How to list only top level directories in Python?

I want to be able to list only the directories inside some folder. This means I don't want filenames listed, nor do I want additional sub-folders.

Let's see if an example helps. In the current directory we have:

>>> os.listdir(os.getcwd())
['cx_Oracle-doc', 'DLLs', 'Doc', 'include', 'Lib', 'libs', 'LICENSE.txt', 'mod_p
ython-wininst.log', 'NEWS.txt', 'pymssql-wininst.log', 'python.exe', 'pythonw.ex
e', 'README.txt', 'Removemod_python.exe', 'Removepymssql.exe', 'Scripts', 'tcl',
 'Tools', 'w9xpopen.exe']

However, I don't want filenames listed. Nor do I want sub-folders such as \Lib\curses. Essentially what I want works with the following:

>>> for root, dirnames, filenames in os.walk('.'):
...     print dirnames
...     break
...
['cx_Oracle-doc', 'DLLs', 'Doc', 'include', 'Lib', 'libs', 'Scripts', 'tcl', 'Tools']

However, I'm wondering if there's a simpler way of achieving the same results. I get the impression that using os.walk only to return the top level is inefficient/too much.

Asked by: Tara506 | Posted: 28-01-2022

Answer 1

os.walk

Use os.walk with next item function:

next(os.walk('.'))[1]

For Python <=2.5 use:

os.walk('.').next()[1]

How this works

os.walk is a generator and calling next will get the first result in the form of a 3-tuple (dirpath, dirnames, filenames). Thus the [1] index returns only the dirnames from that tuple.

Answered by: Miranda716 | Posted: 01-03-2022

Answer 2

Filter the result using os.path.isdir() (and use os.path.join() to get the real path):

>>> [ name for name in os.listdir(thedir) if os.path.isdir(os.path.join(thedir, name)) ]
['ctypes', 'distutils', 'encodings', 'lib-tk', 'config', 'idlelib', 'xml', 'bsddb', 'hotshot', 'logging', 'doc', 'test', 'compiler', 'curses', 'site-packages', 'email', 'sqlite3', 'lib-dynload', 'wsgiref', 'plat-linux2', 'plat-mac']

Answered by: Kelvin244 | Posted: 01-03-2022

Answer 3

Filter the list using os.path.isdir to detect directories.

filter(os.path.isdir, os.listdir(os.getcwd()))

Answered by: Lucas874 | Posted: 01-03-2022

Answer 4

directories=[d for d in os.listdir(os.getcwd()) if os.path.isdir(d)]

Answered by: Kellan391 | Posted: 01-03-2022

Answer 5

Note that, instead of doing os.listdir(os.getcwd()), it's preferable to do os.listdir(os.path.curdir). One less function call, and it's as portable.

So, to complete the answer, to get a list of directories in a folder:

def listdirs(folder):
    return [d for d in os.listdir(folder) if os.path.isdir(os.path.join(folder, d))]

If you prefer full pathnames, then use this function:

def listdirs(folder):
    return [
        d for d in (os.path.join(folder, d1) for d1 in os.listdir(folder))
        if os.path.isdir(d)
    ]

Answered by: Abigail361 | Posted: 01-03-2022

Answer 6

This seems to work too (at least on linux):

import glob, os
glob.glob('*' + os.path.sep)

Answered by: Elise737 | Posted: 01-03-2022

Answer 7

Using list comprehension,

[a for a in os.listdir() if os.path.isdir(a)]

I think It is the simplest way

Answered by: Alissa482 | Posted: 01-03-2022

Answer 8

Just to add that using os.listdir() does not "take a lot of processing vs very simple os.walk().next()[1]". This is because os.walk() uses os.listdir() internally. In fact if you test them together:

>>>> import timeit
>>>> timeit.timeit("os.walk('.').next()[1]", "import os", number=10000)
1.1215229034423828
>>>> timeit.timeit("[ name for name in os.listdir('.') if os.path.isdir(os.path.join('.', name)) ]", "import os", number=10000)
1.0592019557952881

The filtering of os.listdir() is very slightly faster.

Answered by: Chester811 | Posted: 01-03-2022

Answer 9

A very much simpler and elegant way is to use this:

 import os
 dir_list = os.walk('.').next()[1]
 print dir_list

Run this script in the same folder for which you want folder names.It will give you exactly the immediate folders name only(that too without the full path of the folders).

Answered by: Kelvin374 | Posted: 01-03-2022

Answer 10

[x for x in os.listdir(somedir) if os.path.isdir(os.path.join(somedir, x))]

Answered by: Jack346 | Posted: 01-03-2022

Answer 11

Python 3.4 introduced the pathlib module into the standard library, which provides an object oriented approach to handle filesystem paths:

from pathlib import Path

p = Path('./')
[f for f in p.iterdir() if f.is_dir()]

Answered by: Patrick537 | Posted: 01-03-2022

Answer 12

You could also use os.scandir:

with os.scandir(os.getcwd()) as mydir:
    dirs = [i.name for i in mydir if i.is_dir()]

In case you want the full path you can use i.path.

Using scandir() instead of listdir() can significantly increase the performance of code that also needs file type or file attribute information, because os.DirEntry objects expose this information if the operating system provides it when scanning a directory.

Answered by: Alfred125 | Posted: 01-03-2022

Answer 13

being a newbie here i can't yet directly comment but here is a small correction i'd like to add to the following part of ΤΖΩΤΖΙΟΥ's answer :

If you prefer full pathnames, then use this function:

def listdirs(folder):  
  return [
    d for d in (os.path.join(folder, d1) for d1 in os.listdir(folder))
    if os.path.isdir(d)
]

for those still on python < 2.4: the inner construct needs to be a list instead of a tuple and therefore should read like this:

def listdirs(folder):  
  return [
    d for d in [os.path.join(folder, d1) for d1 in os.listdir(folder)]
    if os.path.isdir(d)
  ]

otherwise one gets a syntax error.

Answered by: Lana725 | Posted: 01-03-2022

Answer 14

2021 answer using glob:

import glob, os

p = "/some/path/"
for d in glob.glob(p + "*" + os.path.sep):
  print(d)

Answered by: Rafael723 | Posted: 01-03-2022

Answer 15

For a list of full path names I prefer this version to the other solutions here:

def listdirs(dir):
    return [os.path.join(os.path.join(dir, x)) for x in os.listdir(dir) 
        if os.path.isdir(os.path.join(dir, x))]

Answered by: Richard415 | Posted: 01-03-2022

Answer 16

scanDir = "abc"
directories = [d for d in os.listdir(scanDir) if os.path.isdir(os.path.join(os.path.abspath(scanDir), d))]

Answered by: Miller543 | Posted: 01-03-2022

Answer 17

Like so?

>>>> [path for path in os.listdir(os.getcwd()) if os.path.isdir(path)]

Answered by: Patrick385 | Posted: 01-03-2022

Answer 18

FWIW, the os.walk approach is almost 10x faster than the list comprehension and filter approaches:

In [30]: %timeit [d for d in os.listdir(os.getcwd()) if os.path.isdir(d)]
1.23 ms ± 97.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [31]: %timeit list(filter(os.path.isdir, os.listdir(os.getcwd())))
1.13 ms ± 13.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [32]: %timeit next(os.walk(os.getcwd()))[1]
132 µs ± 9.34 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

Answered by: Alina412 | Posted: 01-03-2022

Answer 19

Using python 3.x with pathlib.Path.iter_dir

$ mkdir tmpdir
$ mkdir -p tmpdir/a/b/c
$ mkdir -p tmpdir/x/y/z

$ touch tmpdir/a/b/c/abc.txt
$ touch tmpdir/a/b/ab.txt
$ touch tmpdir/a/a.txt

$ python --version
Python 3.7.12

>>> from pathlib import Path
>>> tmpdir = Path("./tmpdir")
>>> [d for d in tmpdir.iterdir() if d.is_dir]
[PosixPath('tmpdir/x'), PosixPath('tmpdir/a')]
>>> sorted(d for d in tmpdir.iterdir() if d.is_dir)
[PosixPath('tmpdir/a'), PosixPath('tmpdir/x')]

Answered by: Jack881 | Posted: 01-03-2022

Answer 20

A safer option that does not fail when there is no directory.

def listdirs(folder):
    if os.path.exists(folder):
         return [d for d in os.listdir(folder) if os.path.isdir(os.path.join(folder, d))]
    else:
         return []

Answered by: Lyndon482 | Posted: 01-03-2022

Answer 21

-- This will exclude files and traverse through 1 level of sub folders in the root

def list_files(dir):
    List = []
    filterstr = ' '
    for root, dirs, files in os.walk(dir, topdown = True):
        #r.append(root)
        if (root == dir):
            pass
        elif filterstr in root:
            #filterstr = ' '
            pass
        else:
            filterstr = root
            #print(root)
            for name in files:
                print(root)
                print(dirs)
                List.append(os.path.join(root,name))
            #print(os.path.join(root,name),"\n")
                print(List,"\n")

    return List

Answered by: Edward264 | Posted: 01-03-2022

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