How to get only the name of the path with python? [duplicate]

Question

In Python 3.4+, you can use the pathlib module (included in Python's standard library):

>>> from pathlib import Path
>>> p = Path("/home/user/Downloads/repo/test.txt")
>>> print(p.stem)
test
>>> print(p.name)
test.txt

Use the os.path module to work with paths; the os.path.basename() function gives you the last part after the last path separator, and os.path.splitext() gives you the filename with the extension split off:

import os.path

basename = os.path.splitext(os.path.basename(f.name))[0]

Using the os.path functions ensures that your code will continue to work correctly on different operating systems, even if the path separators are different.

In Python 3.4 or newer (or as a separate backport install), you can also use the pathlib library, which offers a more object-oriented approach to path handling. pathlib.Path() objects have a .stem attribute, which is the final component without the extension suffix:

try:
    import pathlib
except ImportError:
    # older Python version, import the backport instead
    import pathlib2 as pathlib

basename = pathlib.Path(f.name).stem

Demo:

>>> import os.path
>>> a = "/home/user/Downloads/repo/test.txt"
>>> os.path.basename(a)
'test.txt'
>>> os.path.splitext(os.path.basename(a))
('test', '.txt')
>>> os.path.splitext(os.path.basename(a))[0]
'test'
>>> import pathlib
>>> pathlib.Path(a)
PosixPath('/home/user/Downloads/repo/test.txt')
>>> pathlib.Path(a).stem
'test'

It seems that you're either looking for os.path.basename or os.path.splitext:

>>> import os.path
>>> os.path.basename("/var/log/err.log")
'err.log'
>>> os.path.splitext(os.path.basename("/var/log/err.log"))
('err', '.log')
>>> os.path.splitext(os.path.basename("/var/log/err.log"))[0]
'err'
>>>