Extracting extension from filename in Python

Question

Is there a function to extract the extension from a filename?

Answer 1

name_only=file_name[:filename.index(".")

That will give you the file name up to the first ".", which would be the most common.

Answer 2

def NewFileName(fichier):
    cpt = 0
    fic , *ext =  fichier.split('.')
    ext = '.'.join(ext)
    while os.path.isfile(fichier):
        cpt += 1
        fichier = '{0}-({1}).{2}'.format(fic, cpt, ext)
    return fichier

Answer 3

You can find some great stuff in pathlib module.

import pathlib
x = pathlib.PurePosixPath("C:\\Path\\To\\File\\myfile.txt").suffix
print(x)

# Output 
'.txt'

Answer 4

Although it is an old topic, but i wonder why there is none mentioning a very simple api of python called rpartition in this case:

to get extension of a given file absolute path, you can simply type:

filepath.rpartition('.')[-1]

example:

path = '/home/jersey/remote/data/test.csv'
print path.rpartition('.')[-1]

will give you: 'csv'

Answer 5

Just join all pathlib suffixes.

>>> x = 'file/path/archive.tar.gz'
>>> y = 'file/path/text.txt'
>>> ''.join(pathlib.Path(x).suffixes)
'.tar.gz'
>>> ''.join(pathlib.Path(y).suffixes)
'.txt'

Answer 6

One option may be splitting from dot:

>>> filename = "example.jpeg"
>>> filename.split(".")[-1]
'jpeg'

No error when file doesn't have an extension:

>>> "filename".split(".")[-1]
'filename'

But you must be careful:

>>> "png".split(".")[-1]
'png'    # But file doesn't have an extension