Separating file extensions using python os.path module

Question

I'm working in python with os.path.splitext() and curious if it is possible to separate filenames from extensions with multiple "."? e.g. "foobar.aux.xml" using splitext. Filenames vary from [foobar, foobar.xml, foobar.aux.xml]. Is there a better way?

Answer 1

Split with os.extsep.

>>> import os
>>> 'filename.ext1.ext2'.split(os.extsep)
['filename', 'ext1', 'ext2']

If you want everything after the first dot:

>>> 'filename.ext1.ext2'.split(os.extsep, 1)
['filename', 'ext1.ext2']

If you are using paths with directories that may contain dots:

>>> def my_splitext(path):
...     """splitext for paths with directories that may contain dots."""
...     li = []
...     path_without_extensions = os.path.join(os.path.dirname(path), os.path.basename(path).split(os.extsep)[0])
...     extensions = os.path.basename(path).split(os.extsep)[1:]
...     li.append(path_without_extensions)
...     # li.append(extensions) if you want extensions in another list inside the list that is returned.
...     li.extend(extensions)
...     return li
... 
>>> my_splitext('/path.with/dots./filename.ext1.ext2')
['/path.with/dots./filename', 'ext1', 'ext2']

Answer 2

you could try with:

names = pathname.split('.')
filename = names[0]
extensions = names[1:]

if you want to use splitext, you can use something like:

import os

path = 'filename.es.txt'

while True:
    path, ext = os.path.splitext(path)
    if not ext:
        print path
        break
    else:
        print ext

produces:

.txt
.es
filename