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

## Question:

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

"/path/to/some/file.txt"  →  "file"


Getting the name of the file without the extension:

import os
print(os.path.splitext("/path/to/some/file.txt")[0])


Prints:

/path/to/some/file


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

import os
print(os.path.splitext("/path/to/some/file.txt.zip.asc")[0])


Prints:

/path/to/some/file.txt.zip


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

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 os.foo without referring to os.

You can make your own with:

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


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

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

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


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


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

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


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

>>> print(os.path.splitext(os.path.basename(s))[0])
akarmi


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

>>> file = '/root/dir/sub.exten/file.data.1.2.dat'
>>> print ('.').join(file.split('.')[:-1])
/root/dir/sub.exten/file.data.1.2


We could do some simple split / pop magic as seen here (https://stackoverflow.com/a/424006/1250044), to extract the filename (respecting the windows and POSIX differences).

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

getFileNameWithoutExtension('/path/to/file-0.0.1.ext')
# => file-0.0.1

getFileNameWithoutExtension('\path\to\file-0.0.1.ext')
# => file-0.0.1

import os
path = "a/b/c/abc.txt"
print os.path.splitext(os.path.basename(path))[0]


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]
images.tar


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
images


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

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

See https://docs.python.org/3/library/os.path.html
"""
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.

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)


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


my.report

import os
list = []
def getFileName( path ):
for file in os.listdir(path):
#print file
try:
base=os.path.basename(file)
splitbase=os.path.splitext(base)
ext = os.path.splitext(base)[1]
if(ext):
list.append(base)
else:
newpath = path+"/"+file
#print path
getFileName(newpath)
except:
pass
return list

getFileName("/home/weexcel-java3/Desktop/backup")
print list


In Python 3.4+ you can use the pathlib solution

from pathlib import Path

print(Path(your_path).resolve().stem)


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
else:
return file_name

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


Behavior:

>>> path_base_name('file')
'file'
>>> path_base_name(u'file')
u'file'
>>> path_base_name('file.txt')
'file'
>>> path_base_name(u'file.txt')
u'file'
>>> path_base_name('file.tar.gz')
'file'
>>> path_base_name('file.a.b.c.d.e.f.g')
'file'
>>> path_base_name('relative/path/file.ext')
'file'
>>> path_base_name('/absolute/path/file.ext')
'file'
>>> path_base_name('Relative\Windows\Path\file.txt')
'file'
>>> path_base_name('C:\Absolute\Windows\Path\file.txt')
'file'
>>> path_base_name('/path with spaces/file.ext')
'file'
>>> path_base_name('C:\Windows Path With Spaces\file.txt')
'file'
>>> path_base_name('some/path/file name with spaces.tar.gz.zip.rar.7z')
'file name with spaces'


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

Path('/root/dir/sub/file.ext').stem


will return

'file'


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'.

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)

Example:

>>> import os
>>> path = 'my_text_file.txt'
>>> root, ext = os.path.splitext(path)
>>> root
'my_text_file'
>>> ext
'.txt'


https://docs.python.org/3/library/os.path.html

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

>>> from pathlib import Path

>>> p = Path("/a/b/c.txt")
>>> p.with_suffix('')
WindowsPath('/a/b/c')
>>> p.stem
'c'


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 = [
"FileName",
"./FileName",
"../../FileName",
"FileName.txt",
"./FileName.txt.zip.asc",
"/path/to/some/FileName",
"/path/to/some/FileName.txt",
"/path/to/some/FileName.txt.zip.asc"
]

for example_path in example_paths:
print(get_filename_without_extension(example_path))


In every case, the value printed is:

FileName

import os
filename, file_extension =os.path.splitext(os.path.basename('/d1/d2/example.cs'))

• filename is ‘example’

• file_extension is ‘.cs’

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)


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.”

e.g.

"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
else:
return file_name_prefix_match.group("file_name_prefix")

get_file_name_prefix("path/to/file.json")
>> file

get_file_name_prefix("path/to/my.file.json")
>> my.file

get_file_name_prefix("path/to/no_extension")
>> no_extension


## 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',
'thefile.tar',
'./thefile',
'./thefile.tar',
'../../thefile.tar',
'.././thefile.tar',
'rel/pa.th/to/thefile',
'/abs/path/to/thefile.tar']

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',
'/abs/pa.th/to/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('./thefile.tar.gz.bz.7zip')

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

# or, equivalently

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

# or simply run the full loop

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

# In any case:

print(fn)     # thefile

# Explanation
#
# pth.name     -> 'thefile.tar.gz.bz.7zip'
# 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 += ['./thefile.tar.gz.bz.7zip',
'/abs/pa.th/to/thefile.tar.gz.bz.7zip']

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

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, .tar.gz.bz, etc; you can simply rsplit the known extension and take the first element:


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

fn = pth.name.rsplit('.tar')[0]

print(fn)      # thefile


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

print(fn)      # thefile


This works for filenames without extension also

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 != '':
extensions.append(result)
else:
break
extensions.reverse()
return "".join(extensions)


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

# use pathlib. the below works with compound filetypes and normal ones
source_file = 'spaces.tar.gz.zip.rar.7z'
source_path = pathlib.Path(source_file)
source_path.name.replace(''.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

import pathlib
filename = '/path/to/dir/stem.ext.tar.gz'
pathlib.Path(filename).name[:-len(''.join(pathlib.Path(filename).suffixes))]
# -> 'stem'


or this equivalent?

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


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]


Performance:

Python 3.9.10 (tags/v3.9.10:f2f3f53, Jan 17 2022, 15:14:21) [MSC v.1929 64 bit (AMD64)]
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]:


>>>print(os.path.splitext(os.path.basename("/path/to/file/varun.txt"))[0]) varun

Here /path/to/file/varun.txt is the path to file and the output is varun

Assuming you’re already using pathlib and Python 3.9 or newer, here’s a simple one-line approach that removes all extensions.

>>> from pathlib import Path
>>> pth = Path("/path/to.some/file.foo.bar.txt")
>>> pth.name.removesuffix("".join(pth.suffixes))
'file'

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