How can I check the extension of a file?

I’m working on a certain program where I need to do different things depending on the extension of the file. Could I just use this?

if m == *.mp3
   ...
elif m == *.flac
   ...

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

Assuming m is a string, you can use endswith:

if m.endswith('.mp3'):
...
elif m.endswith('.flac'):
...

To be case-insensitive, and to eliminate a potentially large else-if chain:

m.lower().endswith(('.png', '.jpg', '.jpeg'))

Method 2

os.path provides many functions for manipulating paths/filenames. (docs)

os.path.splitext takes a path and splits the file extension from the end of it.

import os

filepaths = ["/folder/soundfile.mp3", "folder1/folder/soundfile.flac"]

for fp in filepaths:
    # Split the extension from the path and normalise it to lowercase.
    ext = os.path.splitext(fp)[-1].lower()

    # Now we can simply use == to check for equality, no need for wildcards.
    if ext == ".mp3":
        print fp, "is an mp3!"
    elif ext == ".flac":
        print fp, "is a flac file!"
    else:
        print fp, "is an unknown file format."

Gives:

/folder/soundfile.mp3 is an mp3!
folder1/folder/soundfile.flac is a flac file!

Method 3

Use pathlib From Python3.4 onwards.

from pathlib import Path
Path('my_file.mp3').suffix == '.mp3'

Method 4

Look at module fnmatch. That will do what you’re trying to do.

import fnmatch
import os

for file in os.listdir('.'):
    if fnmatch.fnmatch(file, '*.txt'):
        print file

Method 5

or perhaps:

from glob import glob
...
for files in glob('path/*.mp3'): 
  do something
for files in glob('path/*.flac'): 
  do something else

Method 6

one easy way could be:

import os

if os.path.splitext(file)[1] == ".mp3":
    # do something

os.path.splitext(file) will return a tuple with two values (the filename without extension + just the extension). The second index ([1]) will therefor give you just the extension. The cool thing is, that this way you can also access the filename pretty easily, if needed!

Method 7

An old thread, but may help future readers…

I would avoid using .lower() on filenames if for no other reason than to make your code more platform independent. (linux is case sensistive, .lower() on a filename will surely corrupt your logic eventually …or worse, an important file!)

Why not use re? (Although to be even more robust, you should check the magic file header of each file…
How to check type of files without extensions in python? )

import re

def checkext(fname):   
    if re.search('.mp3$',fname,flags=re.IGNORECASE):
        return('mp3')
    if re.search('.flac$',fname,flags=re.IGNORECASE):
        return('flac')
    return('skip')

flist = ['myfile.mp3', 'myfile.MP3','myfile.mP3','myfile.mp4','myfile.flack','myfile.FLAC',
     'myfile.Mov','myfile.fLaC']

for f in flist:
    print "{} ==> {}".format(f,checkext(f))

Output:

myfile.mp3 ==> mp3
myfile.MP3 ==> mp3
myfile.mP3 ==> mp3
myfile.mp4 ==> skip
myfile.flack ==> skip
myfile.FLAC ==> flac
myfile.Mov ==> skip
myfile.fLaC ==> flac

Method 8

import os
source = ['test_sound.flac','ts.mp3']

for files in source:
   fileName,fileExtension = os.path.splitext(files)
   print fileExtension   # Print File Extensions
   print fileName   # It print file name

Method 9

You should make sure the “file” isn’t actually a folder before checking the extension. Some of the answers above don’t account for folder names with periods. (folder.mp3 is a valid folder name).


Checking the extension of a file:

import os

file_path = "C:/folder/file.mp3"
if os.path.isfile(file_path):
    file_extension = os.path.splitext(file_path)[1]
    if file_extension.lower() == ".mp3":
        print("It's an mp3")
    if file_extension.lower() == ".flac":
        print("It's a flac")

Output:

It's an mp3

Checking the extension of all files in a folder:

import os

directory = "C:/folder"
for file in os.listdir(directory):
    file_path = os.path.join(directory, file)
    if os.path.isfile(file_path):
        file_extension = os.path.splitext(file_path)[1]
        print(file, "ends in", file_extension)

Output:

abc.txt ends in .txt
file.mp3 ends in .mp3
song.flac ends in .flac

Comparing file extension against multiple types:

import os

file_path = "C:/folder/file.mp3"
if os.path.isfile(file_path):
    file_extension = os.path.splitext(file_path)[1]
    if file_extension.lower() in {'.mp3', '.flac', '.ogg'}:
        print("It's a music file")
    elif file_extension.lower() in {'.jpg', '.jpeg', '.png'}:
        print("It's an image file")

Output:

It's a music file

Method 10

#!/usr/bin/python

import shutil, os

source = ['test_sound.flac','ts.mp3']

for files in source:
  fileName,fileExtension = os.path.splitext(files)

  if fileExtension==".flac" :
    print 'This file is flac file %s' %files
  elif  fileExtension==".mp3":
    print 'This file is mp3 file %s' %files
  else:
    print 'Format is not valid'

Method 11

if (file.split(".")[1] == "mp3"):
    print "its mp3"
elif (file.split(".")[1] == "flac"):
    print "its flac"
else:
    print "not compat"

Method 12

If your file is uploaded then

import os


file= request.FILES['your_file_name']          #Your input file_name for your_file_name
ext = os.path.splitext(file.name)[-1].lower()


if ext=='.mp3':
    #do something

elif ext=='.xls' or '.xlsx' or '.csv':
    #do something

else:
    #The uploaded file is not the required format

Method 13

file='test.xlsx'
if file.endswith('.csv'):
    print('file is CSV')
elif file.endswith('.xlsx'):
    print('file is excel')
else:
    print('none of them')


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

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