How to make a method into a property without using the @property decorator

Peter Otten __peter__ at web.de
Sat Oct 23 17:01:56 CEST 2010


Phlip wrote:

> Pythonistas:
> 
> Here's the property decorator:
> 
>    @property
>    def foo(self):  return 'bar'
> 
> If I generate foo dynamically, how to I make it a property?
> 
>   setattr(self, 'foo', property(lambda: 'bar'))
> 
> Variations of that are apparently not working.

You have to put the property descriptor into the class, not an instance:

>>> class A(object):
...     pass
...
>>> a = A()
>>> setattr(type(a), "foo", property(lambda self: "FOO"))
>>> a.foo
'FOO'
>>> A().foo
'FOO'

> (I'm heading for a proxy pattern, where if you never call the
> generated prop, you never hit the database to load it into memory.)

You may be better off with __getattr__().

Peter



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