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

Question

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?

Asked By: talnicolas

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Source

Answer 1

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.)

Answered By: Josh Lee

Answer 2

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 '.'.

Answered By: Adam Rosenfield

Answer 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

Answered By: cle

Answer 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(".")
]

Answered By: scum

Answer 5

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)

Answered By: Joshmaker

Answer 6

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

Answered By: hrsma2i

Answer 7

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('*')

Answered By: user 923227

Answer 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)

Answered By: Ayaan Imran Saleem

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

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