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?

Answer 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 '.'.

Answer 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)