如何从Python中的路径获取没有扩展名的文件名?

问题描述：

如何从 Python 中的路径获取不带扩展名的文件名?

How to get the filename without the extension from a path in Python?

例如，如果我有 /path/to/some/file.txt"，我会想要 file".

For instance, if I had "/path/to/some/file.txt", I would want "file".

答

获取不带扩展名的文件名:

Getting the name of the file without the extension:

import os
print(os.path.splitext("/path/to/some/file.txt")[0])

打印:

/path/to/some/file

重要说明:如果文件名有多个点，则仅删除最后一个后的扩展名.例如:

Important Note: If the filename has multiple dots, only the extension after the last one is removed. For example:

import os
print(os.path.splitext("/path/to/some/file.txt.zip.asc")[0])

打印:

/path/to/some/file.txt.zip

如果您需要处理这种情况，请参阅下面的其他答案.

See other answers below if you need to handle that case.