Get a filtered list of files in a directory

Question

I am trying to get a list of files in a directory using Python, but I do not want a list of ALL the files.

What I essentially want is the ability to do something like the following but using Python and not executing ls.

ls 145592*.jpg

If there is no built-in method for this, I am currently thinking of writing a for loop to iterate through the results of an os.listdir() and to append all the matching files to a new list.

However, there are a lot of files in that directory and therefore I am hoping there is a more efficient method (or a built-in method).

Answer 1

use os.walk to recursively list your files

import os
root = "/home"
pattern = "145992"
alist_filter = ['jpg','bmp','png','gif'] 
path=os.path.join(root,"mydir_to_scan")
for r,d,f in os.walk(path):
    for file in f:
        if file[-3:] in alist_filter and pattern in file:
            print os.path.join(root,file)

Answer 2

Another option:

>>> import os, fnmatch
>>> fnmatch.filter(os.listdir('.'), '*.py')
['manage.py']

https://docs.python.org/3/library/fnmatch.html

Answer 3

glob.glob() is definitely the way to do it (as per Ignacio). However, if you do need more complicated matching, you can do it with a list comprehension and re.match(), something like so:

files = [f for f in os.listdir('.') if re.match(r'[0-9]+.*\.jpg', f)]

More flexible, but as you note, less efficient.

Answer 4

You can use subprocess.check_ouput() as

import subprocess

list_files = subprocess.check_output("ls 145992*.jpg", shell=True) 

Of course, the string between quotes can be anything you want to execute in the shell, and store the output.

Answer 5

import os

dir="/path/to/dir"
[x[0]+"/"+f for x in os.walk(dir) for f in x[2] if f.endswith(".jpg")]

This will give you a list of jpg files with their full path. You can replace x[0]+"/"+f with f for just filenames. You can also replace f.endswith(".jpg") with whatever string condition you wish.

Answer 6

Preliminary code

import glob
import fnmatch
import pathlib
import os

pattern = '*.py'
path = '.'

Solution 1 - use "glob"

# lookup in current dir
glob.glob(pattern)

In [2]: glob.glob(pattern)
Out[2]: ['wsgi.py', 'manage.py', 'tasks.py']

Solution 2 - use "os" + "fnmatch"

Variant 2.1 - Lookup in current dir

# lookup in current dir
fnmatch.filter(os.listdir(path), pattern)

In [3]: fnmatch.filter(os.listdir(path), pattern)
Out[3]: ['wsgi.py', 'manage.py', 'tasks.py']

Variant 2.2 - Lookup recursive

# lookup recursive
for dirpath, dirnames, filenames in os.walk(path):

    if not filenames:
        continue

    pythonic_files = fnmatch.filter(filenames, pattern)
    if pythonic_files:
        for file in pythonic_files:
            print('{}/{}'.format(dirpath, file))

Result

./wsgi.py
./manage.py
./tasks.py
./temp/temp.py
./apps/diaries/urls.py
./apps/diaries/signals.py
./apps/diaries/actions.py
./apps/diaries/querysets.py
./apps/library/tests/test_forms.py
./apps/library/migrations/0001_initial.py
./apps/polls/views.py
./apps/polls/formsets.py
./apps/polls/reports.py
./apps/polls/admin.py

Solution 3 - use "pathlib"

# lookup in current dir
path_ = pathlib.Path('.')
tuple(path_.glob(pattern))

# lookup recursive
tuple(path_.rglob(pattern))

Notes:

1. Tested on the Python 3.4
2. The module "pathlib" was added only in the Python 3.4
3. The Python 3.5 added a feature for recursive lookup with glob.glob https://docs.python.org/3.5/library/glob.html#glob.glob. Since my machine is installed with Python 3.4, I have not tested that.