Custom namespaces

>>> import __main__
>>> __main__.__dict__ = MyNamespace()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: readonly attribute

Why is __dict__ made read-only?

I next thought I could change the type of the namespace to my class:

>>> __main__.__dict__.__class__ = MyNamespace
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: __class__ assignment: only for heap types

Drat, foiled again!!!

Okay, if I can't do this at the module level, can I at least install a
custom namespace at the class level?

>>> class MyNamespace(dict):
... def __getitem__(self, key):
... print "Looking up key '%s'" % key
... return super(MyNamespace, self).__getitem__(key)
...
>>> namespace = MyNamespace(x=1, y=2, z=3)
>>> namespace['x']
Looking up key 'x'
1
>>> C = new.classobj("C", (object,), namespace)
>>> C.x
1

Apparently not. It looks like the namespace provided to the class
constructor gets copied when the class is made.

Interestingly enough, the namespace argument gets modified *before* it
gets copied, which has an unwanted side-effect:

>>> namespace
{'y': 2, 'x': 1, '__module__': '__main__', 'z': 3, '__doc__': None}

Is there any way to install a custom type as a namespace?

--
Steven

For classes/objects, yes, using metaclasses.
See the __prepare__() method in PEP 3115:
http://www.python.org/dev/peps/pep-3115/

Cheers,
Chris
--
http://blog.rebertia.com

Looks good, but that's Python 3 only, yes?

At least, I can't get the metaclass to change the __dict__ in Python 2.6.
There's obviously no __prepare__ before 3.0, but I tried the following,
and still __dict__ ends up as a regular dict. Am I missing something?

class VerboseDict(dict):
def __getitem__(self, item):
print ("Looking up key '%s'..." % item)
return super(VerboseDict, self).__getitem__(item)

class Meta(type):
def __new__(cls, name, bases, namespace):
namespace = VerboseDict(namespace)
obj = super(Meta, cls).__new__(cls, name, bases, namespace)
return obj

MyClass = Meta('MyClass', (object,), dict(x=1, y=2))

--
Steven

Correct.

I would /guess/ that type.__new__() is internally doing the equivalent
of dict(namespace). Hence why the addition of __prepare__() was
necessary; it's probably impossible to accomplish what you want in
earlier Python versions.

Here is an example of usage:

http://www.artima.com/weblogs/viewpost.jsp?thread=236234
http://www.artima.com/weblogs/viewpost.jsp?thread=236260
(yes, it only works for Python 3+)