanize

Answer 9

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

Flamen

Seriously, use the os.path module. – Brilliantine

Do people try their answers before posting? – Aretta

Answer 10

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

Edelmiraedelson

he wants folderD. not folderC – Woodchopper

It gives "folderD" because the trailing slash makes the final item in the list be "" – Secrete

Seriously, use the os.path module. – Brilliantine