Extracting extension from filename in Python
Question:
Is there a function to extract the extension from a filename?
Answers:
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', '')
import os.path
extension = os.path.splitext(filename)[1]
import os.path
extension = os.path.splitext(filename)[1][1:]
To get only the text of the extension, without the dot.
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()
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
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()
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
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])
filename='ext.tar.gz'
extension = filename[filename.rfind('.'):]
name_only=file_name[:filename.index(".")
That will give you the file name up to the first “.”, which would be the most common.
# 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 ""
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
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
New in version 3.4.
import pathlib
print(pathlib.Path('yourPath.example').suffix) # '.example'
print(pathlib.Path("hello/foo.bar.tar.gz").suffixes) # ['.bar', '.tar', '.gz']
I’m surprised no one has mentioned pathlib
yet, pathlib
IS awesome!
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’
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'
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
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]
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'
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'
This is The Simplest Method to get both Filename & Extension in just a single line.
fName, ext = 'C:/folder name/Flower.jpeg'.split('/')[-1].split('.')
>>> print(fName)
Flower
>>> print(ext)
jpeg
Unlike other solutions, you don’t need to import any package for this.
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)
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
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
- get all file name inside the list
- splitting file name and check the penultimate extension, is it in the pen_ext list or not?
- if yes then join it with the last extension and set it as the file’s extension
- if not then just put the last extension as the file’s extension
- and then check it out
a = ".bashrc"
b = "text.txt"
extension_a = a.split(".")
extension_b = b.split(".")
print(extension_a[-1]) # bashrc
print(extension_b[-1]) # txt
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)
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'
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)
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))
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]
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)
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
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:\UsersAsus-pcDownloads") #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
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 🙂
Is there a function to extract the extension from a filename?
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', '')
import os.path
extension = os.path.splitext(filename)[1]
import os.path
extension = os.path.splitext(filename)[1][1:]
To get only the text of the extension, without the dot.
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()
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
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()
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
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])
filename='ext.tar.gz'
extension = filename[filename.rfind('.'):]
name_only=file_name[:filename.index(".")
That will give you the file name up to the first “.”, which would be the most common.
# 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 ""
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
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
New in version 3.4.
import pathlib
print(pathlib.Path('yourPath.example').suffix) # '.example'
print(pathlib.Path("hello/foo.bar.tar.gz").suffixes) # ['.bar', '.tar', '.gz']
I’m surprised no one has mentioned pathlib
yet, pathlib
IS awesome!
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’
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'
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
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]
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'
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'
This is The Simplest Method to get both Filename & Extension in just a single line.
fName, ext = 'C:/folder name/Flower.jpeg'.split('/')[-1].split('.')
>>> print(fName)
Flower
>>> print(ext)
jpeg
Unlike other solutions, you don’t need to import any package for this.
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)
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
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
- get all file name inside the list
- splitting file name and check the penultimate extension, is it in the pen_ext list or not?
- if yes then join it with the last extension and set it as the file’s extension
- if not then just put the last extension as the file’s extension
- and then check it out
a = ".bashrc"
b = "text.txt"
extension_a = a.split(".")
extension_b = b.split(".")
print(extension_a[-1]) # bashrc
print(extension_b[-1]) # txt
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)
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'
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)
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))
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]
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)
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
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:\UsersAsus-pcDownloads") #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
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 🙂