如何从Python路径中获取不带扩展名的文件名？

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

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.

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.

Using pathlib in Python 3.4+

from pathlib import Path

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

will return

'file'

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

In Python 3.4+ you can use the pathlib solution

from pathlib import Path

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

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")
>>> print(p.with_suffix(''))
\a\b\c
>>> print(p.stem)
c

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

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

@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

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.

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'

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

This returns the filename without the extension(C:\Users\Public\Videos\Sample Videos\wildlife)

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)

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'

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

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

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

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

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

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.

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.

Very very very simpely no other modules !!!

import os
p = r"C:\Users\bilal\Documents\face Recognition python\imgs\northon.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)

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
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

import os filename, file_extension = os.path.splitext(‘/d1/d2/example.cs’) filename is ‘/d1/d2/example’ file_extension is ‘.cs’