Python type() or __class__, == or is

Question:

I want to test whether an object is an instance of a class, and only this class (no subclasses). I could do it either with:

obj.__class__ == Foo
obj.__class__ is Foo
type(obj) == Foo
type(obj) is Foo

Are there reasons to choose one over another? (performance differences, pitfalls, etc)

In other words: a) is there any practical difference between using __class__ and type(x)? b) are class objects always safe for comparison using is?


Update: Thanks all for the feedback. I’m still puzzled by whether or not class objects are singletons, my common sense says they are, but it’s been really hard to get a confirmation (try googling for “python”, “class” and “unique” or “singleton”).

I’d also like to clarify that, for my particular needs, the “cheaper” solution that just works is the best, since I’m trying to optimize the most out of a few, specialized classes (almost reaching the point where the sensible thing to do is to drop Python and develop that particular module in C). But the reason behind the question was to understand better the language, since some of its features are a bit too obscure for me to find that information easily. That’s why I’m letting the discussion extend a little instead of settling for __class__ is, so I can hear the opinion of more experienced people. So far it’s been very fruitful!

I ran a small test to benchmark the performance of the 4 alternatives. The profiler results were:

               Python  PyPy (4x)
type()    is   2.138   2.594
__class__ is   2.185   2.437
type()    ==   2.213   2.625
__class__ ==   2.271   2.453

Unsurprisingly, is performed better than == for all cases. type() performed better in Python (2% faster) and __class__ performed better in PyPy (6% faster). Interesting to note that __class__ == performed better in PyPy than type() is.


Update 2: many people don’t seem to understand what I mean with “a class is a singleton”, so I’ll ilustrate with an example:

>>> class Foo(object): pass
...
>>> X = Foo
>>> class Foo(object): pass
...
>>> X == Foo
False
>>> isinstance(X(), Foo)
False
>>> isinstance(Foo(), X)
False

>>> x = type('Foo', (object,), dict())
>>> y = type('Foo', (object,), dict())
>>> x == y
False
>>> isinstance(x(), y)
False

>>> y = copy.copy(x)
>>> x == y
True
>>> x is y
True
>>> isinstance(x(), y)
True
>>> y = copy.deepcopy(x)
>>> x == y
True
>>> x is y
True
>>> isinstance(x(), y)
True

It doesn’t matter if there are N objects of type type, given an object, only one will be its class, hence it’s safe to compare for reference in this case. And since reference comparison will always be cheaper than value comparison, I wanted to know whether or not my assertion above holds. I’m reaching the conclusion that it does, unless someone presents evidence in contrary.

Asked By: mgibsonbr

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Answers:

The result of type() is equivalent to obj.__class__ in new style classes, and class objects are not safe for comparison using is, use == instead.

For new style classes the preferable way here would be type(obj) == Foo.

As Michael Hoffman pointed out in his answer, there is a difference here between new and old style classes, so for backwards compatible code you may need to use obj.__class__ == Foo.

For those claiming that isinstance(obj, Foo) is preferable, consider the following scenario:

class Foo(object):
    pass

class Bar(Foo):
    pass

>>> obj = Bar()
>>> isinstance(obj, Foo)
True
>>> type(obj) == Foo
False

The OP wants the behavior of type(obj) == Foo, where it will be false even though Foo is a base class of Bar.

Answered By: Andrew Clark

is should only be used for identity checks, not type checks (there is an exception to the rule where you can and should use is for check against singletons).

Note: I would generally not use type and == for type checks, either. The preferable way for type checks is isinstance(obj, Foo). If you ever have a reason to check if something is not an subclass instance, it smells like a fishy design to me. When class Bar(Foo):, then Bar is a Foo, and you should be avoiding any situations where some part of your code has to work on a Foo instance but breaks on a Bar instance.

Answered By: wim

For old-style classes, there is a difference:

>>> class X: pass
... 
>>> type(X)
<type 'classobj'>
>>> X.__class__
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: class X has no attribute '__class__'
>>> x = X()
>>> x.__class__
<class __main__.X at 0x171b5d50>
>>> type(x)
<type 'instance'>

The point of new-style classes was to unify class and type. Technically speaking, __class__ is the only solution that will work both for new and old-style class instances, but it will also throw an exception on old-style class objects themselves. You can call type() on any object, but not every object has __class__. Also, you can muck with __class__ in a way you can’t muck with type().

>>> class Z(object):
...     def __getattribute__(self, name):
...             return "ham"
... 
>>> z = Z()
>>> z.__class__
'ham'
>>> type(z)
<class '__main__.Z'>

Personally, I usually have an environment with new-style classes only, and as a matter of style prefer to use type() as I generally prefer built-in functions when they exist to using magic attributes. For example, I would also prefer bool(x) to x.__nonzero__().

Answered By: Michael Hoffman

Update: Thanks all for the feedback. I’m still puzzled by whether or not class objects are singletons, my common sense says they are, but it’s been really hard to get a confirmation (try googling for “python”, “class” and “unique” or “singleton”).

I can confirm that __instance__ is a singleton. Here is the proof.

>>> t1=File.test()
made class
>>> t2=File.test()
made class
>>> print t1.__class__()
made class
<File.test object at 0x1101bdd10>
>>> print t2.__class__()
made class
<File.test object at 0x1101bdd10>

As you can see both t1 and t2 print out the same value in memory even though t1 and t2 are at different values in memory. Here is the proof of that.

>>> print t1
<File.test object at 0x1101bdc90>
>>> print t2
<File.test object at 0x1101bdcd0>

The __instance__ method only exists if you used class "name"(object):. If you use the classic style class, class "name":, than the __instance__ method doesn’t exist.

What this means is to be the most generic you probably want to use type unless you know for a fact instance does exist.

Answered By: Zachary Kraus
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