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

Question

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'

Asked By: sds

||

Source

Answer 1

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.

Answered By: Martijn Pieters

Answer 2

argparse documentation show various examples of what you’re trying to do:

import argparse
parser = argparse.ArgumentParser()
parser.add_argument("-a")
parser.parse_args(['-a 12'])
>>> Namespace(a=' 12')

Answered By: El Bert

Answer 3

It is now recommended to use SimpleNamespace from the types module. It does the same thing as the accepted answer except for it will be faster and have a few more builtins such as equals and repr.

from types import SimpleNamespace

sn = SimpleNamespace()
sn.a = 'test'
sn.a

# output
'test'

Answered By: Mitchell Walls

Answer 4

First create a dict and then use the dict to create an namespace:

from argparse import Namespace
x = {'a': 1, 'b': 2}
ns = Namespace(**x)
print(ns.a) #output 1

Answered By: Youjun Hu

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

Question:

Answers: