List of objects to JSON with Python

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鱼传尺愫
鱼传尺愫 2020-12-05 02:24

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:
          


        
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  • 2020-12-05 02:57

    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)
    
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  • 2020-12-05 03:03

    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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  • 2020-12-05 03:15

    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__)

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