Python: Extract variables out of namespace

Question

I\'m using argparse in python to parse commandline arguments:

parser = ArgumentParser()
parser.add_argument(\"--a\")
parser.add_argument(\"--b\")
parser.add_         


        

Answer 1

You can add things to the local scope by calling locals(). It returns a dictionary that represents the currently available scope. You can assign values to it as well - locals()['a'] = 12 will result in a being in the local scope with a value of 12.

Answer 2

If you want them as globals, you can do:

globals().update(vars(args))

If you're in a function and want them as local variables of that function, you can do this in Python 2.x as follows:

def foo(args):
   locals().update(vars(args))       
   print a, b, c
   return
   exec ""  # forces Python to use a dict for all local vars
            # does not need to ever be executed!  but assigning
            # to locals() won't work otherwise.

This trick doesn't work in Python 3, where exec is not a statement, nor likely in other Python variants such as Jython or IronPython.

Overall, though, I would recommend just using a shorter name for the args object, or use your clipboard. :-)