How to get the filename without the extension from a path in Python? How to get the filename without the extension from a path in Python? python python

How to get the filename without the extension from a path in Python?


Getting the name of the file without the extension:

import osprint(os.path.splitext("/path/to/some/file.txt")[0])

Prints:

/path/to/some/file

Documentation for os.path.splitext.

Important Note: If the filename has multiple dots, only the extension after the last one is removed. For example:

import osprint(os.path.splitext("/path/to/some/file.txt.zip.asc")[0])

Prints:

/path/to/some/file.txt.zip

See other answers below if you need to handle that case.


Use .stem from pathlib in Python 3.4+

from pathlib import PathPath('/root/dir/sub/file.ext').stem

will return

'file'

Note that if your file has multiple extensions .stem will only remove the last extension. For example, Path('file.tar.gz').stem will return 'file.tar'.


You can make your own with:

>>> import os>>> base=os.path.basename('/root/dir/sub/file.ext')>>> base'file.ext'>>> os.path.splitext(base)('file', '.ext')>>> os.path.splitext(base)[0]'file'

Important note: If there is more than one . in the filename, only the last one is removed. For example:

/root/dir/sub/file.ext.zip -> file.ext/root/dir/sub/file.ext.tar.gz -> file.ext.tar

See below for other answers that address that.


matomo