改写代码: import os import re folder_path = "lab3" file_names = os.listdir(folder_path) answer = "" for file_name in file_names: file_path = os.path.join(folder_path, file_name) with open(file_path, "r") as file: content = file.read() pattern = answer + ".*?(.)(?!" + answer + ").*" diff_chars = re.findall(pattern, content) for char in diff_chars: answer += char print(answer)
时间: 2024-04-10 07:28:48 浏览: 17
import os
import re
folder_path = "lab3"
file_names = os.listdir(folder_path)
answer = ""
for file_name in file_names:
file_path = os.path.join(folder_path, file_name)
with open(file_path, "r") as file:
content = file.read()
pattern = answer + ".*?(.)(?!" + answer + ").*"
diff_chars = re.findall(pattern, content)
for char in diff_chars:
answer += char
print(answer)
相关问题
file_names = os.listdir(folder_path)
您好!对于您的问题,`os.listdir(folder_path)` 是一个用于获取指定文件夹路径下所有文件名的函数。它将返回一个包含文件名的列表。您可以将 `folder_path` 替换为您想要获取文件列表的文件夹的路径。例如,如果您想获取名为 `folder` 的文件夹中的所有文件名,您可以使用以下代码:
```python
import os
folder_path = 'path/to/folder'
file_names = os.listdir(folder_path)
print(file_names)
```
这将打印出 `folder_path` 文件夹中的所有文件名列表。希望这可以帮助到您!如果有任何其他问题,请随时提问。
import os folder_list = ['/path/to/folder1', '/path/to/folder2', '/path/to/folder3'] for folder_path in folder_list: for sub_folder in os.listdir(folder_path): sub_folder_path = os.path.join(folder_path, sub_folder) if os.path.isdir(sub_folder_path): for file_name in os.listdir(sub_folder_path): file_path = os.path.join(sub_folder_path, file_name) if os.path.isfile(file_path): print('{}\t{}'.format(os.path.basename(folder_path), file_name))在这个程序输出结果的基础上,改写,输出结果的上一级文件夹名
可以将 `os.path.basename(folder_path)` 改为 `os.path.basename(os.path.dirname(folder_path))`,这样就可以输出结果的上一级文件夹名了。改写后的程序如下:
```
import os
folder_list = ['/path/to/folder1', '/path/to/folder2', '/path/to/folder3']
for folder_path in folder_list:
for sub_folder in os.listdir(folder_path):
sub_folder_path = os.path.join(folder_path, sub_folder)
if os.path.isdir(sub_folder_path):
for file_name in os.listdir(sub_folder_path):
file_path = os.path.join(sub_folder_path, file_name)
if os.path.isfile(file_path):
print('{}\t{}'.format(os.path.basename(os.path.dirname(folder_path)), file_name))
```
注意,这里的 `/path/to/folder1` 等路径需要根据实际情况进行修改。