List of objects to JSON with Python

Question

I have a problem converting Object instances to JSON:

ob = Object()

list_name = scaping_myObj(base_url, u, number_page)

for ob in list_name:
   json_string = json.dumps(ob.__dict__)
   print json_string

In list_name I have a list of Object instances.

json_string return, for example:

{"city": "rouen", "name": "1, 2, 3 Soleil"}
{"city": "rouen", "name": "Maman, les p'tits bateaux"}

But I would like just 1 JSON string with all the info in a list:

[{"city": "rouen", "name": "1, 2, 3 Soleil"}, {"city": "rouen", "name": "Maman, les p'tits bateaux"}]

Here I have 2 Json so I want only one with all the info like [{"city": "rouen", "name": "1, 2, 3 Soleil"}, {"city": "rouen", "name": "Maman, les p'tits bateaux"}] — pedro
– pedro, Commented Sep 25, 2014 at 7:51

Martijn Pieters · Accepted Answer · 2019-09-19 15:25:00Z

149

You can use a list comprehension to produce a list of dictionaries, then convert that:

json_string = json.dumps([ob.__dict__ for ob in list_name])

or use a default function; json.dumps() will call it for anything it cannot serialise:

def obj_dict(obj):
    return obj.__dict__

json_string = json.dumps(list_name, default=obj_dict)

The latter works for objects inserted at any level of the structure, not just in lists.

Personally, I'd use a project like marshmallow to handle anything more complex; e.g. handling your example data could be done with

from marshmallow import Schema, fields

class ObjectSchema(Schema):
    city = fields.Str()
    name = fields.Str()

object_schema = ObjectSchema()
json_string = object_schema.dumps(list_name, many=True)

edited Sep 19, 2019 at 15:25

answered Sep 25, 2014 at 7:50

Martijn Pieters

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15 Comments

pedro Over a year ago

your first solution works perfectly. Thanks a lot :)

Matthias Over a year ago

That looks like a great solution. At the moment I'm trying to jsonify a complex data structure and this could be the way to go.

Roon13 Over a year ago

I haven't tried second solution Because your first solution worked for me :)

SkyLeach Over a year ago

It should be noted that deserialization in this way becomes tricky (subnested objects will be dicts on json.loads.

Martijn Pieters Apr 5 at 16:59

@alper: yes, but you could have trivially figured that out by looking at the documentation.

|

Eric Romrell · Accepted Answer · 2018-11-29 15:24:13Z

18

Similar to @MartijnPieters' answer, you can use the json.dumps default parameter with a lambda, if you don't want to have to create a separate function: json.dumps(obj, default = lambda x: x.__dict__)

answered Nov 29, 2018 at 15:24

Eric Romrell

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Comments

Steven · Accepted Answer · 2016-12-15 04:58:43Z

Another possible solution to this problem is jsonpickle which can be used to transform any Python object into JSON (not just simple lists).

From the jsonpickle home page:

jsonpickle is a Python library for serialization and deserialization of complex Python objects to and from JSON. The standard Python libraries for encoding Python into JSON, such as the stdlib’s json, simplejson, and demjson, can only handle Python primitives that have a direct JSON equivalent (e.g. dicts, lists, strings, ints, etc.). jsonpickle builds on top of these libraries and allows more complex data structures to be serialized to JSON. jsonpickle is highly configurable and extendable–allowing the user to choose the JSON backend and add additional backends.

Performing a transformation is simple:

import jsonpickle

class JsonTransformer(object):
    def transform(self, myObject):
        return jsonpickle.encode(myObject, unpicklable=False)

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