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
我最近也遇到过类似的问题,我发现python 3.6的最佳答案(用户havlock添加的)是使用os.scandir。由于似乎没有使用它的解决方案,所以我将添加自己的解决方案。首先是一种非递归解决方案,它只列出根目录下的子目录。
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
递归的版本是这样的:
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
记住这一项。Path使用子目录的绝对路径。如果您只需要文件夹名称,您可以使用entry.name代替。参考os。DirEntry获取关于条目对象的其他详细信息。
其他回答
全路径,计算路径为。,.., \\, ..\\…\ \文件夹等:
import os, pprint
pprint.pprint([os.path.join(os.path.abspath(path), x[0]) \
for x in os.walk(os.path.abspath(path))])
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"
]
]
函数返回给定文件路径内所有子目录的List。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
这个答案似乎并不存在。
directories = [ x for x in os.listdir('.') if os.path.isdir(x) ]
import os
d = '.'
[os.path.join(d, o) for o in os.listdir(d)
if os.path.isdir(os.path.join(d,o))]