How to ignore hidden files using os.listdir()?

My python script executes an os.listdir(path) where the path is a queue containing archives that I need to treat one by one.

The problem is that I’m getting the list in an array and then I just do a simple array.pop(0). It was working fine until I put the project in subversion. Now I get the .svn folder in my array and of course it makes my application crash.

So here is my question: is there a function that ignores hidden files when executing an os.listdir() and if not what would be the best way?

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

You can write one yourself:

import os

def listdir_nohidden(path):
    for f in os.listdir(path):
        if not f.startswith('.'):
            yield f

Or you can use a glob:

import glob
import os

def listdir_nohidden(path):
    return glob.glob(os.path.join(path, '*'))

Either of these will ignore all filenames beginning with '.'.

Method 2

This is an old question, but seems like it is missing the obvious answer of using list comprehension, so I’m adding it here for completeness:

[f for f in os.listdir(path) if not f.startswith('.')]

As a side note, the docs state listdir will return results in ‘arbitrary order’ but a common use case is to have them sorted alphabetically. If you want the directory contents alphabetically sorted without regards to capitalization, you can use:

sorted((f for f in os.listdir() if not f.startswith(".")), key=str.lower)

(Edited to use key=str.lower instead of a lambda)

Method 3

On Windows, Linux and OS X:

if os.name == 'nt':
    import win32api, win32con


def folder_is_hidden(p):
    if os.name== 'nt':
        attribute = win32api.GetFileAttributes(p)
        return attribute & (win32con.FILE_ATTRIBUTE_HIDDEN | win32con.FILE_ATTRIBUTE_SYSTEM)
    else:
        return p.startswith('.') #linux-osx

Method 4

Joshmaker has the right solution to your question.
How to ignore hidden files using os.listdir()?

In Python 3 however, it is recommended to use pathlib instead of os.

from pathlib import Path 
visible_files = [
    file for file in Path(".").iterdir() if not file.name.startswith(".")
]

Method 5

glob:

>>> import glob
>>> glob.glob('*')

(glob claims to use listdir and fnmatch under the hood, but it also checks for a leading '.', not by using fnmatch.)

Method 6

I think it is too much of work to go through all of the items in a loop. I would prefer something simpler like this:

lst = os.listdir(path)
if '.DS_Store' in lst:
    lst.remove('.DS_Store')

If the directory contains more than one hidden files, then this can help:

all_files = os.popen('ls -1').read()
lst = all_files.split('n')

for platform independence as @Josh mentioned the glob works well:

import glob
glob.glob('*')

Method 7

filenames = (f.name for f in os.scandir() if not f.name.startswith('.'))

Method 8

You can just use a simple for loop that will exclude any file or directory that has “.” in the front.

Code for professionals:

import os

directory_things = [i for i in os.listdir() if i[0] != "."] # Exclude all with . in the start

Code for noobs

items_in_directory = os.listdir()
final_items_in_directory = []

for i in items_in_directory:
    if i[0] != ".": # If the item doesn't have any '.' in the start
        final_items_in_directory.append(i)

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