Python os.path.join() on a list

The problem is, os.path.join doesn't take a list as argument, it has to be separate arguments.

This is where *, the 'splat' operator comes into play...

I can do

>>> s = "c:/,home,foo,bar,some.txt".split(",")
>>> os.path.join(*s)
'c:/home\\foo\\bar\\some.txt'

Assuming join wasn't designed that way (which it is, as ATOzTOA pointed out), and it only took two parameters, you could still use the built-in reduce:

>>> reduce(os.path.join,["c:/","home","foo","bar","some.txt"])
'c:/home\\foo\\bar\\some.txt'

Same output like:

>>> os.path.join(*["c:/","home","foo","bar","some.txt"])
'c:/home\\foo\\bar\\some.txt' 

Just for completeness and educational reasons (and for other situations where * doesn't work).

Hint for Python 3

reduce was moved to the functools module.

It's just the method. You're not missing anything. The official documentation shows that you can use list unpacking to supply several paths:

s = "c:/,home,foo,bar,some.txt".split(",")
os.path.join(*s)

Note the *s intead of just s in os.path.join(*s). Using the asterisk will trigger the unpacking of the list, which means that each list argument will be supplied to the function as a separate argument.

I stumbled over the situation where the list might be empty. In that case:

os.path.join('', *the_list_with_path_components)

Note the first argument, which will not alter the result.