How to list only top level directories in Python?

fuentesjr_10708 0
Viewed 216253

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.

filesystems python
21 Answers

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.

Alex_Coventry_19412130

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']
Thomas_Wouters_176240

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

filter(os.path.isdir, os.listdir(os.getcwd()))
Colin_Jensen_98840 
directories=[d for d in os.listdir(os.getcwd()) if os.path.isdir(d)]
Mark_Roddy_99400

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)
    ]
user_68990

Using list comprehension,

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

I think It is the simplest way

user_96051620 
[x for x in os.listdir(somedir) if os.path.isdir(os.path.join(somedir, x))]
Moe_30510

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()]
user_21668230

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.

user_21321570

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.

user_2949300

2021 answer using glob:

import glob, os

p = "/some/path/"
for d in glob.glob(p + "*" + os.path.sep):
  print(d)
user_7974950

Like so?

>>>> [path for path in os.listdir(os.getcwd()) if os.path.isdir(path)]
user_325380

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)
user_101340770

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')]
user_11726850
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