List all subdirectories on given level

Question

I have backup directory structure like this (all directories are not empty):

/home/backups/mysql/
    2012/
        12/
           15/
    2013/
        04/
         


        

Answer 1

from glob import iglob

level3 = iglob('/home/backups/mysql/*/*/*')

(This will skip "hidden" directories with names starting with .)

If there may be non-directories at level 3, skip them using:

from itertools import ifilter
import os.path

l3_dirs = ifilter(os.path.isdir, level3)

In Python 3, use filter instead of ifilter.

Answer 2

You can use glob to search down a directory tree, like this:

import os, glob
def get_all_backup_paths(dir, level):
   pattern = dir + level * '/*'
   return [d for d in glob.glob(pattern) if os.path.isdir(d)]

I included a check for directories as well, in case there might be files mixed in with the directories.

Answer 3

import functools, os

def deepdirs(directory, depth = 0):
    if depth == 0:
        return list(filter(os.path.isdir, [os.path.join(directory, d) for d in os.listdir(directory)]))
    else:
        return functools.reduce(list.__add__, [deepdirs(d) for d in deepdirs(directory, depth-1)], [])