CODEWITHSUNDEEP
pythoncommand-line-argumentsargparse
Lemming
Lemming

Reputation: 4205

Python: Extract variables out of namespace

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.

Upvotes: 8

Views: 6715

Answers (2)

kindall
kindall

Reputation: 184101

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. :-)

Upvotes: 9

g.d.d.c
g.d.d.c

Reputation: 47978

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.

Upvotes: 0

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