How to convert Object with Properties to JSON without "_" in Python 3?

In Python, you'd normally not use properties for basic attributes. You'd leave name and age to be directly accessible attributes. There is no need to wrap those in property objects unless you need to transform the data when getting or setting.

If you have good reasons to use attributes with underscores but reflect them as JSON dictionaries, you can transform your dictionary when converting to a dictionary:

object_dict = lambda o: {key.lstrip('_'): value for key, value in o.__dict__.items()}
return json.dumps(self, default=object_dict, allow_nan=False, sort_keys=False, indent=4)

Note that this does nothing to prevent collisions. If you have both a _name and a name attribute on your instance, you'll clobber one or the other.


If the example code you posted mirrors your real code, there really isn't any reason for the properties at all. You could just do:

class User(object):
    def __init__(self):
        self.name = None
        self.age = None

since you're not really hiding anything from the user behind the underscores and properties anyway.

If you do need to do the transformation, I like to do it in a custom encoder:

class MyEncoder(json.JSONEncoder):
    def default(self, o):
        return {k.lstrip('_'): v for k, v in vars(o).items()}

json_encoded_user = json.dumps(some_user, cls=MyEncoder)

Tags:

Python

Json