有没有办法在Python中返回当前目录中所有子目录的列表?
我知道您可以对文件执行此操作,但我需要获得目录列表。
有没有办法在Python中返回当前目录中所有子目录的列表?
我知道您可以对文件执行此操作,但我需要获得目录列表。
当前回答
下面这个类将能够获得一个给定目录中的文件,文件夹和所有子文件夹的列表
import os
import json
class GetDirectoryList():
def __init__(self, path):
self.main_path = path
self.absolute_path = []
self.relative_path = []
def get_files_and_folders(self, resp, path):
all = os.listdir(path)
resp["files"] = []
for file_folder in all:
if file_folder != "." and file_folder != "..":
if os.path.isdir(path + "/" + file_folder):
resp[file_folder] = {}
self.get_files_and_folders(resp=resp[file_folder], path= path + "/" + file_folder)
else:
resp["files"].append(file_folder)
self.absolute_path.append(path.replace(self.main_path + "/", "") + "/" + file_folder)
self.relative_path.append(path + "/" + file_folder)
return resp, self.relative_path, self.absolute_path
@property
def get_all_files_folder(self):
self.resp = {self.main_path: {}}
all = self.get_files_and_folders(self.resp[self.main_path], self.main_path)
return all
if __name__ == '__main__':
mylib = GetDirectoryList(path="sample_folder")
file_list = mylib.get_all_files_folder
print (json.dumps(file_list))
而样本目录看起来像
sample_folder/
lib_a/
lib_c/
lib_e/
__init__.py
a.txt
__init__.py
b.txt
c.txt
lib_d/
__init__.py
__init__.py
d.txt
lib_b/
__init__.py
e.txt
__init__.py
结果
[
{
"files": [
"__init__.py"
],
"lib_b": {
"files": [
"__init__.py",
"e.txt"
]
},
"lib_a": {
"files": [
"__init__.py",
"d.txt"
],
"lib_c": {
"files": [
"__init__.py",
"c.txt",
"b.txt"
],
"lib_e": {
"files": [
"__init__.py",
"a.txt"
]
}
},
"lib_d": {
"files": [
"__init__.py"
]
}
}
},
[
"sample_folder/lib_b/__init__.py",
"sample_folder/lib_b/e.txt",
"sample_folder/__init__.py",
"sample_folder/lib_a/lib_c/lib_e/__init__.py",
"sample_folder/lib_a/lib_c/lib_e/a.txt",
"sample_folder/lib_a/lib_c/__init__.py",
"sample_folder/lib_a/lib_c/c.txt",
"sample_folder/lib_a/lib_c/b.txt",
"sample_folder/lib_a/lib_d/__init__.py",
"sample_folder/lib_a/__init__.py",
"sample_folder/lib_a/d.txt"
],
[
"lib_b/__init__.py",
"lib_b/e.txt",
"sample_folder/__init__.py",
"lib_a/lib_c/lib_e/__init__.py",
"lib_a/lib_c/lib_e/a.txt",
"lib_a/lib_c/__init__.py",
"lib_a/lib_c/c.txt",
"lib_a/lib_c/b.txt",
"lib_a/lib_d/__init__.py",
"lib_a/__init__.py",
"lib_a/d.txt"
]
]
其他回答
只列出目录
print("\nWe are listing out only the directories in current directory -")
directories_in_curdir = list(filter(os.path.isdir, os.listdir(os.curdir)))
print(directories_in_curdir)
只列出当前目录中的文件
files = list(filter(os.path.isfile, os.listdir(os.curdir)))
print("\nThe following are the list of all files in the current directory -")
print(files)
这个答案似乎并不存在。
directories = [ x for x in os.listdir('.') if os.path.isdir(x) ]
import os
path = "test/"
files = [x[0] + "/" + y for x in os.walk(path) if len(x[-1]) > 0 for y in x[-1]]
以Eli Bendersky的解决方案为基础,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"
>是要遍历的目录的路径。
最简单的方法:
from pathlib import Path
from glob import glob
current_dir = Path.cwd()
all_sub_dir_paths = glob(str(current_dir) + '/*/') # returns list of sub directory paths
all_sub_dir_names = [Path(sub_dir).name for sub_dir in all_sub_dir_paths]