Getting a list of all subdirectories in the current directory

孤街浪徒 提交于 2019-11-25 23:38:47

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:


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5:


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:


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10:


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:


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13:


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:


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17:


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18:


This answer didn't seem to exist already.

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



回答19:


This is how I do it.

    import os
    for x in os.listdir(os.getcwd()):
        if os.path.isdir(x):
            print(x)



回答20:


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21:


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22:


This will list all subdirectories right down the file tree.

import pathlib


def list_dir(dir):
    path = pathlib.Path(dir)
    dir = []
    try:
        for item in path.iterdir():
            if item.is_dir():
                dir.append(item)
                dir = dir + list_dir(item)
        return dir
    except FileNotFoundError:
        print('Invalid directory')

pathlib is new in version 3.4




回答23:


This function, with a given parent directory iterates over all its directories recursively and prints all the filenames which it founds inside. Too useful.

import os

def printDirectoryFiles(directory):
   for filename in os.listdir(directory):  
        full_path=os.path.join(directory, filename)
        if not os.path.isdir(full_path): 
            print( full_path + "\n")


def checkFolders(directory):

    dir_list = next(os.walk(directory))[1]

    #print(dir_list)

    for dir in dir_list:           
        print(dir)
        checkFolders(directory +"/"+ dir) 

    printDirectoryFiles(directory)       

main_dir="C:/Users/S0082448/Desktop/carpeta1"

checkFolders(main_dir)


input("Press enter to exit ;")




回答24:


Function to return a List of all subdirectories within a given file path. Will search through the entire file tree.

import os

def get_sub_directory_paths(start_directory, sub_directories):
    """
    This method iterates through all subdirectory paths of a given 
    directory to collect all directory paths.

    :param start_directory: The starting directory path.
    :param sub_directories: A List that all subdirectory paths will be 
        stored to.
    :return: A List of all sub-directory paths.
    """

    for item in os.listdir(start_directory):
        full_path = os.path.join(start_directory, item)

        if os.path.isdir(full_path):
            sub_directories.append(full_path)

            # Recursive call to search through all subdirectories.
            get_sub_directory_paths(full_path, sub_directories)

return sub_directories


来源:https://stackoverflow.com/questions/973473/getting-a-list-of-all-subdirectories-in-the-current-directory

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