python – Check if argparse optional argument is set or not

I think that optional arguments (specified with --) are initialized to None if they are not supplied. So you can test with is not None. Try the example below:

import argparse

def main():
    parser = argparse.ArgumentParser(description=My Script)
    parser.add_argument(--myArg)
    args, leftovers = parser.parse_known_args()

    if args.myArg is not None:
        print myArg has been set (value is %s) % args.myArg

As @Honza notes is None is a good test. Its the default default, and the user cant give you a string that duplicates it.

You can specify another default=mydefaultvalue, and test for that. But what if the user specifies that string? Does that count as setting or not?

You can also specify default=argparse.SUPPRESS. Then if the user does not use the argument, it will not appear in the args namespace. But testing that might be more complicated:

parser.add_argument(--foo, default=argparse.SUPPRESS)

# ...

args.foo # raises an AttributeError
hasattr(args, foo)  # returns False
getattr(args, foo, other) # returns other

Internally the parser keeps a list of seen_actions, and uses it for required and mutually_exclusive testing. But it isnt available to you out side of parse_args.

python – Check if argparse optional argument is set or not

I think using the option default=argparse.SUPPRESS makes most sense. Then, instead of checking if the argument is not None, one checks if the argument is in the resulting namespace.

Example:

import argparse

parser = argparse.ArgumentParser()
parser.add_argument(--foo, default=argparse.SUPPRESS)
ns = parser.parse_args()

print(Parsed arguments: {}.format(ns))
print(foo in namespace?: {}.format(foo in ns))

Usage:

$ python argparse_test.py --foo 1
Parsed arguments: Namespace(foo=1)
foo in namespace?: True

Argument is not supplied:

$ python argparse_test.py
Parsed arguments: Namespace()
foo in namespace?: False