[英]how to get a list of sub directory names, the file names in each and the paths to each of these (Python)
我想在 python 中创建一个子目录列表。该列表将由列表组成。 每个子列表中的第一项是子目录名称及其作为元组的路径。 然后是该子目录中的文件及其路径。
示例:文件结构
assets/
location1/
file1.png
file2.png
file3.png
location2/
file4.png
file5.png
会返回:
[
[
('location1', 'assets/location1'),
('file1.png', 'assets/location1/file1.png'),
('file2.png', 'assets/location1/file2.png'),
('file3.png', 'assets/location1/file3.png')
],
[
('location2', 'assets/location2'),
('file4.png', 'assets/location2/file4.png'),
('file5.png', 'assets/location2/file5.png')
]
]
希望这是有道理的,提前感谢您的宝贵时间!
这是我过去用过的一个例子:
import os
import re
def crawler(path: str, ignore_hidden: bool = True) -> list[dict]:
"""
It crawls a directory and returns a list of dictionaries, each dictionary representing a file or
directory
Args:
path (str): The path to the directory you want to crawl.
ignore_hidden (bool): If True, ignore hidden files and directories. Defaults to True
Returns:
A list of dictionaries.
"""
files = []
for obj in os.listdir(path):
obj_path = os.path.normpath(os.path.join(path, obj))
file = {
"name": obj,
"path": os.path.relpath(obj_path),
}
pattern = r"^\.\w+$"
match = re.search(pattern, obj, re.IGNORECASE)
if match and ignore_hidden:
continue
if os.path.isfile(obj_path):
file["type"] = "file"
else:
file["type"] = "dir"
file["files"] = crawler(obj_path)
files.append(file)
return files
例如 output:
In [1]: crawler(os.getcwd())
Out[1]:
[{'name': 'about.txt', 'path': 'about.txt', 'type': 'file'},
{'name': 'android-chrome-192x192.png',
'path': 'android-chrome-192x192.png',
'type': 'file'},
{'name': 'android-chrome-512x512.png',
'path': 'android-chrome-512x512.png',
'type': 'file'},
{'name': 'apple-touch-icon.png',
'path': 'apple-touch-icon.png',
'type': 'file'},
{'name': 'favicon-16x16.png', 'path': 'favicon-16x16.png', 'type': 'file'},
{'name': 'favicon-32x32.png', 'path': 'favicon-32x32.png', 'type': 'file'},
{'name': 'favicon.ico', 'path': 'favicon.ico', 'type': 'file'},
{'name': 'New folder',
'path': 'New folder',
'type': 'dir',
'files': [{'name': 'New Text Document.txt',
'path': 'New folder\\New Text Document.txt',
'type': 'file'}]},
{'name': 'site.webmanifest', 'path': 'site.webmanifest', 'type': 'file'}]
你可以使用 python os function 你可以使用 os 路径和 os listdir 和 os 路径 isdir
# is function return all the subdirectories as you ask at the question
def get_subdirectories(directory):
subdir = []
for item in os.listdir(directory):
path = os.path.join(directory, item)
if os.path.isdir(path):
subdir.append([(item, path)])
subdir += get_subdirectories(path)
else:
subdir[-1].append((item, path))
return subdir
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