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 "
@IceAdor's refers to rsplit in a comment to @user2902201's solution. rsplit is the simplest solution that supports multiple periods.
Here it is spelt out:
file = 'my.report.txt'
print file.rsplit('.', 1)[0]
my.report
On Windows system I used drivername prefix as well, like:
>>> s = 'c:\\temp\\akarmi.txt'
>>> print(os.path.splitext(s)[0])
c:\temp\akarmi
So because I do not need drive letter or directory name, I use:
>>> print(os.path.splitext(os.path.basename(s))[0])
akarmi
import os
path = "a/b/c/abc.txt"
print os.path.splitext(os.path.basename(path))[0]
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.
os.path.splitext() won't work if there are multiple dots in the extension.
For example, images.tar.gz
>>> import os
>>> file_path = '/home/dc/images.tar.gz'
>>> file_name = os.path.basename(file_path)
>>> print os.path.splitext(file_name)[0]
images.tar
You can just find the index of the first dot in the basename and then slice the basename to get just the filename without extension.
>>> import os
>>> file_path = '/home/dc/images.tar.gz'
>>> file_name = os.path.basename(file_path)
>>> index_of_dot = file_name.index('.')
>>> file_name_without_extension = file_name[:index_of_dot]
>>> print file_name_without_extension
images
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.