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?

For instance, if I had "/path/to/some/file.txt", I would want "file".

Solution 1:

Getting the name of the file without the extension:

import os
print(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 os
print(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.

Solution 2:

Use .stem from pathlib in Python 3.4+

from pathlib import Path

Path('/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'.

Solution 3:

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.

Solution 4:

>>> print(os.path.splitext(os.path.basename("/path/to/file/hemanth.txt"))[0])
hemanth