How to list only top level directories in Python?

Question

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.

Osterman

os.walk() uses a generator function and would not be ineffective if you are using it just for top-level. – Siegbahn

[d for d in a_path.iterdir() if d.is_dir] uses pathlib.Path – Peripeteia

@DarrenWeber: I think it should be: [d for d in a_path.iterdir() if d.is_dir()] – Oleoresin

Answer 1

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']

Undesigning

This take a lot of processing vs very simple os.walk().next()[1] – Neurophysiology

Doge, to easy! Had to use next(os.walk(dpath))[1] but same concept. Thanks for the suggestion @PhyoArkarLwin – Ordinarily

Answer 2

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.

Dilution

A little more description on this is that this is a generator, it won't be walking the other dirs unless you tell it to. So .next()[1] does in one line what all the list comprehensions do. I'd probably do something like DIRNAMES=1 and then next()[DIRNAMES] to make it easier to understand for future code maintainers. – Coppola

+1 amazing solution. To specify a directory to browse, use: os.walk( os.path.join(mypath,'.')).next()[1] – Gullett

for python v3: next(os.walk('.'))[1] – Castilian

if your going to do more then text processing; ie processing in the actual folders then full paths might be needed: sorted( [os.path.join(os.getcwd(), item) for item in os.walk(os.curdir).next()[1]] ) – Arjuna

why the dot? next(os.walk('.'))[1] – Rifkin

. indicates the current directory. It's equivalent to os.getcwd(). – Dilution

Answer 3

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

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

Markhor

I think this is by far the best combination of readability and conciseness in any of these answers. – Dozen

This didn't work. My guess is that os.listdir returns a file/folder name, passed on to os.path.isdir, but the latter needs a full path. – Gullett

filter is faster than os.walk timeit(os.walk(os.getcwd()).next()[1]) 1000 loops, best of 3: 734 µs per loop timeit(filter(os.path.isdir, os.listdir(os.getcwd()))) 1000 loops, best of 3: 477 µs per loop – Frias

A comment here is that os.listdir does not return the absolute path so filter(os.path.isdir, [os.path.join(base, f) for f in os.listdir(base)]) solved it for me. – Gujarati

Answer 4

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

Gonyea

This can be shortened to filter (os.path.isdir, os.listdir (os.getcwd ()) – Inelegancy

Does anyone have any information on whether filter or a list comprehension is faster? Otherwise its just a subjective argument. This of course assumes there's 10 million directories in the cwd and performance is an issue. – Gonyea

Answer 5

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

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

Darvon

+1 for glob. It can save you a lot of code, especially iterations, and is very much akin to UNIX terminal usage (ls) – Ives

Rather than glob.glob('*' + os.path.sep) you might want to write [dir for dir in glob.glob("*") if os.path.isdir( dir)] – Mcgaw

Note that the strings glob("./*"), glob("*"), glob("*/"), or glob("*" + os.path.sep) returns can be different, such as ".\\" or "\\" can be attached to head or tail of the directory name, in Windows. glob("*") returns exactly the directory name, without ".\\" or "\\". – Cahan

Answer 6

Using list comprehension,

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

I think It is the simplest way

Tarr

to list all subdirs in currentdir: [a for a in os.listdir(currentdir) if os.path.isdir(os.path.join(currentdir, a))] – Shipway

Answer 7

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)
    ]

Duvall

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.

Loram

Coming in Python 3.5 is a faster way of getting directory contents: python.org/dev/peps/pep-0471 – Loram

pep-0471 - the scandir package - is happily available for Python 2.6 onwards as an installable package on PyPI. It offers replacements for os.walk and os.listdir that are much faster. – Loram

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).

Axon

Answer 10

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()]

Thermit

Answer 11

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.

Dallis

Answer 12

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

Merete

Answer 13

2021 answer using glob:

import glob, os

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

Skedaddle

Answer 14

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.

Arbitrate

i know its been a while, but this first example really helped me. – Exscind

You get a syntax error because your version doesn't support generator expressions. These were introduced in Python 2.4 whereas list comprehensions have been available since Python 2.0. – Phyte

Answer 15

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)

Day

What python were you on? – Vilayet

Probably Python 3.6 – Day

Answer 16

Like so?

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

Brasilin

Answer 17

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))]

Fissiparous

Answer 18

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

Pretend

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')]

Peripeteia

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 []

Gnostic

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

Archaeology