What is the difference between SimpleNamespace and empty class definition?

Question

The following seems to work either way. What is the advantage (other than the nice repr ) of using types.SimpleNamespace ? Or is it the same thing?

>>> import types
>>> class Cls():
...     pass
... 
>>> foo = types.SimpleNamespace() # or foo = Cls()
>>> foo.bar = 42
>>> foo.bar
42
>>> del foo.bar
>>> foo.bar
AttributeError: 'types.SimpleNamespace' object has no attribute 'bar'

Answer 1

This is explained pretty well in the types module description. It shows you that types.SimpleNamespace is roughly equivalent to this:

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

    def __repr__(self):
        keys = sorted(self.__dict__)
        items = ("{}={!r}".format(k, self.__dict__[k]) for k in keys)
        return "{}({})".format(type(self).__name__, ", ".join(items))

    def __eq__(self, other):
        return self.__dict__ == other.__dict__

This provides the following advantages over an empty class:

1. It allows you to initialize attributes while constructing the object: sn = SimpleNamespace(a=1, b=2)
2. It provides a readable repr() : eval(repr(sn)) == sn
3. It overrides the default comparison. Instead of comparing by id() , it compares attribute values instead.

Answer 2

A class types.SimpleNamespace provides a mechanism to instantiate an object that can hold attributes and nothing else. It is, in effect, an empty class with a fancier __init__() and a helpful __repr__() :

>>> from types import SimpleNamespace
>>> sn = SimpleNamespace(x = 1, y = 2)
>>> sn
namespace(x=1, y=2)
>>> sn.z = 'foo'
>>> del(sn.x)
>>> sn
namespace(y=2, z='foo')

or

from types import SimpleNamespace

sn = SimpleNamespace(x = 1, y = 2)
print(sn)

sn.z = 'foo'
del(sn.x)
print(sn)

output:

namespace(x=1, y=2)
namespace(y=2, z='foo')

---

This answer may also be helpful.