"isinstance" question
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From: John Nagle on 22 Jun 2010 22:45 I want to test whether an object is an instance of any user-defined class. "isinstance" is less helpful than one would expect. >>> import types >>> class foo() : # define dummy class .... pass .... >>> x = foo() >>> >>> type(x) <type 'instance'> >>> >>> isinstance(x, types.ClassType) False >>> isinstance(x, types.InstanceType) True >>> foo <class __main__.foo at 0x004A2BD0> >>> x <__main__.foo instance at 0x020080A8> So far, so good. x is an InstanceType. But let's try a class with a constructor: >>> class bar(object) : .... def __init__(self, val) : .... self.val = val .... >>> b = bar(100) >>> b <__main__.bar object at 0x01FF50D0> >>> isinstance(b, types.InstanceType) False >>> isinstance(b, types.ClassType) False >>>>>> bar <class '__main__.bar'> Without a constructor, we get an "instance". With a constructor, we get an "object", one which is not an InstanceType. One might think that testing for types.ObjectType would help. But no, everything is an ObjectType: >>> isinstance(1, types.ObjectType) True >>> isinstance(None, types.ObjectType) True So that's useless. I have to be missing something obvious here. (CPython 2.6) John Nagle
From: Ben Finney on 22 Jun 2010 23:13 John Nagle <nagle(a)animats.com> writes: > I want to test whether an object is an instance of any user-defined > class. "isinstance" is less helpful than one would expect. Right. The type hierarchy is now unified; there's essentially no difference in later Python versions between user-defined types and built-in types. > So that's useless. For what purpose? > I have to be missing something obvious here. Maybe if we know what you're trying to do, we can suggest a better way. -- \ “Human reason is snatching everything to itself, leaving | `\ nothing for faith.” —Saint Bernard, 1090–1153 | _o__) | Ben Finney
From: John Nagle on 22 Jun 2010 23:58 On 6/22/2010 8:13 PM, Ben Finney wrote: > John Nagle<nagle(a)animats.com> writes: > >> I want to test whether an object is an instance of any user-defined >> class. "isinstance" is less helpful than one would expect. > > Right. The type hierarchy is now unified; there's essentially no > difference in later Python versions between user-defined types and > built-in types. OK, now that I know that, there's no problem. I'n doing something that involves program analysis through introspection. More on this later. John Nagle
From: Gabriel Genellina on 23 Jun 2010 01:06 En Tue, 22 Jun 2010 23:45:07 -0300, John Nagle <nagle(a)animats.com> escribi�: > I want to test whether an object is an instance of any user-defined > class. "isinstance" is less helpful than one would expect. > > >>> import types > >>> class foo() : # define dummy class > ... pass > ... > >>> x = foo() > >>> > >>> type(x) > <type 'instance'> > >>> > >>> isinstance(x, types.ClassType) > False > >>> isinstance(x, types.InstanceType) > True > >>> foo > <class __main__.foo at 0x004A2BD0> > >>> x > <__main__.foo instance at 0x020080A8> > > So far, so good. x is an InstanceType. But let's try a > class with a constructor: > > >>> class bar(object) : > ... def __init__(self, val) : > ... self.val = val > ... > >>> b = bar(100) > >>> b > <__main__.bar object at 0x01FF50D0> > >>> isinstance(b, types.InstanceType) > False > >>> isinstance(b, types.ClassType) > False > >>>>>> bar > <class '__main__.bar'> > > Without a constructor, we get an "instance". With a constructor, > we get an "object", one which is not an InstanceType. That's not the relevant difference. In the first case, you don't inherit from object; in the second one, you do. foo is a "classic" class (or "old-style" class); x is an instance of foo, its *type* is InstanceType, its *class* is foo. All instances of any other classic class have the same type (InstanceType). bar is a "new-style" class, b is an instance of bar, its type is bar, its class is bar. class and type are equivalent for new style classes; things are a lot more regular and predictable. In Python 3.x classic classes are gone. -- Gabriel Genellina
From: Thomas Jollans on 23 Jun 2010 05:50
On 06/23/2010 08:39 AM, Satish Eerpini wrote: > > > I want to test whether an object is an instance of any > user-defined > class. "isinstance" is less helpful than one would expect. > > >>> import types > >>> class foo() : # define dummy class > ... pass > ... > >>> x = foo() > >>> > >>> type(x) > <type 'instance'> > >>> > >>> isinstance(x, types.ClassType) > False > >>> isinstance(x, types.InstanceType) > True > >>> foo > <class __main__.foo at 0x004A2BD0> > >>> x > <__main__.foo instance at 0x020080A8> > > So far, so good. x is an InstanceType. But let's try a > class with a constructor: > > >>> class bar(object) : > ... def __init__(self, val) : > ... self.val = val > ... > >>> b = bar(100) > >>> b > <__main__.bar object at 0x01FF50D0> > >>> isinstance(b, types.InstanceType) > False > >>> isinstance(b, types.ClassType) > False > >>>>>> bar > <class '__main__.bar'> > > well the same code on my side returns true when you run isinstance(b, > types.InstanceType) even when the class has a constructor. Why is there > a difference in the output when we are both using Cython 2.6 ?? (2.6.4 > to be exact) I assume you mean CPython, not Cython. As Gabriel said, the difference is that bar is a subclass of object. If isinstance(bar(100), types.InstanceType), then you did not subclass object, and get an old-style class. |