This question already has answers here:
What are the differences between type() and isinstance()?
(8 answers)
Closed 2 years ago.
From what I read googling, it seems that isinstanceof() is always better than type().
What are some situations when using type() is better than isinstanceof() in python?
I am using python 3.7.
They do two different things, you can't really compare them directly. What you've probably read is that you should prefer isinstance when checking the type of an object at runtime. But that isn't the only use-case for type (that is the use-case for isinstance, as its name implies).
What may not be obvious is that type is a class. You can think of "type" and "class" as synonymous. Indeed, it is the class of class objects, a metaclass. But it is a class just like int, float, list, dict etc. Or just like a use-defined class, class Foo: pass.
In its single argument form, it returns the class of whatever object you pass in. This is the form that can be used for type-checking. It is essentially equivalent to some_object.__class__.
>>> "a string".__class__
<class 'str'>
>>> type("a string")
<class 'str'>
Note:
>>> type(type) is type
True
You might also find this form useful if you ever wanted access to the type of an object itself for other reasons.
In its three-argument form, type(name, bases, namespace) it returns a new type object, a new class. Just like any other type constructor, just like list() returns a new list.
So instead of:
class Foo:
bar = 42
def __init__(self, val):
self.val = val
You could write:
def _foo_init(self, val):
self.val = val
Foo = type('Foo', (object,), {'bar':42, '__init__': _foo_init})
isinstance is a function which checks if... an object is an instance of some type. It is a function used for introspection.
When you want to introspect on the type of an object, usually you will probably use isintance(some_object, SomeType), but you might also use type(some_object) is SomeType. The key difference is that isinstance will return True if some_object.__class__ is precisely SomeType or any of the other types SomeType inherits from (i.e. in the method resolution order of SomeType, SomeType.mro()).
So, isinstance(some_object, SomeType) is essentially equivalent to some_object.__class__ is SomeType or some_object.__class__ in SomeType.mro()
Whereas if you use type(some_object) is SomeType, you are only asking some_object.__class__ is SomeType.
Here's a practical example of when you might want to use type instead of isinstance, suppose you wanted to distinguish between int and bool objects. In Python, bool inherits from int, so:
>>> issubclass(bool, int)
True
So that means:
>>> some_boolean = True
>>> isinstance(some_boolean, int)
True
but
>>> type(some_boolean) is int
False
type says the type of variable:
a = 10
type(a)
It will give its type as 'int'
isinstance() says if variable is related to specified type
class b:
def __init__(self):
print('Hi')
c = b()
m = isinstance(c, b)
It will return True because object c is of class type a otherwise it will return False.
As I know, subclass should be subclass of object, but why it also is an instance of object?
hmank ~ » python
Python 3.7.3 (default, Mar 26 2019, 21:43:19)
[GCC 8.2.1 20181127] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> class A:
... pass
...
>>> isinstance(A, object)
True
>>> issubclass(A, object)
True
>>>
What's more, I quote from #prosti answer that isinstance
Returns a Boolean stating whether the object is an instance or subclass of another object.
But the examples on the same link shows
>>> class Foo: pass
...
>>> class Bar(Foo): pass
...
>>> isinstance(Bar, Foo)
False
>
Seems it means a class1 is an instance of class2, if and only if class2 is object, right?
Not a bad question.
class A:
pass
a = A()
print(isinstance(a, A)) #True
print(isinstance(object, A)) #False
print(isinstance(A, object)) #True
print(issubclass(A, A)) #True
print(issubclass(A, object)) #True
By definition, isinstance:
Returns a Boolean stating whether the object is an instance or subclass of another object.
On the other hand issubclass:
Returns a Bool type indicating whether an object is a subclass of a class.
With additional remark that a class is considered a subclass of itself.
Update:
Seems it means a class1 is an instance of class2, if and only if class2 is object, right?
You get answers by testing, and logic is super simple. A class is a class and object is an instance of a class.
You can check the code in case you really need to understand the implementation.
Also you may find the test cases if you are a geek.
The object must be instantiated in order to classify for True in the following examples:
class Foo: pass
class Bar(Foo): pass
print(isinstance(Bar(), Foo)) #True
print(isinstance(Bar(), Bar)) #True
print(Bar) #<class '__main__.Bar'>
print(Bar()) #<__main__.Bar object at 0x7f9fc8f32828>
Also, some examples in here are Python3 specific, if you are Python2 guy, you must know that you should be more explicit and write:
class Bar(object): pass
The (object) part is a must if you write Python agnostic code.
Lastly check Standard on Overloading isinstance() and issubclass() but have in mind standards are "live" and may update in the future.
Lastly you may check this on classes objects relation.
Because everything in python is treated as an object, which is a nicer way to say, every object is an instance of object in python!
If you think about it, it makes total sence,since python is an object oriented language, and because of that, it is be normal and expected that every value is an object
In [9]: class A:
...: pass
...:
In [10]: isinstance(A, object)
Out[10]: True
In [11]: isinstance(A(), object)
Out[11]: True
In [12]: isinstance(1, object)
Out[12]: True
In [13]: isinstance([2,3], object)
Out[13]: True
In [14]: isinstance('hello', object)
Out[14]: True
In python3 all classes derive from object thus
class A(object):
pass
and
class A:
pass
are identical.
Regarding why isinstance(A, object) returns True
see the following code
class A:
#classmethod
def do(cls):
print(f"hello from {cls.__name__}")
A.do()
# hello from A
I would like to create a dictionary which based on a string keyword returns a subclass of Foo which I can later instantiate. Is this possible or is it an incorrect approach to the problem?
Pseudo code:
subclasses_of_foo = {"foo1": Foo1, "foo2": Foo2}
subclass_of_foo = subclasses_of_foo["foo1"]
instance = subclass_of_foo()
Sure, you can do that. Give it a go.
Your approach is entirely correct and works. Classes are just objects, just like everything else in Python. They can be stored as values in a dictionary.
Demo:
>>> class Foo:
... pass
...
>>> class Bar:
... pass
...
>>> classmap = {'foo': Foo, 'bar': Bar}
>>> classmap['foo']
<class __main__.Foo at 0x107eee1f0>
>>> classmap['foo']()
<__main__.Foo instance at 0x107eefcb0>
Note that duck typing is something else entirely; it is the practice of treating any object as the correct type provided it implements the attributes and methods you expected (if it walks like a duck, it is a duck).
This question already has answers here:
What's the canonical way to check for type in Python?
(15 answers)
Closed 6 months ago.
Is there a simple way to determine if a variable is a list, dictionary, or something else?
There are two built-in functions that help you identify the type of an object. You can use type() if you need the exact type of an object, and isinstance() to check an object’s type against something. Usually, you want to use isinstance() most of the times since it is very robust and also supports type inheritance.
To get the actual type of an object, you use the built-in type() function. Passing an object as the only parameter will return the type object of that object:
>>> type([]) is list
True
>>> type({}) is dict
True
>>> type('') is str
True
>>> type(0) is int
True
This of course also works for custom types:
>>> class Test1 (object):
pass
>>> class Test2 (Test1):
pass
>>> a = Test1()
>>> b = Test2()
>>> type(a) is Test1
True
>>> type(b) is Test2
True
Note that type() will only return the immediate type of the object, but won’t be able to tell you about type inheritance.
>>> type(b) is Test1
False
To cover that, you should use the isinstance function. This of course also works for built-in types:
>>> isinstance(b, Test1)
True
>>> isinstance(b, Test2)
True
>>> isinstance(a, Test1)
True
>>> isinstance(a, Test2)
False
>>> isinstance([], list)
True
>>> isinstance({}, dict)
True
isinstance() is usually the preferred way to ensure the type of an object because it will also accept derived types. So unless you actually need the type object (for whatever reason), using isinstance() is preferred over type().
The second parameter of isinstance() also accepts a tuple of types, so it’s possible to check for multiple types at once. isinstance will then return true, if the object is of any of those types:
>>> isinstance([], (tuple, list, set))
True
Use type():
>>> a = []
>>> type(a)
<type 'list'>
>>> f = ()
>>> type(f)
<type 'tuple'>
It might be more Pythonic to use a try...except block. That way, if you have a class which quacks like a list, or quacks like a dict, it will behave properly regardless of what its type really is.
To clarify, the preferred method of "telling the difference" between variable types is with something called duck typing: as long as the methods (and return types) that a variable responds to are what your subroutine expects, treat it like what you expect it to be. For example, if you have a class that overloads the bracket operators with getattr and setattr, but uses some funny internal scheme, it would be appropriate for it to behave as a dictionary if that's what it's trying to emulate.
The other problem with the type(A) is type(B) checking is that if A is a subclass of B, it evaluates to false when, programmatically, you would hope it would be true. If an object is a subclass of a list, it should work like a list: checking the type as presented in the other answer will prevent this. (isinstance will work, however).
On instances of object you also have the:
__class__
attribute. Here is a sample taken from Python 3.3 console
>>> str = "str"
>>> str.__class__
<class 'str'>
>>> i = 2
>>> i.__class__
<class 'int'>
>>> class Test():
... pass
...
>>> a = Test()
>>> a.__class__
<class '__main__.Test'>
Beware that in python 3.x and in New-Style classes (aviable optionally from Python 2.6) class and type have been merged and this can sometime lead to unexpected results. Mainly for this reason my favorite way of testing types/classes is to the isinstance built in function.
Determine the type of a Python object
Determine the type of an object with type
>>> obj = object()
>>> type(obj)
<class 'object'>
Although it works, avoid double underscore attributes like __class__ - they're not semantically public, and, while perhaps not in this case, the builtin functions usually have better behavior.
>>> obj.__class__ # avoid this!
<class 'object'>
type checking
Is there a simple way to determine if a variable is a list, dictionary, or something else? I am getting an object back that may be either type and I need to be able to tell the difference.
Well that's a different question, don't use type - use isinstance:
def foo(obj):
"""given a string with items separated by spaces,
or a list or tuple,
do something sensible
"""
if isinstance(obj, str):
obj = str.split()
return _foo_handles_only_lists_or_tuples(obj)
This covers the case where your user might be doing something clever or sensible by subclassing str - according to the principle of Liskov Substitution, you want to be able to use subclass instances without breaking your code - and isinstance supports this.
Use Abstractions
Even better, you might look for a specific Abstract Base Class from collections or numbers:
from collections import Iterable
from numbers import Number
def bar(obj):
"""does something sensible with an iterable of numbers,
or just one number
"""
if isinstance(obj, Number): # make it a 1-tuple
obj = (obj,)
if not isinstance(obj, Iterable):
raise TypeError('obj must be either a number or iterable of numbers')
return _bar_sensible_with_iterable(obj)
Or Just Don't explicitly Type-check
Or, perhaps best of all, use duck-typing, and don't explicitly type-check your code. Duck-typing supports Liskov Substitution with more elegance and less verbosity.
def baz(obj):
"""given an obj, a dict (or anything with an .items method)
do something sensible with each key-value pair
"""
for key, value in obj.items():
_baz_something_sensible(key, value)
Conclusion
Use type to actually get an instance's class.
Use isinstance to explicitly check for actual subclasses or registered abstractions.
And just avoid type-checking where it makes sense.
You can use type() or isinstance().
>>> type([]) is list
True
Be warned that you can clobber list or any other type by assigning a variable in the current scope of the same name.
>>> the_d = {}
>>> t = lambda x: "aight" if type(x) is dict else "NOPE"
>>> t(the_d) 'aight'
>>> dict = "dude."
>>> t(the_d) 'NOPE'
Above we see that dict gets reassigned to a string, therefore the test:
type({}) is dict
...fails.
To get around this and use type() more cautiously:
>>> import __builtin__
>>> the_d = {}
>>> type({}) is dict
True
>>> dict =""
>>> type({}) is dict
False
>>> type({}) is __builtin__.dict
True
be careful using isinstance
isinstance(True, bool)
True
>>> isinstance(True, int)
True
but type
type(True) == bool
True
>>> type(True) == int
False
While the questions is pretty old, I stumbled across this while finding out a proper way myself, and I think it still needs clarifying, at least for Python 2.x (did not check on Python 3, but since the issue arises with classic classes which are gone on such version, it probably doesn't matter).
Here I'm trying to answer the title's question: how can I determine the type of an arbitrary object? Other suggestions about using or not using isinstance are fine in many comments and answers, but I'm not addressing those concerns.
The main issue with the type() approach is that it doesn't work properly with old-style instances:
class One:
pass
class Two:
pass
o = One()
t = Two()
o_type = type(o)
t_type = type(t)
print "Are o and t instances of the same class?", o_type is t_type
Executing this snippet would yield:
Are o and t instances of the same class? True
Which, I argue, is not what most people would expect.
The __class__ approach is the most close to correctness, but it won't work in one crucial case: when the passed-in object is an old-style class (not an instance!), since those objects lack such attribute.
This is the smallest snippet of code I could think of that satisfies such legitimate question in a consistent fashion:
#!/usr/bin/env python
from types import ClassType
#we adopt the null object pattern in the (unlikely) case
#that __class__ is None for some strange reason
_NO_CLASS=object()
def get_object_type(obj):
obj_type = getattr(obj, "__class__", _NO_CLASS)
if obj_type is not _NO_CLASS:
return obj_type
# AFAIK the only situation where this happens is an old-style class
obj_type = type(obj)
if obj_type is not ClassType:
raise ValueError("Could not determine object '{}' type.".format(obj_type))
return obj_type
using type()
x='hello this is a string'
print(type(x))
output
<class 'str'>
to extract only the str use this
x='this is a string'
print(type(x).__name__)#you can use__name__to find class
output
str
if you use type(variable).__name__ it can be read by us
In many practical cases instead of using type or isinstance you can also use #functools.singledispatch, which is used to define generic functions (function composed of multiple functions implementing the same operation for different types).
In other words, you would want to use it when you have a code like the following:
def do_something(arg):
if isinstance(arg, int):
... # some code specific to processing integers
if isinstance(arg, str):
... # some code specific to processing strings
if isinstance(arg, list):
... # some code specific to processing lists
... # etc
Here is a small example of how it works:
from functools import singledispatch
#singledispatch
def say_type(arg):
raise NotImplementedError(f"I don't work with {type(arg)}")
#say_type.register
def _(arg: int):
print(f"{arg} is an integer")
#say_type.register
def _(arg: bool):
print(f"{arg} is a boolean")
>>> say_type(0)
0 is an integer
>>> say_type(False)
False is a boolean
>>> say_type(dict())
# long error traceback ending with:
NotImplementedError: I don't work with <class 'dict'>
Additionaly we can use abstract classes to cover several types at once:
from collections.abc import Sequence
#say_type.register
def _(arg: Sequence):
print(f"{arg} is a sequence!")
>>> say_type([0, 1, 2])
[0, 1, 2] is a sequence!
>>> say_type((1, 2, 3))
(1, 2, 3) is a sequence!
As an aside to the previous answers, it's worth mentioning the existence of collections.abc which contains several abstract base classes (ABCs) that complement duck-typing.
For example, instead of explicitly checking if something is a list with:
isinstance(my_obj, list)
you could, if you're only interested in seeing if the object you have allows getting items, use collections.abc.Sequence:
from collections.abc import Sequence
isinstance(my_obj, Sequence)
if you're strictly interested in objects that allow getting, setting and deleting items (i.e mutable sequences), you'd opt for collections.abc.MutableSequence.
Many other ABCs are defined there, Mapping for objects that can be used as maps, Iterable, Callable, et cetera. A full list of all these can be seen in the documentation for collections.abc.
value = 12
print(type(value)) # will return <class 'int'> (means integer)
or you can do something like this
value = 12
print(type(value) == int) # will return true
type() is a better solution than isinstance(), particularly for booleans:
True and False are just keywords that mean 1 and 0 in python. Thus,
isinstance(True, int)
and
isinstance(False, int)
both return True. Both booleans are an instance of an integer. type(), however, is more clever:
type(True) == int
returns False.
In general you can extract a string from object with the class name,
str_class = object.__class__.__name__
and using it for comparison,
if str_class == 'dict':
# blablabla..
elif str_class == 'customclass':
# blebleble..
For the sake of completeness, isinstance will not work for type checking of a subtype that is not an instance. While that makes perfect sense, none of the answers (including the accepted one) covers it. Use issubclass for that.
>>> class a(list):
... pass
...
>>> isinstance(a, list)
False
>>> issubclass(a, list)
True
What is the difference between type(obj) and obj.__class__? Is there ever a possibility of type(obj) is not obj.__class__?
I want to write a function that works generically on the supplied objects, using a default value of 1 in the same type as another parameter. Which variation, #1 or #2 below, is going to do the right thing?
def f(a, b=None):
if b is None:
b = type(a)(1) # #1
b = a.__class__(1) # #2
This is an old question, but none of the answers seems to mention that. in the general case, it IS possible for a new-style class to have different values for type(instance) and instance.__class__:
class ClassA(object):
def display(self):
print("ClassA")
class ClassB(object):
__class__ = ClassA
def display(self):
print("ClassB")
instance = ClassB()
print(type(instance))
print(instance.__class__)
instance.display()
Output:
<class '__main__.ClassB'>
<class '__main__.ClassA'>
ClassB
The reason is that ClassB is overriding the __class__ descriptor, however the internal type field in the object is not changed. type(instance) reads directly from that type field, so it returns the correct value, whereas instance.__class__ refers to the new descriptor replacing the original descriptor provided by Python, which reads the internal type field. Instead of reading that internal type field, it returns a hardcoded value.
Old-style classes are the problem, sigh:
>>> class old: pass
...
>>> x=old()
>>> type(x)
<type 'instance'>
>>> x.__class__
<class __main__.old at 0x6a150>
>>>
Not a problem in Python 3 since all classes are new-style now;-).
In Python 2, a class is new-style only if it inherits from another new-style class (including object and the various built-in types such as dict, list, set, ...) or implicitly or explicitly sets __metaclass__ to type.
type(obj) and type.__class__ do not behave the same for old style classes:
>>> class a(object):
... pass
...
>>> class b(a):
... pass
...
>>> class c:
... pass
...
>>> ai=a()
>>> bi=b()
>>> ci=c()
>>> type(ai) is ai.__class__
True
>>> type(bi) is bi.__class__
True
>>> type(ci) is ci.__class__
False
There's an interesting edge case with proxy objects (that use weak references):
>>> import weakref
>>> class MyClass:
... x = 42
...
>>> obj = MyClass()
>>> obj_proxy = weakref.proxy(obj)
>>> obj_proxy.x # proxies attribute lookup to the referenced object
42
>>> type(obj_proxy) # returns type of the proxy
weakproxy
>>> obj_proxy.__class__ # returns type of the referenced object
__main__.MyClass
>>> del obj # breaks the proxy's weak reference
>>> type(obj_proxy) # still works
weakproxy
>>> obj_proxy.__class__ # fails
ReferenceError: weakly-referenced object no longer exists
FYI - Django does this.
>>> from django.core.files.storage import default_storage
>>> type(default_storage)
django.core.files.storage.DefaultStorage
>>> default_storage.__class__
django.core.files.storage.FileSystemStorage
As someone with finite cognitive capacity who's just trying to figure out what's going in order to get work done... it's frustrating.