How do I get the filename without the extension from a path in Python?
"/path/to/some/file.txt" → "file"
Answers:
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Method 1
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
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
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.
Method 2
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'.
Method 3
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
See below for other answers that address that.
Method 4
>>> print(os.path.splitext(os.path.basename("/path/to/file/hemanth.txt"))[0])
hemanth
Method 5
In Python 3.4+ you can use the pathlib solution
from pathlib import Path print(Path(your_path).resolve().stem)
Method 6
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'
Method 7
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
Method 8
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
Method 9
@IceAdor’s refers to rsplit in a comment to @user2902201’s solution. rsplit is the simplest solution that supports multiple periods.
Here it is spelt out:
file = 'my.report.txt'
print file.rsplit('.', 1)[0]
my.report
Method 10
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'
Method 11
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.
Method 12
import os
filename, file_extension =os.path.splitext(os.path.basename('/d1/d2/example.cs'))
- filename is ‘example’
- file_extension is ‘.cs’
‘
Method 13
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
Method 14
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
Method 15
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)
Method 16
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'
Method 17
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)
Method 18
import os path = "a/b/c/abc.txt" print os.path.splitext(os.path.basename(path))[0]
Method 19
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
Method 20
Improving upon @spinup answer:
fn = pth.name
for s in pth.suffixes:
fn = fn.rsplit(s)[0]
break
print(fn) # thefile
This works for filenames without extension also
Method 21
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
Method 22
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
Method 23
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.
Method 24
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
Method 25
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.
Method 26
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
Method 27
What about the following?
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))]
Method 28
# 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
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0