Python determine if isintance without checking superclass

Question

class A:
   pass

class B(A):
   pass

How can I check that var is instance of B, not A, without referencing A. I have around 50 classes that subclass some class. I want to check whether instance is a child. Something like not isinstance(var, A).

Is there any way?

`B` specifically, or any subclass of `A`? (If any subclass of `A`, why do you want to avoid referring to `A` explicitly?) — user2357112, Oct 22 '17 at 17:29
Please clarify. To me it's not quite clear whether you want to check if it's an instance of `B` specifically, or of any class derived from `A` (but not `A` itself). What's the point of _"without referencing `A`"_ ? Have you considered implementing `A` as an _abstract base class_? — moooeeeep, Oct 22 '17 at 17:51

score 6 · Accepted Answer · answered Oct 23 '17 at 06:45

isinstance(object, classinfo) checks if the object is an instance of (or subclass of) the class given as second parameter.

type(object) returns the exact type of the object.

So you can check whether the object is an instance of a subclass of A, but not A itself, you can do something like the following.

if isinstance(obj, A) and not type(obj) == A:
    print("is subclass of A")
else:
    print("is either A or completely unrelated")

For reference:

score 0 · Answer 2 · answered Oct 22 '17 at 17:27

So I assume that B does something particular that A can't do. Ie it has a method that A doesn't. So one way to check that the instance is from B is to use the following.

if hasattr(my_instance, 'method_name_particular_to_b'):
    # do this

This will check if that method exists on my_instance if it does. You can treat it as an instance from B, otherwise treat it as an instance from A.

score 0 · Answer 3 · answered Oct 22 '17 at 17:36

It is really hard to use isinstance and judge whether a particular variable var is an instance of a subclass or superclass. What you can in fact do is specify a special property __class__ which can tell you about the instance's class.

for example:

class A:
    @property
    def __class__(self):
        return '__main__.A'

class B(A):
    @property
    def __class__(self):
        return '__main__.B'

b = B()
a = A()

print b.__class__
> '__main__.B'

print a.__class__
> '__main__.A'

If you however inherit class A with object you can check the behaviour without specifying these properties.

class A(object):
    pass

class B(A):
    pass

b = B()
a = A()

print b.__class__
> '__main__.B'

print a.__class__
> '__main__.A'

PS: I referred this SO question although it is in Java

score 0 · Answer 4 · answered Oct 22 '17 at 17:38

0

Not elegant, but seems working: not var.__class__.__name__=='A'

answered Oct 22 '17 at 17:38

Paul R

2,631
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@user2357112 That wont work. If `var = B()`, `not isinstance(var, A)` returns `False`. – pstatix Oct 22 '17 at 17:42
@pstatix: Bleh, right. Got mixed up there. Should be `var.__class__ is not A`, which is a solution the questioner probably wasn't aware of. – user2357112 Oct 22 '17 at 17:46
Is `var.__class__` better than `type(var)`? – Stefan Pochmann Oct 22 '17 at 17:47
1

@StefanPochmann: It handles old-style classes, which the questioner may be using. – user2357112 Oct 22 '17 at 17:47
1

(`isinstance(var, A) and var.__class__ is not A` to check specifically for subclasses and reject superclasses and unrelated classes.) – user2357112 Oct 22 '17 at 17:48

score 0 · Answer 5 · answered Oct 22 '17 at 17:38

0

You can find the type of an instance using __class__.

var.__class__ will give you the specific type of the object.

answered Oct 22 '17 at 17:38

salmanwahed

9,450
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Python determine if isintance without checking superclass

5 Answers5