Python os.path.join on Windows

To be even more pedantic, the most python doc consistent answer would be:

mypath = os.path.join('c:', os.sep, 'sourcedir')

Since you also need os.sep for the posix root path:

mypath = os.path.join(os.sep, 'usr', 'lib')

Windows has a concept of current directory for each drive. Because of that, "c:sourcedir" means "sourcedir" inside the current C: directory, and you'll need to specify an absolute directory.

Any of these should work and give the same result, but I don't have a Windows VM fired up at the moment to double check:

"c:/sourcedir"
os.path.join("/", "c:", "sourcedir")
os.path.join("c:/", "sourcedir")

To be pedantic, it's probably not good to hardcode either / or \ as the path separator. Maybe this would be best?

mypath = os.path.join('c:%s' % os.sep, 'sourcedir')

or

mypath = os.path.join('c:' + os.sep, 'sourcedir')

The reason os.path.join('C:', 'src') is not working as you expect is because of something in the documentation that you linked to:

Note that on Windows, since there is a current directory for each drive, os.path.join("c:", "foo") represents a path relative to the current directory on drive C: (c:foo), not c:\foo.

As ghostdog said, you probably want mypath=os.path.join('c:\\', 'sourcedir')

For a system-agnostic solution that works on both Windows and Linux, no matter what the input path, one could use os.path.join(os.sep, rootdir + os.sep, targetdir)

On WIndows:

>>> os.path.join(os.sep, "C:" + os.sep, "Windows")
'C:\\Windows'

On Linux:

>>> os.path.join(os.sep, "usr" + os.sep, "lib")
'/usr/lib'