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_argument("--c")
args = parser.parse_args()

Now I want to do some calculations with a, b, and c. However, I find it tiresome to write args.a + args.b + args.c all the time.

Therefore, I'm extracting those variables:

a, b, c = [args.a, args.b, args.c]

Such that I can write a + b + c.

Is there a more elegant way of doing that?

Manual extraction gets very tedious and error prone when adding many arguments.

kindall · Accepted Answer · 2011-11-29 05:43:24Z

9

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. :-)

edited Nov 29, 2011 at 5:43

answered Nov 29, 2011 at 5:35

kindall

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3 Comments

Lemming Over a year ago

Thanks for the answer. It works, but the exec "" part feels like a very ugly hack. Why is that necessary?

Lemming Over a year ago

Correction: It does not work if the function contains nested functions. In this case the following error message pops up:

exec ""     SyntaxError: unqualified exec is not allowed in function 'main' it contains a nested function with free variables

kindall Over a year ago

Ick. I did not know that, but it makes sense. Closures require Python's standard local variable mechanism, which accesses locals by index. The use of the exec statement anywhere in a function forces Python to use an alternate method of accessing locals by name, because exec could define or modify locals (this is also why you can update locals() if exec is seen in the function, but not otherwise).

g.d.d.c · Accepted Answer · 2011-11-29 05:31:30Z

0

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.

answered Nov 29, 2011 at 5:31

g.d.d.c

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2 Comments

Michael Hoffman Over a year ago

You should not modify the return value of locals(). It is not guaranteed to have an effect, and usually doesn't in CPython. docs.python.org/library/functions.html#locals

Lemming Over a year ago

Seems that this doesn't work without the exec "" trick, that kindall mentioned.

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