ArgumentParser: Optional Argument With Optional Value

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.Argument

Solution 1:

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]

Solution 2:

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.

Solution 3:

Not sure if this is the standard way, but you can set default argument to something , and then that value would be used in case --abc is not in the argument list.

Example code -

import argparse

parser = argparse.ArgumentParser()
parser.add_argument('--abc', nargs='?', default="-1")
args = parser.parse_args()
print(args)

Result -

>python a.py
Namespace(abc='-1')

>python a.py --abc
Namespace(abc=None)

>python a.py --abc something
Namespace(abc='something')