Determine the type of an object?


Is there a simple way to determine if a variable is a list, dictionary, or something else?

Asked By: Justin Ethier



Use type():

>>> a = []
>>> type(a)
<type 'list'>
>>> f = ()
>>> type(f)
<type 'tuple'>
Answered By: brettkelly

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
>>> type({}) is dict
>>> type('') is str
>>> type(0) is int

This of course also works for custom types:

>>> class Test1 (object):
>>> class Test2 (Test1):
>>> a = Test1()
>>> b = Test2()
>>> type(a) is Test1
>>> type(b) is Test2

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

To cover that, you should use the isinstance function. This of course also works for built-in types:

>>> isinstance(b, Test1)
>>> isinstance(b, Test2)
>>> isinstance(a, Test1)
>>> isinstance(a, Test2)
>>> isinstance([], list)
>>> isinstance({}, dict)

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))
Answered By: poke

It might be more Pythonic to use a tryexcept 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).

Answered By: Seth Johnson

On instances of object you also have the:


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.

Answered By: Lorenzo Persichetti

You can use type() or isinstance().

>>> type([]) is list

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


To get around this and use type() more cautiously:

>>> import __builtin__
>>> the_d = {}
>>> type({}) is dict
>>> dict =""
>>> type({}) is dict
>>> type({}) is __builtin__.dict
Answered By: deed02392

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:

class Two:

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
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
Answered By: Alan Franzoni

As an aside to the previous answers, it’s worth mentioning the existence of 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

from 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

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

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)


  • 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.

be careful using isinstance

isinstance(True, bool)
>>> isinstance(True, int)

but type

type(True) == bool
>>> type(True) == int
Answered By: tnusraddinov

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)


isinstance(False, int)

both return True. Both booleans are an instance of an integer. type(), however, is more clever:

type(True) == int

returns False.

Answered By: Alec

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..

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

def say_type(arg):
    raise NotImplementedError(f"I don't work with {type(arg)}")

def _(arg: int):
    print(f"{arg} is an integer")

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 import Sequence

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!
Answered By: Georgy

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)
>>> issubclass(a, list)
Answered By: Adnan Y
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
Answered By: Yash Marmat

using type()

x='hello this is a string'


<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



if you use type(variable).__name__ it can be read by us

Answered By: vkv-onfire
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