How to check if a variable is a dictionary in Python? How to check if a variable is a dictionary in Python? python python

How to check if a variable is a dictionary in Python?

You could use if type(ele) is dict or use isinstance(ele, dict) which would work if you had subclassed dict:

d = {'abc': 'abc', 'def': {'ghi': 'ghi', 'jkl': 'jkl'}}for element in d.values():    if isinstance(element, dict):       for k, v in element.items():           print(k,' ',v)

How would you check if a variable is a dictionary in Python?

This is an excellent question, but it is unfortunate that the most upvoted answer leads with a poor recommendation, type(obj) is dict.

(Note that you should also not use dict as a variable name - it's the name of the builtin object.)

If you are writing code that will be imported and used by others, do not presume that they will use the dict builtin directly - making that presumption makes your code more inflexible and in this case, create easily hidden bugs that would not error the program out.

I strongly suggest, for the purposes of correctness, maintainability, and flexibility for future users, never having less flexible, unidiomatic expressions in your code when there are more flexible, idiomatic expressions.

is is a test for object identity. It does not support inheritance, it does not support any abstraction, and it does not support the interface.

So I will provide several options that do.

Supporting inheritance:

This is the first recommendation I would make, because it allows for users to supply their own subclass of dict, or a OrderedDict, defaultdict, or Counter from the collections module:

if isinstance(any_object, dict):

But there are even more flexible options.

Supporting abstractions:

from import Mappingif isinstance(any_object, Mapping):

This allows the user of your code to use their own custom implementation of an abstract Mapping, which also includes any subclass of dict, and still get the correct behavior.

Use the interface

You commonly hear the OOP advice, "program to an interface".

This strategy takes advantage of Python's polymorphism or duck-typing.

So just attempt to access the interface, catching the specific expected errors (AttributeError in case there is no .items and TypeError in case items is not callable) with a reasonable fallback - and now any class that implements that interface will give you its items (note .iteritems() is gone in Python 3):

try:    items = any_object.items()except (AttributeError, TypeError):    non_items_behavior(any_object)else: # no exception raised    for item in items: ...

Perhaps you might think using duck-typing like this goes too far in allowing for too many false positives, and it may be, depending on your objectives for this code.


Don't use is to check types for standard control flow. Use isinstance, consider abstractions like Mapping or MutableMapping, and consider avoiding type-checking altogether, using the interface directly.

The OP did not exclude the starting variable, so for completeness here is how to handle the generic case of processing a supposed dictionary that may include items as dictionaries.

Also following the pure Python(3.8) recommended way to test for dictionary in the above comments.

from import Mappingmy_dict = {'abc': 'abc', 'def': {'ghi': 'ghi', 'jkl': 'jkl'}}def parse_dict(in_dict):     if isinstance(in_dict, Mapping):        for k_outer, v_outer in in_dict.items():            if isinstance(v_outer, Mapping):                for k_inner, v_inner in v_outer.items():                    print(k_inner, v_inner)            else:                print(k_outer, v_outer)parse_dict(my_dict)