Python class instance dict doesn't contain all instance variables. Why?

Question

Here's a bit of a newbie Python question about instance variables.

Consider the following Python 2.7 class definition:

class Foo(object):
    a = 1

    def __init__(self):
            self.b = 2

    def __repr__(self):
            return "%s" % self.__dict__

Now, when I create an instance of Foo, Foo.__dict__ contains b, but not a.

>>> x=Foo()
>>> x
{'b': 2}
>>> dir(x)
['__class__', '__delattr__', '__dict__', '__doc__', '__format__', 
 '__getattribute__', '__hash__', '__init__', '__module__', '__new__', 
 '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__sizeof__',
 '__str__', '__subclasshook__', '__weakref__', 'a', 'b']
>>> x.__dict__
{'b': 2}

And here I thought I had a pretty good grasp on the Way of the Python.

What's the difference between x.a and x.b? As far as I can tell they're both instance variables.

Edit: OK, re-reading the Python docs I see that Foo.a is a class attribute rather than an instance variable. Hm... I guess the confusion comes from the fact that I can assign a new value to x.a and the new value only affects the x instance -- I suppose I'm now aliasing a member variable over the top of the Foo.a attribute:

>>> y=Foo()
>>> y.a = 2
>>> y
{'a': 2, 'b': 2}
>>> x
{'b': 2}
>>> x.a
1
>>> z=Foo()
>>> z
{'b': 2}
>>> z.a
1
>>> Foo.a
1
>>> x.a
1
>>> y.a
2

So, now I overwrite the previous value of Foo.a, and it affects all instances of Foo that haven't aliased Foo.a:

>>> Foo.a=999
>>> x.a
999
>>> y.a
2

Note also that it's not good practice to rely on __dict__ too much: clases using __slots__ and extension classes (from C or Cython) don't put everything in there either. Most uses can be replaced by some combination of getattr, setattr, hasattr, and dir. — Danica
– Danica, Commented May 12, 2012 at 19:36
@Dougal: Thanks for that advice -- I saw that on another answer here on StackOverflow, but it's good reinforcement. — Inactivist
– Inactivist, Commented May 12, 2012 at 20:01

Mark Byers · Accepted Answer · 2012-05-12 19:33:28Z

8

Your a isn't an instance variable. You defined it as part of the class.

>>> class Foo(object):
...    a = 1
...
>>> Foo.a
1

If you want an instance variable you should put it inside the __init__ method, because this method is called when your object is created.

answered May 12, 2012 at 19:33

Mark Byers

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Cat Plus Plus · Accepted Answer · 2012-05-12 19:33:20Z

4

a is not an instance attribute, it's a class attribute.

answered May 12, 2012 at 19:33

Cat Plus Plus

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3 Comments

Inactivist Over a year ago

Yes, but if I assign a new value to x.a it affects only the x instance. If it's a class attribute, why doesn't changing the value of x.a affect all instances of Foo?

kindall Over a year ago

Because you assigned it on the instance.

Inactivist Over a year ago

Yes, I see that now. I missed that on my initial read of the docs. Thanks for the assist!

User · Accepted Answer · 2012-05-12 19:54:01Z

0

May this help you further?

>>> class X(object):

    def __getattribute__(self, name):
        print name
        return object.__getattribute__(self, name)


>>> l = dir(X())
__dict__
__members__
__methods__
__class__
>>> l
['__class__', '__delattr__', '__dict__', '__doc__', '__format__', '__getattribute__', '__hash__', '__init__', '__module__', '__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__sizeof__', '__str__', '__subclasshook__', '__weakref__']

answered May 12, 2012 at 19:54

User

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