If I have an optional argument with optional argument value, is there a way to validate if the argument is set when the value is not given?
For instance:
parser = argparse.ArgumentParser()
parser.add_argument('--abc', nargs='?')
args = parser.parse_args()
Would correctly give me:
optional arguments:
--abc [ABC]
How do I distinguish between 1 and 2 below?
...
Update:
Found a trick to solve this problem: you can use "nargs='*'" instead of "nargs='?'". This way #1 would return None, and #2 would return an empty list. The downside is this will allow multiple values for the arguments to be accepted too; so you'd need to add a check for it if appropriate.
Alternatively you can also set a default value for the argument; see answer from chepner and Anand S Kumar.
With nargs='?'
, you can supply both a default
and const
.
In [791]: parser=argparse.ArgumentParser()
In [792]: parser.add_argument('--abc', nargs='?', default='default', const='const')
If the argument is not given it uses the default:
In [793]: parser.parse_args([])
Out[793]: Namespace(abc='default')
If given, but without an argument string, it uses the const:
In [794]: parser.parse_args(['--abc'])
Out[794]: Namespace(abc='const')
Otherwise it uses the argument string:
In [795]: parser.parse_args(['--abc','test'])
Out[795]: Namespace(abc='test')
In [796]: parser.print_help()
usage: ipython3 [-h] [--abc [ABC]]
optional arguments:
-h, --help show this help message and exit
--abc [ABC]
Use a different default value for the option. Compare
>>> parser = argparse.ArgumentParser()
>>> parser.add_argument('--abc', nargs='?', default="default")
>>> parser.parse_args()
Namespace(abc='default')
>>> parser.parse_args(['--abc'])
Namespace(abc=None)
>>> parser.parse_args(['--abc', 'value'])
Namespace(abc='value')
I'm not sure how you would provide a different value for when --abc
is used without an argument, short of using a custom action instead of the nargs
argument.
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