How do I get the filename without the extension from a path in Python?


How do I get the filename without the extension from a path in Python?

"/path/to/some/file.txt"  →  "file"
Asked By: Joan Venge



Getting the name of the file without the extension:

import os



Documentation for os.path.splitext.

Important Note: If the filename has multiple dots, only the extension after the last one is removed. For example:

import os



See other answers below if you need to handle that case.

Answered By: Geo

But even when I import os, I am not able to call it path.basename. Is it possible to call it as directly as basename?

import os, and then use os.path.basename

importing os doesn’t mean you can use without referring to os.

Answered By: Devin Jeanpierre

You can make your own with:

>>> import os
>>> base=os.path.basename('/root/dir/sub/file.ext')
>>> base
>>> os.path.splitext(base)
('file', '.ext')
>>> os.path.splitext(base)[0]

Important note: If there is more than one . in the filename, only the last one is removed. For example:

/root/dir/sub/ -> file.ext

/root/dir/sub/file.ext.tar.gz -> file.ext.tar

See below for other answers that address that.

Answered By: gimel
>>> print(os.path.splitext(os.path.basename("/path/to/file/hemanth.txt"))[0])
Answered By:

On Windows system I used drivername prefix as well, like:

>>> s = 'c:\temp\akarmi.txt'
>>> print(os.path.splitext(s)[0])

So because I do not need drive letter or directory name, I use:

>>> print(os.path.splitext(os.path.basename(s))[0])
Answered By: Zéiksz

If you want to keep the path to the file and just remove the extension

>>> file = '/root/dir/sub.exten/'
>>> print ('.').join(file.split('.')[:-1])
Answered By: user2902201

We could do some simple split / pop magic as seen here (, to extract the filename (respecting the windows and POSIX differences).

def getFileNameWithoutExtension(path):
  return path.split('\').pop().split('/').pop().rsplit('.', 1)[0]

# => file-0.0.1

# => file-0.0.1
Answered By: yckart
import os
path = "a/b/c/abc.txt"
print os.path.splitext(os.path.basename(path))[0]
Answered By: user4949344

os.path.splitext() won’t work if there are multiple dots in the extension.

For example, images.tar.gz

>>> import os
>>> file_path = '/home/dc/images.tar.gz'
>>> file_name = os.path.basename(file_path)
>>> print os.path.splitext(file_name)[0]

You can just find the index of the first dot in the basename and then slice the basename to get just the filename without extension.

>>> import os
>>> file_path = '/home/dc/images.tar.gz'
>>> file_name = os.path.basename(file_path)
>>> index_of_dot = file_name.index('.')
>>> file_name_without_extension = file_name[:index_of_dot]
>>> print file_name_without_extension

For convenience, a simple function wrapping the two methods from os.path :

def filename(path):
  """Return file name without extension from path.

  import os.path
  b = os.path.split(path)[1]  # path, *filename*
  f = os.path.splitext(b)[0]  # *file*, ext
  #print(path, b, f)
  return f

Tested with Python 3.5.

Answered By: handle

import os

filename = C:\Users\Public\Videos\Sample Videos\wildlife.wmv

This returns the filename without the extension(C:UsersPublicVideosSample Videoswildlife)

temp = os.path.splitext(filename)[0]  

Now you can get just the filename from the temp with

os.path.basename(temp)   #this returns just the filename (wildlife)
Answered By: learncode

As noted by @IceAdor in a comment to @user2902201’s solution, rsplit is the simplest solution robust to multiple periods (by limiting the number of splits to maxsplit of just 1 (from the end of the string)).

Here it is spelt out:

file = ''
print file.rsplit('.', maxsplit=1)[0]

Answered By: dlink
import os
list = []
def getFileName( path ):
for file in os.listdir(path):
    #print file
        ext = os.path.splitext(base)[1]
            newpath = path+"/"+file
            #print path
return list

print list
Answered By: shivendra singh

In Python 3.4+ you can use the pathlib solution

from pathlib import Path

Answered By: Morgoth

A multiple extension aware procedure. Works for str and unicode paths. Works in Python 2 and 3.

import os

def file_base_name(file_name):
    if '.' in file_name:
        separator_index = file_name.index('.')
        base_name = file_name[:separator_index]
        return base_name
        return file_name

def path_base_name(path):
    file_name = os.path.basename(path)
    return file_base_name(file_name)


>>> path_base_name('file')
>>> path_base_name(u'file')
>>> path_base_name('file.txt')
>>> path_base_name(u'file.txt')
>>> path_base_name('file.tar.gz')
>>> path_base_name('file.a.b.c.d.e.f.g')
>>> path_base_name('relative/path/file.ext')
>>> path_base_name('/absolute/path/file.ext')
>>> path_base_name('Relative\Windows\Path\file.txt')
>>> path_base_name('C:\Absolute\Windows\Path\file.txt')
>>> path_base_name('/path with spaces/file.ext')
>>> path_base_name('C:\Windows Path With Spaces\file.txt')
>>> path_base_name('some/path/file name with')
'file name with spaces'
Answered By: user6798019

the easiest way to resolve this is to

import ntpath 
print('Base name is ',ntpath.basename('/path/to/the/file/'))

this saves you time and computation cost.

Use .stem from pathlib in Python 3.4+

from pathlib import Path


will return


Note that if your file has multiple extensions .stem will only remove the last extension. For example, Path('file.tar.gz').stem will return 'file.tar'.

Answered By: mxdbld

Thought I would throw in a variation to the use of the os.path.splitext without the need to use array indexing.

The function always returns a (root, ext) pair so it is safe to use:

root, ext = os.path.splitext(path)


>>> import os
>>> path = 'my_text_file.txt'
>>> root, ext = os.path.splitext(path)
>>> root
>>> ext
Answered By: ScottMcC

In python 3 pathlib "The pathlib module offers high-level path objects."

>>> from pathlib import Path

>>> p = Path("/a/b/c.txt")
>>> p.with_suffix('')
>>> p.stem
Answered By: jjisnow

The other methods don’t remove multiple extensions. Some also have problems with filenames that don’t have extensions. This snippet deals with both instances and works in both Python 2 and 3. It grabs the basename from the path, splits the value on dots, and returns the first one which is the initial part of the filename.

import os

def get_filename_without_extension(file_path):
    file_basename = os.path.basename(file_path)
    filename_without_extension = file_basename.split('.')[0]
    return filename_without_extension

Here’s a set of examples to run:

example_paths = [

for example_path in example_paths:

In every case, the value printed is:

Answered By: Alan W. Smith
import os
filename, file_extension =os.path.splitext(os.path.basename('/d1/d2/example.cs'))
  • filename is ‘example’

  • file_extension is ‘.cs’

Answered By: Antonio Ramasco

Very very very simpely no other modules !!!

import os
p = r"C:UsersbilalDocumentsface Recognition pythonimgsnorthon.jpg"

# Get the filename only from the initial file path.
filename = os.path.basename(p)

# Use splitext() to get filename and extension separately.
(file, ext) = os.path.splitext(filename)

# Print outcome.
print("Filename without extension =", file)
print("Extension =", ext)
Answered By: Bilal

I didn’t look very hard but I didn’t see anyone who used regex for this problem.

I interpreted the question as “given a path, return the basename without the extension.”


"path/to/file.json" => "file"

"path/to/my.file.json" => "my.file"

In Python 2.7, where we still live without pathlib

def get_file_name_prefix(file_path):
    basename = os.path.basename(file_path)

    file_name_prefix_match = re.compile(r"^(?P<file_name_pre fix>.*)..*$").match(basename)

    if file_name_prefix_match is None:
        return file_name
>> file

>> my.file

>> no_extension
Answered By: John Carrell

Answers using Pathlib for Several Scenarios

Using Pathlib, it is trivial to get the filename when there is just one extension (or none), but it can be awkward to handle the general case of multiple extensions.

Zero or One extension

from pathlib import Path

pth = Path('./thefile.tar')

fn = pth.stem

print(fn)      # thefile

# Explanation:
# the `stem` attribute returns only the base filename, stripping
# any leading path if present, and strips the extension after
# the last `.`, if present.

# Further tests

eg_paths = ['thefile',

for p in eg_paths:
    print(Path(p).stem)  # prints thefile every time

Two or fewer extensions

from pathlib import Path

pth = Path('./thefile.tar.gz')

fn = pth.with_suffix('').stem

print(fn)      # thefile

# Explanation:
# Using the `.with_suffix('')` trick returns a Path object after
# stripping one extension, and then we can simply use `.stem`.

# Further tests

eg_paths += ['./thefile.tar.gz',

for p in eg_paths:
    print(Path(p).with_suffix('').stem)  # prints thefile every time

Any number of extensions (0, 1, or more)

from pathlib import Path

pth = Path('./')

fn =
if len(pth.suffixes) > 0:
    s = pth.suffixes[0]
    fn = fn.rsplit(s)[0]

# or, equivalently

fn =
for s in pth.suffixes:
    fn = fn.rsplit(s)[0]

# or simply run the full loop

fn =
for _ in pth.suffixes:
    fn = fn.rsplit('.')[0]

# In any case:

print(fn)     # thefile

# Explanation
#     -> ''
# pth.suffixes -> ['.tar', '.gz', '.bz', '.7zip']
# If there may be more than two extensions, we can test for
# that case with an if statement, or simply attempt the loop
# and break after rsplitting on the first extension instance.
# Alternatively, we may even run the full loop and strip one 
# extension with every pass.

# Further tests

eg_paths += ['./',

for p in eg_paths:
    pth = Path(p)
    fn =
    for s in pth.suffixes:
        fn = fn.rsplit(s)[0]

    print(fn)  # prints thefile every time

Special case in which the first extension is known

For instance, if the extension could be .tar, .tar.gz,, etc; you can simply rsplit the known extension and take the first element:

pth = Path('foo/bar/baz.baz/thefile.tar.gz')

fn ='.tar')[0]

print(fn)      # thefile
Answered By: SpinUp

Improving upon @spinup answer:

fn =
for s in pth.suffixes:
    fn = fn.rsplit(s)[0]
print(fn)      # thefile 

This works for filenames without extension also

Answered By: M Ganesh

I’ve read the answers, and I notice that there are many good solutions.
So, for those who are looking to get either (name or extension), here goes another solution, using the os module, both methods support files with multiple extensions.

import os

def get_file_name(path):
    if not os.path.isdir(path):
        return os.path.splitext(os.path.basename(path))[0].split(".")[0]

def get_file_extension(path):
    extensions = []
    copy_path = path
    while True:
        copy_path, result = os.path.splitext(copy_path)
        if result != '':
    return "".join(extensions)

Note: this solution on windows does not support file names with the "" character

Answered By: esteban21
# use pathlib. the below works with compound filetypes and normal ones
source_file = ''
source_path = pathlib.Path(source_file)''.join(source_path.suffixes), '')
>>> 'spaces'

despite the many working implementations described above I added this ^ as it uses pathlib only and works for compound filetypes and normal ones

Answered By: John—

What about the following?

import pathlib
filename = '/path/to/dir/stem.ext.tar.gz'
# -> 'stem'

or this equivalent?

pathlib.Path(filename).name[:-sum(map(len, pathlib.Path(filename).suffixes))]
Answered By: wolfrevo

Using pathlib.Path.stem is the right way to go, but here is an ugly solution that is way more efficient than the pathlib based approach.

You have a filepath whose fields are separated by a forward slash /, slashes cannot be present in filenames, so you split the filepath by /, the last field is the filename.

The extension is always the last element of the list created by splitting the filename by dot ., so if you reverse the filename and split by dot once, the reverse of the second element is the file name without extension.

name = path.split('/')[-1][::-1].split('.', 1)[1][::-1]


Python 3.9.10 (tags/v3.9.10:f2f3f53, Jan 17 2022, 15:14:21) [MSC v.1929 64 bit (AMD64)]
Type 'copyright', 'credits' or 'license' for more information
IPython 7.28.0 -- An enhanced Interactive Python. Type '?' for help.

In [1]: from pathlib import Path

In [2]: file = 'D:/ffmpeg/ffmpeg.exe'

In [3]: Path(file).stem
Out[3]: 'ffmpeg'

In [4]: file.split('/')[-1][::-1].split('.', 1)[1][::-1]
Out[4]: 'ffmpeg'

In [5]: %timeit Path(file).stem
6.15 µs ± 433 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [6]: %timeit file.split('/')[-1][::-1].split('.', 1)[1][::-1]
671 ns ± 37.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [7]:
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