what if argparse input greater than metavar code example

Example 1: argeparse can it take a type list

import argparse

parser = argparse.ArgumentParser()

# By default it will fail with multiple arguments.
parser.add_argument('--default')

# Telling the type to be a list will also fail for multiple arguments,
# but give incorrect results for a single argument.
parser.add_argument('--list-type', type=list)

# This will allow you to provide multiple arguments, but you will get
# a list of lists which is not desired.
parser.add_argument('--list-type-nargs', type=list, nargs='+')

# This is the correct way to handle accepting multiple arguments.
# '+' == 1 or more.
# '*' == 0 or more.
# '?' == 0 or 1.
# An int is an explicit number of arguments to accept.
parser.add_argument('--nargs', nargs='+')

# To make the input integers
parser.add_argument('--nargs-int-type', nargs='+', type=int)

# An alternate way to accept multiple inputs, but you must
# provide the flag once per input. Of course, you can use
# type=int here if you want.
parser.add_argument('--append-action', action='append')

# To show the results of the given option to screen.
for _, value in parser.parse_args()._get_kwargs():
    if value is not None:
        print(value)

Example 2: argparse accept only few options

...
parser.add_argument('--val',
                    choices=['a', 'b', 'c'],
                    help='Special testing value')

args = parser.parse_args(sys.argv[1:])