How do I create a Python namespace (argparse.parse_args value)?

To interactively test my python script, I would like to create a Namespace object, similar to what would be returned by argparse.parse_args(). The obvious way,

>>> import argparse
>>> parser = argparse.ArgumentParser()
>>> parser.parse_args()
Namespace()
>>> parser.parse_args("-a")
usage: [-h]
: error: unrecognized arguments: - a

Process Python exited abnormally with code 2

may result in Python repl exiting (as above) on a silly error.

So, what is the easiest way to create a Python namespace with a given set of attributes?

E.g., I can create a dict on the fly (dict([("a",1),("b","c")])) but I cannot use it as a Namespace:

AttributeError: 'dict' object has no attribute 'a'

PS. The new exit_on_error option looks like a promising alternative to creating a Namespace object, but it is severely broken, apparently by design.

Solution

You can create a simple class:

class Namespace:
    def __init__(self, **kwargs):
        self.__dict__.update(kwargs)

and it'll work the exact same way as the argparse Namespace class when it comes to attributes:

>>> args = Namespace(a=1, b='c')
>>> args.a
1
>>> args.b
'c'

Alternatively, just import the class; it is available from the argparse module:

from argparse import Namespace

args = Namespace(a=1, b='c')

As of Python 3.3, there is also types.SimpleNamespace, which essentially does the same thing:

>>> from types import SimpleNamespace
>>> args = SimpleNamespace(a=1, b='c')
>>> args.a
1
>>> args.b
'c'

The two types are distinct; SimpleNamespace is primarily used for the sys.implementation attribute and the return value of time.get_clock_info().

Further comparisons:

Both classes support equality testing; for two instances of the same class, instance_a == instance_b is true if they have the same attributes with the same values.
Both classes have a helpful __repr__ to show what attributes they have.
Namespace() objects support containment testing; 'attrname' in instance is true if the namespace instance has an attribute namend attrname. SimpleNamespace does not.
Namespace() objects have an undocumented ._get_kwargs() method that returns a sorted list of (name, value) attributes for that instance. You can get the same for either class using sorted(vars(instance).items()).
While SimpleNamespace() is implemented in C and Namespace() is implemented in Python, attribute access is no faster because both use the same __dict__ storage for the attributes. Equality testing and producing the representation are a little faster for SimpleNamespace() instances.