目录中的 Python 文件夹名称

Python folder names in the directory

提问by HightronicDesign

how can i get the folder names existing in a directory using Python ?

如何使用 Python 获取目录中存在的文件夹名称？

I want to save all the subfolders into a list to work with the names after that but i dont know how to read the subfolder names ?

我想将所有子文件夹保存到一个列表中以处理之后的名称，但我不知道如何读取子文件夹名称？

Thanks for you help

谢谢你的帮助

采纳答案by Christian Eichelmann

You can use os.walk()

您可以使用 os.walk()

# !/usr/bin/python

import os

directory_list = list()
for root, dirs, files in os.walk("/path/to/your/dir", topdown=False):
    for name in dirs:
        directory_list.append(os.path.join(root, name))

print directory_list

EDIT

编辑

If you only want the first level and not actually "walk" through the subdirectories, it is even less code:

如果您只想要第一级而不是真正“遍历”子目录，则代码更少：

import os

root, dirs, files = os.walk("/path/to/your/dir").next()
print dirs

This is not really what os.walkis made for. If you really only want one level of subdirectories, you can also use os.listdir()like Yannik Ammann suggested:

这不是真正的os.walk目的。如果你真的只想要一级子目录，你也可以os.listdir()像 Yannik Ammann 建议的那样使用：

root='/path/to/my/dir'
dirlist = [ item for item in os.listdir(root) if os.path.isdir(os.path.join(root, item)) ]
print dirlist

回答by yamm

You can use os.listdir()here a link to the docs

您可以os.listdir()在此处使用指向文档的链接

Warning returns files and directories

警告返回文件和目录

example:

例子：

import os

path = 'pyth/to/dir/'
dir_list = os.listdir(path)

update:you need to check if the returned names are directories or files

更新：您需要检查返回的名称是目录还是文件

import os

path = 'pyth/to/dir/'
# list of all content in a directory, filtered so only directories are returned
dir_list = [directory for directory in os.listdir(path) if os.path.isdir(path+directory)]

回答by srekcahrai

Use os.walk(path)

用 os.walk(path)

import os

path = 'C:\'

for root, directories, files in os.walk(path):
    for directory in directories:
        print os.path.join(root, directory)

回答by poke

Starting with Python 3.4, you can also use the new pathlibmodule:

从 Python 3.4 开始，您还可以使用新pathlib模块：

from pathlib import Path

p = Path('some/folder')
subdirectories = [x for x in p.iterdir() if x.is_dir()]

print(subdirectories)

回答by titto.sebastian

You should import os first.

您应该先导入 os。

import os
files=[]
files = [f for f in sorted(os.listdir(FileDirectoryPath))]

This would give you list with all files in the FileDirectoryPath sorted.

这将为您提供在 FileDirectoryPath 中排序的所有文件的列表。

回答by oxidworks

I use os.listdir

我使用os.listdir

Get all folder names of a directory

获取目录的所有文件夹名称

folder_names = []
for entry_name in os.listdir(MYDIR):
    entry_path = os.path.join(MYDIR, entry_name)
    if os.path.isdir(entry_path):
        folder_names.append(entry_name)

Get all folder paths of a directory

获取目录的所有文件夹路径

folder_paths = []
for entry_name in os.listdir(MYDIR):
    entry_path = os.path.join(MYDIR, entry_name)
    if os.path.isdir(entry_path):
        folder_paths.append(entry_path)

Get all file names of a directory

获取目录的所有文件名

file_names = []
for file_name in os.listdir(MYDIR):
    file_path = os.path.join(MYDIR, file_name)
    if os.path.isfile(file_path):
        file_names.append(file_name)

Get all file paths of a directory

获取目录的所有文件路径

file_paths = []
for file_name in os.listdir(MYDIR):
    file_path = os.path.join(MYDIR, file_name)
    if os.path.isfile(file_path):
        file_paths.append(file_path)

回答by Traveler_3994

Python 3.x: If you want only the directories in a given directory, try:

Python 3.x：如果您只想要给定目录中的目录，请尝试：

import os
search_path = '.'   # set your path here.
root, dirs, files = next(os.walk(search_path), ([],[],[]))
print(dirs)

The above example will print out a list of the directories in the current directory like this:

上面的例子将打印出当前目录中的目录列表，如下所示：

['dir1', 'dir2', 'dir3']

The output contains only the sub-directory names.
If the directory does not have sub-directories, it will print:

输出仅包含子目录名称。
如果目录没有子目录，它会打印：

[]

os.walk()is a generator method, so use next()to only call it once. The 3-tuple of empty strings is for the error condition when the directory does not contain any sub-directories because the os.walk() generator returns 3-tuples for each layer in the directory tree. Without those, if the directory is empty, next() will raise a StopIteration exception.

os.walk()是一个生成器方法，所以使用next()只调用一次。空字符串的三元组用于当目录不包含任何子目录时的错误情况，因为 os.walk() 生成器为目录树中的每一层返回三元组。如果没有这些，如果目录为空，next() 将引发 StopIteration 异常。

For a more compact version:

对于更紧凑的版本：

dirs = next(os.walk(search_path), ([],[],[]))[1]

回答by Tech

For python 3 I'm using this script

对于 python 3，我正在使用这个脚本

import os

root='./'
dirlist = [ item for item in os.listdir(root) if os.path.isdir(os.path.join(root, item)) ]
for dir in dirlist:
        print(dir)