优化这段代码并说出你优化的点: 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 12:28:48 浏览: 16
优化的代码如下:
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 = f"{answer}.*?(.)(?!{answer}).*"
diff_chars = re.findall(pattern, content)
answer += "".join(diff_chars)
print(answer)
优化点:
1. 使用 f-string 格式化字符串,使代码更简洁易读。
2. 将字符串拼接操作改为使用 join() 方法,提高效率。
3. 修复代码缩进错误,保持代码的正确性。
相关问题
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)
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)
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。