How to get the filename without the extension from a path in Python?
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.
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'
.
>>> print(os.path.splitext(os.path.basename("/path/to/file/hemanth.txt"))[0])
hemanth
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.