Getting a list of all subdirectories in the curren

问题:

Is there a way to return a list of all the subdirectories in the current directory in Python?

I know you can do this with files, but I need to get the list of directories instead.

回答1:

Do you mean immediate subdirectories, or every directory right down the tree?

Either way, you could use os.walk to do this:

os.walk(directory)

will yield a tuple for each subdirectory. Ths first entry in the 3-tuple is a directory name, so

[x[0] for x in os.walk(directory)]

should give you all of the subdirectories, recursively.

Note that the second entry in the tuple is the list of child directories of the entry in the first position, so you could use this instead, but it\'s not likely to save you much.

However, you could use it just to give you the immediate child directories:

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

Or see the other solutions already posted, using os.listdir and os.path.isdir, including those at \"How to get all of the immediate subdirectories in Python\".

回答2:

import os

d = \'.\'
[os.path.join(d, o) for o in os.listdir(d) 
                    if os.path.isdir(os.path.join(d,o))]

回答3:

You could just use glob.glob

from glob import glob
glob(\"/path/to/directory/*/\")

Don\'t forget the trailing / after the *.

回答4:

If you need a recursive solution that will find all the subdirectories in the subdirectories, use walk as proposed before.

If you only need the current directory\'s child directories, combine os.listdir with os.path.isdir

回答5:

Much nicer than the above, because you don\'t need several os.path.join() and you will get the full path directly (if you wish), you can do this in Python 3.5+

subfolders = [f.path for f in os.scandir(folder) if f.is_dir() ]    

This will give the complete path to the subdirectory. If you only want the name of the subdirectory use f.name instead of f.path

https://docs.python.org/3/library/os.html#os.scandir

回答6:

I prefer using filter (https://docs.python.org/2/library/functions.html#filter), but this is just a matter of taste.

d=\'.\'
filter(lambda x: os.path.isdir(os.path.join(d, x)), os.listdir(d))

回答7:

Implemented this using python-os-walk. (http://www.pythonforbeginners.com/code-snippets-source-code/python-os-walk/)

import os

print(\"root prints out directories only from what you specified\")
print(\"dirs prints out sub-directories from root\")
print(\"files prints out all files from root and directories\")
print(\"*\" * 20)

for root, dirs, files in os.walk(\"/var/log\"):
    print(root)
    print(dirs)
    print(files)

回答8:

You can get the list of subdirectories (and files) in Python 2.7 using os.listdir(path)

import os
os.listdir(path)  # list of subdirectories and files

回答9:

Since I stumbled upon this problem using Python 3.4 and Windows UNC paths, here\'s a variant for this environment:

from pathlib import WindowsPath

def SubDirPath (d):
    return [f for f in d.iterdir() if f.is_dir()]

subdirs = SubDirPath(WindowsPath(r\'\\\\file01.acme.local\\home$\'))
print(subdirs)

Pathlib is new in Python 3.4 and makes working with paths under different OSes much easier: https://docs.python.org/3.4/library/pathlib.html

回答10:

Thanks for the tips, guys. I ran into an issue with softlinks (infinite recursion) being returned as dirs. Softlinks? We don\'t want no stinkin\' soft links! So...

This rendered just the dirs, not softlinks:

>>> import os
>>> inf = os.walk(\'.\')
>>> [x[0] for x in inf]
[\'.\', \'./iamadir\']

回答11:

Although this question is answered a long time ago. I want to recommend to use the pathlib module since this is a robust way to work on Windows and Unix OS.

So to get all paths in a specific directory including subdirectories:

from pathlib import Path
paths = list(Path(\'myhomefolder\', \'folder\').glob(\'**/*.txt\'))

# all sorts of operations
file = paths[0]
file.name
file.stem
file.parent
file.suffix

etc.

回答12:

Listing Out only directories

print(\"\\nWe are listing out only the directories in current directory -\")
directories_in_curdir = filter(os.path.isdir, os.listdir(os.curdir))
print(directories_in_curdir)

Listing Out only files in current directory

files = filter(os.path.isfile, os.listdir(os.curdir))
print(\"\\nThe following are the list of all files in the current directory -\")
print(files)

回答13:

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(\'./\')

# List comprehension
[f for f in p.iterdir() if f.is_dir()]

# The trailing slash to glob indicated directories
# This will also include the current directory \'.\'
list(p.glob(\'**/\'))

Pathlib is also available on Python 2.7 via the pathlib2 module on PyPi.

回答14:

Building upon Eli Bendersky\'s solution, use the following example:

import os
test_directory = <your_directory>
for child in os.listdir(test_directory):
    test_path = os.path.join(test_directory, child)
    if os.path.isdir(test_path):
        print test_path
        # Do stuff to the directory \"test_path\"

where <your_directory> is the path to the directory you want to traverse.

回答15:

Here are a couple of simple functions based on @Blair Conrad\'s example -

import os

def get_subdirs(dir):
    \"Get a list of immediate subdirectories\"
    return next(os.walk(dir))[1]

def get_subfiles(dir):
    \"Get a list of immediate subfiles\"
    return next(os.walk(dir))[2]

回答16:

With full path and accounting for path being ., .., \\\\, ..\\\\..\\\\subfolder, etc:

import os, pprint
pprint.pprint([os.path.join(os.path.abspath(path), x[0]) \\
    for x in os.walk(os.path.abspath(path))])

回答17:

This answer didn\'t seem to exist already.

directories = [ x for x in os.listdir(\'.\') if os.path.isdir(x) ]

回答18:

I\'ve had a similar question recently, and I found out that the best answer for python 3.6 (as user havlock added) is to use os.scandir. Since it seems there is no solution using it, I\'ll add my own. First, a non-recursive solution that lists only the subdirectories directly under the root directory.

def get_dirlist(rootdir):

    dirlist = []

    with os.scandir(rootdir) as rit:
        for entry in rit:
            if not entry.name.startswith(\'.\') and entry.is_dir():
                dirlist.append(entry.path)

    dirlist.sort() # Optional, in case you want sorted directory names
    return dirlist

The recursive version would look like this:

def get_dirlist(rootdir):

    dirlist = []

    with os.scandir(rootdir) as rit:
        for entry in rit:
            if not entry.name.startswith(\'.\') and entry.is_dir():
                dirlist.append(entry.path)
                dirlist += get_dirlist(entry.path)

    dirlist.sort() # Optional, in case you want sorted directory names
    return dirlist

keep in mind that entry.path wields the absolute path to the subdirectory. In case you only need the folder name, you can use entry.name instead. Refer to os.DirEntry for additional details about the entry object.

回答19:

use a filter function os.path.isdir over os.listdir() something like this filter(os.path.isdir,[os.path.join(os.path.abspath(\'PATH\'),p) for p in os.listdir(\'PATH/\')])

回答20:

Copy paste friendly in ipython:

import os
d=\'.\'
folders = list(filter(lambda x: os.path.isdir(os.path.join(d, x)), os.listdir(d)))

Output from print(folders):

[\'folderA\', \'folderB\']

回答21:

If you want just the top list folder, please use listdir as walk take too much time.