问题:获取当前目录中所有子目录的列表
有没有办法在Python中返回当前目录中所有子目录的列表?
我知道您可以使用文件来执行此操作,但是我需要获取目录列表。
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 0
您是指直接子目录,还是树下的每个目录?
无论哪种方式,您都可以使用os.walk
以下方法:
os.walk(directory)
将为每个子目录生成一个元组。该三元组中的第一个条目是目录名称,因此
[x[0] for x in os.walk(directory)]
应该递归地给你所有的子目录。
请注意,该元组中的第二个条目是该条目的子目录列表的第一个位置,因此您可以改用它,但是不太可能为您节省很多。
但是,您可以使用它来为您提供直接的子目录:
next(os.walk('.'))[1]
或者使用os.listdir
和查看已经发布的其他解决方案os.path.isdir
,包括“ 如何在Python中获取所有直接子目录 ”中的解决方案。
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 1
import os
d = '.'
[os.path.join(d, o) for o in os.listdir(d)
if os.path.isdir(os.path.join(d,o))]
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 2
你可以用 glob.glob
from glob import glob
glob("/path/to/directory/*/")
别忘了/
之后的尾随*
。
You could just use glob.glob
from glob import glob
glob("/path/to/directory/*/")
Don’t forget the trailing /
after the *
.
回答 3
比上面的要好得多,因为您不需要几个os.path.join(),并且可以直接获取完整路径(如果需要),因此可以在Python 3.5及更高版本中执行此操作。
subfolders = [ f.path for f in os.scandir(folder) if f.is_dir() ]
这将提供到子目录的完整路径。如果只希望使用子目录的名称,f.name
而不是f.path
https://docs.python.org/3/library/os.html#os.scandir
OT:如果您需要递归所有子文件夹和/或所有文件,请看一下此功能,它比os.walk
&快,glob
并且将返回所有子文件夹以及这些(子)子文件夹中的所有文件的列表:https://stackoverflow.com/a/59803793/2441026
如果您只需要递归所有子文件夹:
def fast_scandir(dirname):
subfolders= [f.path for f in os.scandir(dirname) if f.is_dir()]
for dirname in list(subfolders):
subfolders.extend(fast_scandir(dirname))
return subfolders
返回所有子文件夹及其完整路径的列表。这又比os.walk
和快得多glob
。
所有功能分析
tl; dr:
-如果要获取文件夹使用的所有直接子目录os.scandir
。
-如果要获取所有子目录,甚至嵌套的子目录,请使用os.walk
或-稍微快一点- fast_scandir
上面的函数。
-从来不使用os.walk
只顶级子目录,因为它可以是数百倍慢于(!) os.scandir
。
- 如果您运行下面的代码,请确保运行一次,以便您的操作系统可以访问该文件夹,丢弃结果并运行测试,否则结果将被弄乱。
- 您可能想混淆函数调用,但是我对其进行了测试,但这并不重要。
- 所有示例都将提供文件夹的完整路径。pathlib示例作为(Windows)Path对象。
- 的第一个元素
os.walk
将是基本文件夹。因此,您将不会仅获得子目录。您可以使用fu.pop(0)
将其删除。 - 所有结果都不会使用自然排序。这意味着将对结果进行如下排序:1、10、2。要进行自然排序(1、2、10),请查看https://stackoverflow.com/a/48030307/2441026
结果:
os.scandir took 1 ms. Found dirs: 439
os.walk took 463 ms. Found dirs: 441 -> it found the nested one + base folder.
glob.glob took 20 ms. Found dirs: 439
pathlib.iterdir took 18 ms. Found dirs: 439
os.listdir took 18 ms. Found dirs: 439
已在W7x64,Python 3.8.1中测试。
# -*- coding: utf-8 -*-
# Python 3
import time
import os
from glob import glob
from pathlib import Path
directory = r"<insert_folder>"
RUNS = 1
def run_os_walk():
a = time.time_ns()
for i in range(RUNS):
fu = [x[0] for x in os.walk(directory)]
print(f"os.walk\t\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")
def run_glob():
a = time.time_ns()
for i in range(RUNS):
fu = glob(directory + "/*/")
print(f"glob.glob\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")
def run_pathlib_iterdir():
a = time.time_ns()
for i in range(RUNS):
dirname = Path(directory)
fu = [f for f in dirname.iterdir() if f.is_dir()]
print(f"pathlib.iterdir\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")
def run_os_listdir():
a = time.time_ns()
for i in range(RUNS):
dirname = Path(directory)
fu = [os.path.join(directory, o) for o in os.listdir(directory) if os.path.isdir(os.path.join(directory, o))]
print(f"os.listdir\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")
def run_os_scandir():
a = time.time_ns()
for i in range(RUNS):
fu = [f.path for f in os.scandir(directory) if f.is_dir()]
print(f"os.scandir\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms.\tFound dirs: {len(fu)}")
if __name__ == '__main__':
run_os_scandir()
run_os_walk()
run_glob()
run_pathlib_iterdir()
run_os_listdir()
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 and above.
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
Slightly OT: In case you need all subfolder recursively and/or all files recursively, have a look at this function, that is faster than os.walk
& glob
and will return a list of all subfolders as well as all files inside those (sub-)subfolders: https://stackoverflow.com/a/59803793/2441026
In case you want only all subfolders recursively:
def fast_scandir(dirname):
subfolders= [f.path for f in os.scandir(dirname) if f.is_dir()]
for dirname in list(subfolders):
subfolders.extend(fast_scandir(dirname))
return subfolders
Returns a list of all subfolders with their full paths. This again is faster than os.walk
and a lot faster than glob
.
An analysis of all functions
tl;dr:
– If you want to get all immediate subdirectories for a folder use os.scandir
.
– If you want to get all subdirectories, even nested ones, use os.walk
or – slightly faster – the fast_scandir
function above.
– Never use os.walk
for only top-level subdirectories, as it can be hundreds(!) of times slower than os.scandir
.
- If you run the code below, make sure to run it once so that your OS will have accessed the folder, discard the results and run the test, otherwise results will be screwed.
- You might want to mix up the function calls, but I tested it, and it did not really matter.
- All examples will give the full path to the folder. The pathlib example as a (Windows)Path object.
- The first element of
os.walk
will be the base folder. So you will not get only subdirectories. You can use fu.pop(0)
to remove it. - None of the results will use natural sorting. This means results will be sorted like this: 1, 10, 2. To get natural sorting (1, 2, 10), please have a look at https://stackoverflow.com/a/48030307/2441026
Results:
os.scandir took 1 ms. Found dirs: 439
os.walk took 463 ms. Found dirs: 441 -> it found the nested one + base folder.
glob.glob took 20 ms. Found dirs: 439
pathlib.iterdir took 18 ms. Found dirs: 439
os.listdir took 18 ms. Found dirs: 439
Tested with W7x64, Python 3.8.1.
# -*- coding: utf-8 -*-
# Python 3
import time
import os
from glob import glob
from pathlib import Path
directory = r"<insert_folder>"
RUNS = 1
def run_os_walk():
a = time.time_ns()
for i in range(RUNS):
fu = [x[0] for x in os.walk(directory)]
print(f"os.walk\t\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")
def run_glob():
a = time.time_ns()
for i in range(RUNS):
fu = glob(directory + "/*/")
print(f"glob.glob\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")
def run_pathlib_iterdir():
a = time.time_ns()
for i in range(RUNS):
dirname = Path(directory)
fu = [f for f in dirname.iterdir() if f.is_dir()]
print(f"pathlib.iterdir\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")
def run_os_listdir():
a = time.time_ns()
for i in range(RUNS):
dirname = Path(directory)
fu = [os.path.join(directory, o) for o in os.listdir(directory) if os.path.isdir(os.path.join(directory, o))]
print(f"os.listdir\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")
def run_os_scandir():
a = time.time_ns()
for i in range(RUNS):
fu = [f.path for f in os.scandir(directory) if f.is_dir()]
print(f"os.scandir\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms.\tFound dirs: {len(fu)}")
if __name__ == '__main__':
run_os_scandir()
run_os_walk()
run_glob()
run_pathlib_iterdir()
run_os_listdir()
回答 4
如果您需要一个可在子目录中找到所有子目录的递归解决方案,请按照之前的建议使用walk。
如果仅需要当前目录的子目录,请os.listdir
与os.path.isdir
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
回答 6
使用python-os-walk实现了这一点。(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)
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 7
您可以使用os.listdir(path)获取Python 2.7中的子目录(和文件)列表
import os
os.listdir(path) # list of subdirectories and files
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 8
仅列出目录
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)
仅列出当前目录中的文件
files = filter(os.path.isfile, os.listdir(os.curdir))
print("\nThe following are the list of all files in the current directory -")
print(files)
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 9
蟒3.4引入的pathlib
模块到标准库,它提供了一个面向对象的方法来处理的文件系统的路径:
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('**/'))
也可以通过PyPi上的pathlib2模块在 Python 2.7 上使用Pathlib。
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 10
由于我使用Python 3.4和Windows UNC路径偶然发现了此问题,因此以下是此环境的变体:
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$'))
print(subdirs)
Pathlib是Python 3.4中的新增功能,它使在不同操作系统下使用路径变得更加容易:https ://docs.python.org/3.4/library/pathlib.html
Since I stumbled upon this problem using Python 3.4 and Windows UNC paths, here’s a variant for this environment:
from pathlib import WindowsPath
def SubDirPath (d):
return [f for f in d.iterdir() if f.is_dir()]
subdirs = SubDirPath(WindowsPath(r'\\file01.acme.local\home$'))
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
尽管很久以前就回答了这个问题。我想建议使用该pathlib
模块,因为这是在Windows和Unix OS上工作的可靠方法。
因此,要获取特定目录(包括子目录)中的所有路径:
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
等等
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
谢谢提醒伙计。我遇到了一个以dirs返回的软链接(无限递归)的问题。软链接?我们不希望没有臭味的软链接!所以…
这仅显示目录,而不显示软链接:
>>> import os
>>> inf = os.walk('.')
>>> [x[0] for x in inf]
['.', './iamadir']
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 13
复制粘贴友好ipython
:
import os
d='.'
folders = list(filter(lambda x: os.path.isdir(os.path.join(d, x)), os.listdir(d)))
来自的输出print(folders)
:
['folderA', 'folderB']
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 14
这就是我的方法。
import os
for x in os.listdir(os.getcwd()):
if os.path.isdir(x):
print(x)
This is how I do it.
import os
for x in os.listdir(os.getcwd()):
if os.path.isdir(x):
print(x)
回答 15
这是基于@Blair Conrad的示例的几个简单函数-
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]
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
在Eli Bendersky解决方案的基础上,使用以下示例:
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"
<your_directory>
您要遍历的目录的路径在哪里。
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 17
有了完整路径和占路感.
,..
,\\
,..\\..\\subfolder
,等:
import os, pprint
pprint.pprint([os.path.join(os.path.abspath(path), x[0]) \
for x in os.walk(os.path.abspath(path))])
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
这个答案似乎还不存在。
directories = [ x for x in os.listdir('.') if os.path.isdir(x) ]
This answer didn’t seem to exist already.
directories = [ x for x in os.listdir('.') if os.path.isdir(x) ]
回答 19
我最近有一个类似的问题,我发现python 3.6(作为用户havlock添加)的最佳答案是使用os.scandir
。由于似乎没有使用它的解决方案,因此我将添加自己的解决方案。首先,一种非递归解决方案,仅列出直接位于根目录下的子目录。
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
递归版本如下所示:
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
请记住,这entry.path
将使用子目录的绝对路径。如果只需要文件夹名称,则可以entry.name
改用。有关该对象的更多详细信息,请参考os.DirEntryentry
。
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 20
os.path.isdir
对os.listdir()
这样的东西使用过滤器功能filter(os.path.isdir,[os.path.join(os.path.abspath('PATH'),p) for p in os.listdir('PATH/')])
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 21
这将在文件树的右边列出所有子目录。
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
是3.4版的新功能
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 22
返回给定文件路径中所有子目录的列表的函数。将搜索整个文件树。
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
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
回答 23
我们可以使用os.walk()获取所有文件夹的列表
import os
path = os.getcwd()
pathObject = os.walk(path)
这个pathObject是一个对象,我们可以通过
arr = [x for x in pathObject]
arr is of type [('current directory', [array of folder in current directory], [files in current directory]),('subdirectory', [array of folder in subdirectory], [files in subdirectory]) ....]
我们可以通过遍历arr并打印中间数组来获取所有子目录的列表
for i in arr:
for j in i[1]:
print(j)
这将打印所有子目录。
要获取所有文件:
for i in arr:
for j in i[2]:
print(i[0] + "/" + j)
we can get list of all the folders by using os.walk()
import os
path = os.getcwd()
pathObject = os.walk(path)
this pathObject is a object and we can get an array by
arr = [x for x in pathObject]
arr is of type [('current directory', [array of folder in current directory], [files in current directory]),('subdirectory', [array of folder in subdirectory], [files in subdirectory]) ....]
We can get list of all the subdirectory by iterating through the arr and printing the middle array
for i in arr:
for j in i[1]:
print(j)
This will print all the subdirectory.
To get all the files:
for i in arr:
for j in i[2]:
print(i[0] + "/" + j)
回答 24
具有给定父级的该函数以递归方式directory
遍历其所有函数directories
以及prints
在filenames
其中找到的所有函数。太有用了。
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 ;")
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 25
通过从此处加入多个解决方案,这就是我最终使用的方法:
import os
import glob
def list_dirs(path):
return [os.path.basename(x) for x in filter(
os.path.isdir, glob.glob(os.path.join(path, '*')))]
By joining multiple solutions from here, this is what I ended up using:
import os
import glob
def list_dirs(path):
return [os.path.basename(x) for x in filter(
os.path.isdir, glob.glob(os.path.join(path, '*')))]
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