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

Answer 1

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)

Answer 2

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.

Answer 3

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

Answer 4

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.

Answer 5

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

Answer 6

Like so?

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