Get basename of a Windows path in Linux

Question

Assume I have a string that hold a Windows file address, let's say

local_address = "C:\\TEMP\\filename.txt"

to retrieve the filename from the address above I use

import os
filename = os.path.basename(local_address)    

In Windows, when I run the code the output is

>>> print filename
filename.txt

But when running the code in linux I get

>>> print filename
C:\TEMP\filename.txt

The reason is (what I think is) that when the Linux implementation of Python expects Linux local file address formats and has no idea about Windows addresses. Letting manual parsing of the address alone, are there other solutions so that I get uniform results?

Answer 1

Python have multiple implementations of os.path module, and if you're lucky, your python may be equipped with ntpath module:

>>> import ntpath
>>> ntpath.basename(r'C:\TEMP\filename.txt')
'filename.txt'

According to os.path documentation:

Since different operating systems have different path name conventions, there are several versions of this module in the standard library. The os.path module is always the path module suitable for the operating system Python is running on, and therefore usable for local paths. However, you can also import and use the individual modules if you want to manipulate a path that is always in one of the different formats. They all have the same interface:

• posixpath for UNIX-style paths
• ntpath for Windows paths

Answer 2

• On windows, the file separator (os.sep) is backslash (\). Slash is also accepted (Windows OS functions accept it)
• On Linux, the file separator (os.sep) is slash (/). Backslash is not accepted.

what would work would be os.path.basename(local_address.replace('\\',os.sep))

to turn the backslashes into slashes so basename can process it (and it would also work on windows: you'd replace something by the same thing)