Argparse: How To Separate Unknown(and Optional) Args When Subparsers Are Present.(subparsers Are Also Optional)

I have the following code parser = argparse.ArgumentParser(allow_abbrev=False, add_help=False) parser.add_argument('--conf', nargs=1) parser.add_argument('-h', '--help', nargs='?',

Solution 1:

As I wrote in a comment, subparsers is a positional argument.

To illustrate with a plain positional:

In [307]: parser = argparse.ArgumentParser()
In [308]: a1 = parser.add_argument('foo')

In [309]: parser.parse_known_args(['one','two'])
Out[309]: (Namespace(foo='one'), ['two'])

'one' is allocated to the first positional. Now give foo choices:

In [310]: a1.choices = ['bar','test']
In [311]: parser.parse_known_args(['one','two'])
usage: ipython3 [-h] {bar,test}
ipython3: error: argument foo: invalid choice: 'one' (choose from 'bar', 'test')

It is still trying to allocate the first string to foo. Since it doesn't match choices, it raises an error.

In [312]: parser.parse_known_args(['bar','one','two'])
Out[312]: (Namespace(foo='bar'), ['one', 'two'])

Strings are assigned to positionals based on position, not on value. Any value checking, such as with type or choices is done after assignment.

Change the choices to a type test:

In [313]: a1.choices = None
In [314]: a1.type = int
In [315]: parser.parse_known_args(['bar','one','two'])
usage: ipython3 [-h] foo
ipython3: error: argument foo: invalid int value: 'bar'

In [316]: parser.parse_known_args(['12','one','two'])
Out[316]: (Namespace(foo=12), ['one', 'two'])