How can I list the contents of a directory in Python?

Each Answer to this Q is separated by one/two green lines.

Can’t be hard, but I’m having a mental block.

import os
os.listdir("path") # returns list

One way:

import os

Another way:


Examples found here.

The glob.glob method above will not list hidden files.

Since I originally answered this question years ago, pathlib has been added to Python. My preferred way to list a directory now usually involves the iterdir method on Path objects:

from pathlib import Path
print(*Path("/home/username/www/").iterdir(), sep="\n")

os.walk can be used if you need recursion:

import os
start_path="." # current directory
for path,dirs,files in os.walk(start_path):
    for filename in files:
        print os.path.join(path,filename)

glob.glob or os.listdir will do it.

The os module handles all that stuff.


Return a list containing the names of the entries in the directory given by path.
The list is in arbitrary order. It does not include the special entries ‘.’ and
‘..’ even if they are present in the directory.

Availability: Unix, Windows.

In Python 3.4+, you can use the new pathlib package:

from pathlib import Path
for path in Path('.').iterdir():

Path.iterdir() returns an iterator, which can be easily turned into a list:

contents = list(Path('.').iterdir())

Since Python 3.5, you can use os.scandir.

The difference is that it returns file entries not names. On some OSes like windows, it means that you don’t have to os.path.isdir/file to know if it’s a file or not, and that saves CPU time because stat is already done when scanning dir in Windows:

example to list a directory and print files bigger than max_value bytes:

for dentry in os.scandir("/path/to/dir"):
    if dentry.stat().st_size > max_value:
       print("{} is biiiig".format(

(read an extensive performance-based answer of mine here)

Below code will list directories and the files within the dir. The other one is os.walk

def print_directory_contents(sPath):
        import os                                       
        for sChild in os.listdir(sPath):                
            sChildPath = os.path.join(sPath,sChild)
            if os.path.isdir(sChildPath):

The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .

Leave a Reply

Your email address will not be published.