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

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Use os.path.splitext:

>>> 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', '')
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11 Comments

the use of basename is a little confusing here since os.path.basename("/path/to/somefile.ext") would return "somefile.ext"
wouldn't endswith() not be more portable and pythonic?
You can't rely on that if you have files with "double extensions", like .mp3.asd for example, because it will return you only the "last" extension!
@klingt.net Well, in that case, .asd is really the extension!! If you think about it, foo.tar.gz is a gzip-compressed file (.gz) which happens to be a tar file (.tar). But it is a gzip file in first place. I wouldn't expect it to return the dual extension at all.
The standard Python function naming convention is really annoying - almost every time I re-look this up, I mistake it as being splittext. If they would just do anything to signify the break between parts of this name, it'd be much easier to recognize that it's splitExt or split_ext. Surely I can't be the only person who has made this mistake?
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New in version 3.4.

import pathlib print(pathlib.Path('/foo/bar.txt').suffix) # Outputs: .txt print(pathlib.Path('/foo/bar.txt').stem) # Outputs: bar print(pathlib.Path("hello/foo.bar.tar.gz").suffixes) # Outputs: ['.bar', '.tar', '.gz'] print(''.join(pathlib.Path("hello/foo.bar.tar.gz").suffixes)) # Outputs: .bar.tar.gz print(pathlib.Path("hello/foo.bar.tar.gz").stem) # Outputs: foo.bar.tar

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

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5 Comments

example for getting .tar.gz: ''.join(pathlib.Path('somedir/file.tar.gz').suffixes)
Great answer. I found this tutorial more useful than the documentation: zetcode.com/python/pathlib
@user3780389 Wouldn't a "foo.bar.tar.gz" still be a valid ".tar.gz"? If so your snippet should be using .suffixes[-2:] to ensure only getting .tar.gz at most.
there are still cases when this does not work as expected like "filename with.a dot inside.tar". This is the solution i am using currently: "".join([s for s in pathlib.Path('somedir/file.tar.gz').suffixes if not " " in s])
this one should be acceptable answer
501 
import os.path extension = os.path.splitext(filename)[1]
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10 Comments

Out of curiosity, why import os.path instead of from os import path?
@kiswa - I suppose you could do it that way. I've seen more code using import os.path though.
it depends really, if you use from os import path then the name path is taken up in your local scope, also others looking at the code may not immediately know that path is the path from the os module. Where as if you use import os.path it keeps it within the os namespace and wherever you make the call people know it's path() from the os module immediately.
I know it's not semantically any different, but I personally find the construction _, extension = os.path.splitext(filename) to be much nicer-looking.
If you want the extension as part of a more complex expression the [1] may be more useful: if check_for_gzip and os.path.splitext(filename)[1] == '.gz':
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import os.path extension = os.path.splitext(filename)[1][1:]

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

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1 Comment

This will return empty for both file names end with . and file names without an extension.
116

For simple use cases 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

Also will not work with hidden files in Unix systems:

>>> ".bashrc".split(".")[-1] 'bashrc' # But this is not an extension

For general use, prefer os.path.splitext

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13 Comments

This would get upset if you're uploading x.tar.gz
Not actually. Extension of a file named "x.tar.gz" is "gz" not "tar.gz". os.path.splitext gives ".os" as extension too.
can we use [1] rather than [-1]. I could not understand [-1] with split
[-1] to get last item of items that splitted by dot. Example: "my.file.name.js".split('.') => ['my','file','name','js]
@BenjaminR ah ok, you are making an optimisation about result list. ['file', 'tar', 'gz'] with 'file.tar.gz'.split('.') vs ['file.tar', 'gz'] with 'file.tar.gz'.rsplit('.', 1). yeah, could be.
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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()
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2 Comments

The strip() will break in rare edge-cases where the filename extension includes whitespace.
Some filesystems are case-sensitive (like the ones on Linux), and even NTFS is case-sensitive, although Windows tries to treat it in a case-insensitive manner. Be careful with case.
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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

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6 Comments

do it twice to get the 2 extensions ?
@maazza yep. gunzip somefile.tar.gz what's the output filename?
This is why we have the extension 'tgz' which means: tar+gzip ! :D
@FlipMcF The filename should obviously be somefile.tar. For tar -xzvf somefile.tar.gz the filename should be somefile.
@peterhil I don't think you want your python script to be aware of the application used to create the filename. It's a bit out of scope of the question. Don't pick on the example, 'filename.csv.gz' is also quite valid.
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You can find some great stuff in pathlib module (available in python 3.x).

import pathlib x = pathlib.PurePosixPath("C:\\Path\\To\\File\\myfile.txt").suffix print(x) # Output '.txt'
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Using PosixPath for a windows path is wrong.
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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()
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3 Comments

To aid my understanding, please could you explain what additional behaviour the second index/slice guards against? (i.e. the [1:] in .splittext(filename)[1][1:]) - thank you in advance
Figured it out for myself: splittext() (unlike if you split a string using '.') includes the '.' character in the extension. The additional [1:] gets rid of it.
This will break if the file extension contains whitespace. That's a rare case, I know, but it's still an edge-case that should be considered.
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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'
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17

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'

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For those not familiar with the API, rpartition returns a tuple: ("string before the right-most occurrence of the separator", "the separator itself", "the rest of the string"). If there's no separator found, the returned tuple will be: ("", "", "the original string").
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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
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4 Comments

This results in an exception when the file doesn't have any extension.
This answer absolutely ignore a variant if a filename contains many points in name. Example get_extension('cmocka-1.1.0.tar.xz') => '.1.0.tar.xz' - wrong.
@PADYMKO, IMHO one should not create filenames with full stops as part of the filename. The code above is not supposed to result in 'tar.xz'
Just change to [-1] then.
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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

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12

Extracting extension from filename in Python

Python os module splitext()

splitext() function splits the file path into a tuple having two values – root and extension.

import os # unpacking the tuple file_name, file_extension = os.path.splitext("/Users/Username/abc.txt") print(file_name) print(file_extension)

Get File Extension using Pathlib Module

Pathlib module to get the file extension

import pathlib pathlib.Path("/Users/pankaj/abc.txt").suffix #output:'.txt'
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11 
filename='ext.tar.gz' extension = filename[filename.rfind('.'):]
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1 Comment

This results in the last char of filename being returned if the filename has no . at all. This is because rfind returns -1 if the string is not found.
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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'
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1 Comment

Or \.[0-9a-z]+$ as in this post.
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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]
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1 Comment

rpartition was already suggested by @weiyixie.
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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])
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5

you can use following code to split file name and extension.

 import os.path filenamewithext = os.path.basename(filepath) filename, ext = os.path.splitext(filenamewithext) #print file name print(filename) #print file extension print(ext)
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4

Well , i know im late

that's my simple solution

file = '/foo/bar/whatever.ext' extension = file.split('.')[-1] print(extension) #output will be ext
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@NsaNinja but the malware.pdf.exe is [exe] type ! also for tar.gz !
I agree that there are drawbacks for completeness, HOWEVER, this is a "simple" solution and for simple uses it works. In my case, for example, I've already confirmed that the file exists and is one of several filtered file types. I just need to know which one. For that application, this works.
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You can use endswith to identify the file extension in python

like bellow example

for file in os.listdir(): if file.endswith('.csv'): df1 =pd.read_csv(file) frames.append(df1) result = pd.concat(frames)
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If you are working with a DirEntry, which is what os.scandir returns, you can call file.name.endswith("ext") instead.
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A true one-liner, if you like regex. And it doesn't matter even if you have additional "." in the middle

import re file_ext = re.search(r"\.([^.]+)$", filename).group(1)

See here for the result: Click Here

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1

try this:

files = ['file.jpeg','file.tar.gz','file.png','file.foo.bar','file.etc'] pen_ext = ['foo', 'tar', 'bar', 'etc'] for file in files: #1 if (file.split(".")[-2] in pen_ext): #2 ext = file.split(".")[-2]+"."+file.split(".")[-1]#3 else: ext = file.split(".")[-1] #4 print (ext) #5
  1. get all file name inside the list
  2. splitting file name and check the penultimate extension, is it in the pen_ext list or not?
  3. if yes then join it with the last extension and set it as the file's extension
  4. if not then just put the last extension as the file's extension
  5. and then check it out
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5 Comments

This breaks for a bunch of special cases. See the accepted answer. It's reinventing the wheel, only in a buggy way.
Hello! While this code may solve the question, including an explanation of how and why this solves the problem would really help to improve the quality of your post, and probably result in more up-votes. Remember that you are answering the question for readers in the future, not just the person asking now. Please edit your answer to add explanations and give an indication of what limitations and assumptions apply.
@Brian like that?
You're only making it worse, breaking it in new ways. foo.tar is a valid file name. What happens if I throw that at your code? What about .bashrc or foo? There is a library function for this for a reason...
just create a list of extension file for the penultimate extension, if not in list then just put the last extension as the file's extension
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This method will require a dictonary, list, or set. you can just use ".endswith" using built in string methods. This will search for name in list at end of file and can be done with just str.endswith(fileName[index]). This is more for getting and comparing extensions.

https://docs.python.org/3/library/stdtypes.html#string-methods

Example 1:

dictonary = {0:".tar.gz", 1:".txt", 2:".exe", 3:".js", 4:".java", 5:".python", 6:".ruby",7:".c", 8:".bash", 9:".ps1", 10:".html", 11:".html5", 12:".css", 13:".json", 14:".abc"} for x in dictonary.values(): str = "file" + x str.endswith(x, str.index("."), len(str))

Example 2:

set1 = {".tar.gz", ".txt", ".exe", ".js", ".java", ".python", ".ruby", ".c", ".bash", ".ps1", ".html", ".html5", ".css", ".json", ".abc"} for x in set1: str = "file" + x str.endswith(x, str.index("."), len(str))

Example 3:

fileName = [".tar.gz", ".txt", ".exe", ".js", ".java", ".python", ".ruby", ".c", ".bash", ".ps1", ".html", ".html5", ".css", ".json", ".abc"]; for x in range(0, len(fileName)): str = "file" + fileName[x] str.endswith(fileName[x], str.index("."), len(str))

Example 4

fileName = [".tar.gz", ".txt", ".exe", ".js", ".java", ".python", ".ruby", ".c", ".bash", ".ps1", ".html", ".html5", ".css", ".json", ".abc"]; str = "file.txt" str.endswith(fileName[1], str.index("."), len(str))

Examples 5, 6, 7 with output enter image description here

Example 8

fileName = [".tar.gz", ".txt", ".exe", ".js", ".java", ".python", ".ruby", ".c", ".bash", ".ps1", ".html", ".html5", ".css", ".json", ".abc"]; exts = [] str = "file.txt" for x in range(0, len(x)): if str.endswith(fileName[1]) == 1: exts += [x]
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1

The easiest way to get is to use mimtypes, below is the example:

import mimetypes mt = mimetypes.guess_type("file name") file_extension = mt[0] print(file_extension)
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1

I'm definitely late to the party, but in case anyone wanted to achieve this without the use of another library:

file_path = "example_tar.tar.gz" file_name, file_ext = [file_path if "." not in file_path else file_path.split(".")[0], "" if "." not in file_path else file_path[file_path.find(".") + 1:]] print(file_name, file_ext)

The 2nd line is basically just the following code but crammed into one line:

def name_and_ext(file_path): if "." not in file_path: file_name = file_path else: file_name = file_path.split(".")[0] if "." not in file_path: file_ext = "" else: file_ext = file_path[file_path.find(".") + 1:] return [file_name, file_ext]

Even though this works, it might not work will all types of files, specifically .zshrc, I would recomment using os's os.path.splitext function, example below:

import os file_path = "example.tar.gz" file_name, file_ext = os.path.splitext(file_path) print(file_name, file_ext)

Cheers :)

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0

For funsies... just collect the extensions in a dict, and track all of them in a folder. Then just pull the extensions you want. 

import os search = {} for f in os.listdir(os.getcwd()): fn, fe = os.path.splitext(f) try: search[fe].append(f) except: search[fe]=[f,] extensions = ('.png','.jpg') for ex in extensions: found = search.get(ex,'') if found: print(found)
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1 Comment

That's a terrible idea. Your code breaks for any file extension you haven't previously added!
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Here if you want to extract the last file extension if it has multiple

class functions: def listdir(self, filepath): return os.listdir(filepath) func = functions() os.chdir("C:\\Users\Asus-pc\Downloads") #absolute path, change this to your directory current_dir = os.getcwd() for i in range(len(func.listdir(current_dir))): #i is set to numbers of files and directories on path directory if os.path.isfile((func.listdir(current_dir))[i]): #check if it is a file fileName = func.listdir(current_dir)[i] #put the current filename into a variable rev_fileName = fileName[::-1] #reverse the filename currentFileExtension = rev_fileName[:rev_fileName.index('.')][::-1] #extract from beginning until before . print(currentFileExtension) #output can be mp3,pdf,ini,exe, depends on the file on your absolute directory

Output is mp3, even works if has only 1 extension name

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-2 
# 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 ""
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-2 
a = ".bashrc" b = "text.txt" extension_a = a.split(".") extension_b = b.split(".") print(extension_a[-1]) # bashrc print(extension_b[-1]) # txt
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1 Comment

Please add explanation of the code, rather than simply just the code snippets.
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