Extracting extension from filename in Python

Question

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

Answer 1

Yes. Use os.path.splitext(see Python 2.X documentation or Python 3.X documentation):

>>> import os
>>> filename, file_extension = os.path.splitext('/path/to/somefile.ext')
>>> filename
'/path/to/somefile'
>>> file_extension
'.ext'

Unlike most manual string-splitting attempts, os.path.splitext will correctly treat /a/b.c/d as having no extension instead of having extension .c/d, and it will treat .bashrc as having no extension instead of having extension .bashrc:

>>> os.path.splitext('/a/b.c/d')
('/a/b.c/d', '')
>>> os.path.splitext('.bashrc')
('.bashrc', '')

Answer 2

import os.path
extension = os.path.splitext(filename)[1]

Answer 3

New in version 3.4.

import pathlib

print(pathlib.Path('yourPathGoesHere').suffix)

I'm surprised no one has mentioned pathlib yet, pathlib IS awesome!

If you need all the suffixes (eg if you have a .tar.gz), .suffixes will return a list of them!

Answer 4

import os.path
extension = os.path.splitext(filename)[1][1:]

To get only the text of the extension, without the dot.

Answer 5

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

Answer 6

worth adding a lower in there so you don't find yourself wondering why the JPG's aren't showing up in your list.

os.path.splitext(filename)[1][1:].strip().lower()

Answer 7

Any of the solutions above work, but on linux I have found that there is a newline at the end of the extension string which will prevent matches from succeeding. Add the strip() method to the end. For example:

import os.path
extension = os.path.splitext(filename)[1][1:].strip() 

Answer 8

With splitext there are problems with files with double extension (e.g. file.tar.gz, file.tar.bz2, etc..)

>>> fileName, fileExtension = os.path.splitext('/path/to/somefile.tar.gz')
>>> fileExtension 
'.gz'

but should be: .tar.gz

The possible solutions are here

Answer 9

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 10

filename='ext.tar.gz'
extension = filename[filename.rfind('.'):]

Answer 11

Surprised this wasn't mentioned yet:

import os
fn = '/some/path/a.tar.gz'

basename = os.path.basename(fn)  # os independent
Out[] a.tar.gz

base = basename.split('.')[0]
Out[] a

ext = '.'.join(basename.split('.')[1:])   # <-- main part

# if you want a leading '.', and if no result `None`:
ext = '.' + ext if ext else None
Out[] .tar.gz

Benefits:

• Works as expected for anything I can think of
• No modules
• No regex
• Cross-platform
• Easily extendible (e.g. no leading dots for extension, only last part of extension)

As function:

def get_extension(filename):
    basename = os.path.basename(filename)  # os independent
    ext = '.'.join(basename.split('.')[1:])
    return '.' + ext if ext else None

Answer 12

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 13

You can use a split on a filename:

f_extns = filename.split(".")
print ("The extension of the file is : " + repr(f_extns[-1]))

This does not require additional library

Answer 14

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 15

This is a direct string representation techniques : I see a lot of solutions mentioned, but I think most are looking at split. Split however does it at every occurrence of "." . What you would rather be looking for is partition.

string = "folder/to_path/filename.ext"
extension = string.rpartition(".")[-1]

Answer 16

Another solution with right split:

# to get extension only

s = 'test.ext'

if '.' in s: ext = s.rsplit('.', 1)[1]

# or, to get file name and extension

def split_filepath(s):
    """
    get filename and extension from filepath 
    filepath -> (filename, extension)
    """
    if not '.' in s: return (s, '')
    r = s.rsplit('.', 1)
    return (r[0], r[1])

Answer 17

Even this question is already answered I'd add the solution in Regex.

>>> import re
>>> file_suffix = ".*(\..*)"
>>> result = re.search(file_suffix, "somefile.ext")
>>> result.group(1)
'.ext'

Answer 18

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 19

# try this, it works for anything, any length of extension
# e.g www.google.com/downloads/file1.gz.rs -> .gz.rs

import os.path

class LinkChecker:

    @staticmethod
    def get_link_extension(link: str)->str:
        if link is None or link == "":
            return ""
        else:
            paths = os.path.splitext(link)
            ext = paths[1]
            new_link = paths[0]
            if ext != "":
                return LinkChecker.get_link_extension(new_link) + ext
            else:
                return ""

Answer 20

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

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