How to 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?

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.

Get Filename Without Extension in Python

In most cases, you shouldn't use a regex for that.

os.path.splitext(filename)[0]

This will also handle a filename like .bashrc correctly by keeping the whole name.

get filenames in a directory without extension - Python

You can use os's splitext.

import os

path = '/Users/vivek/Desktop/buffer/xmlTest/' 
files = [os.path.splitext(filename)[0] for filename in os.listdir(path)]
print (files)

Just a heads up: basename won't work for this. basename doesn't remove the extension.

Extract file name from path, no matter what the os/path format

Using os.path.split or os.path.basename as others suggest won't work in all cases: if you're running the script on Linux and attempt to process a classic windows-style path, it will fail.

Windows paths can use either backslash or forward slash as path separator. Therefore, the ntpath module (which is equivalent to os.path when running on windows) will work for all⁽¹⁾ paths on all platforms.

import ntpath
ntpath.basename("a/b/c")

Of course, if the file ends with a slash, the basename will be empty, so make your own function to deal with it:

def path_leaf(path):
    head, tail = ntpath.split(path)
    return tail or ntpath.basename(head)

Verification:

>>> paths = ['a/b/c/', 'a/b/c', '\\a\\b\\c', '\\a\\b\\c\\', 'a\\b\\c', 
...     'a/b/../../a/b/c/', 'a/b/../../a/b/c']
>>> [path_leaf(path) for path in paths]
['c', 'c', 'c', 'c', 'c', 'c', 'c']

^{(1) There's one caveat: Linux filenames may contain backslashes. So on linux, r'a/b\c' always refers to the file b\c in the a folder, while on Windows, it always refers to the c file in the b subfolder of the a folder. So when both forward and backward slashes are used in a path, you need to know the associated platform to be able to interpret it correctly. In practice it's usually safe to assume it's a windows path since backslashes are seldom used in Linux filenames, but keep this in mind when you code so you don't create accidental security holes.}

How to get only filename without extension?

If the column values are always the filename/filepath, split it from right on . with maxsplit parameter as 1 and take the first value after splitting.

>>> df['relfilepath'].str.rsplit('.', n=1).str[0]

0                  20210322636
12              factuur-f23622
14                ingram micro
19    upfront.nl domein - Copy
21           upfront.nl domein
Name: relfilepath, dtype: object

How to get a specific file name from a path without the extensions python

may be funny and dirty, but works :)

sample_name = os.path.splitext(os.path.splitext(os.path.basename(f_file))[0])[0]

also can use shorter, nicer version:

sample_name = os.path.basename(f_file).split('.')[0]

Get file name only (without extension and directory) from file path

Your code works exactly as it should. In example a, the \n is not treated as a backslash character because it is a newline character. In example b, the extension is .mpg, which is properly removed. A file can never have more than one extension, or an extension containing a period.

To only get the bit before the first period, you could use ntpath.basename(filepath).split('.')[0], but this is probably NOT what you want as it is perfectly legal for filenames to contain periods.

Extracting extension from filename in Python

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', '')

How can I replace (or strip) an extension from a filename in Python?

Try os.path.splitext it should do what you want.

import os
print os.path.splitext('/home/user/somefile.txt')[0]+'.jpg'  # /home/user/somefile.jpg

os.path.splitext('/home/user/somefile.txt')  # returns ('/home/user/somefile', '.txt')

How to get filename using os.path.basename without the '> extension

You need to use filedialog.askopenfilename instead of filedialog.askopenfile to obtain the filename without side effects (like opening a file). This returns the full path; you can extract the filename from the fullpath using os.path.basename

import os
import tkinter as tk
from tkinter import filedialog

def browse():
    root = tk.Tk()
    root.withdraw()
    fullpath = filedialog.askopenfilename(parent=root, title='Choose a file')
    filename = os.path.basename(fullpath)
    root.destroy()

    return filename

print(browse())

---

Related Topics