how to make condition with if to check type of variables in python?

Question

I'm looking for a way that convert all my outputs of a program (that I don't know what the types of them is) to dictionary... I think about a solution like this:

if type(body) != dict:
    body = {'result': body}

I know it is wrong! Is there any way that if the type of a variable was not dictionary, ourselves make a dictionary with that ?!

wrong why? it does the job. maybe using `isinstance` allows to check for child classes of `dict` though — Jean-François Fabre, Jul 24 '17 at 07:14
https://stackoverflow.com/questions/1549801/differences-between-isinstance-and-type-in-python — perigon, Jul 24 '17 at 07:16
Well hey you can actually put a dictionary in a dictionary. But can you specify a little more what you want, and post more code if you have more? — Shinra tensei, Jul 24 '17 at 07:19
it's a big program in role of server that response to soap web services. i give wsdl and it's method, this program return answer of webservices on client side. and because of UI team i have to change the answer of web services to a correct json... @Shinra tensei — mahshid.r, Jul 24 '17 at 07:31

score 1 · Accepted Answer · answered Jul 24 '17 at 08:11

If your object isn't a dict but could be converted to a dict easily (e.g. list of pairs), you could use:

def make_a_dict(some_object):
    if isinstance(some_object, dict):
        return some_object
    else:
        try:
            return dict(some_object)
        except:
            return {'return': some_object}

print(make_a_dict({'a': 'b'}))
# {'a': 'b'}
print(make_a_dict([(1,2),(3,4)]))
# {1: 2, 3: 4}
print(make_a_dict(2))
# {'return': 2}

how to make condition with if to check type of variables in python?

1 Answers1