How to make a class JSON serializable
How to make a Python class serializable?
A simple class:
class FileItem:
def __init__(self, fname):
self.fname = fname
What should I do to be able to get output of:
>>> import json
>>> my_file = FileItem('/foo/bar')
>>> json.dumps(my_file)
TypeError: Object of type 'FileItem' is not JSON serializable
Without the error
Here is a simple solution for a simple feature:
.toJSON()
Method
Instead of a JSON serializable class, implement a serializer method:
import json
class Object:
def toJSON(self):
return json.dumps(self, default=lambda o: o.__dict__,
sort_keys=True, indent=4)
So you just call it to serialize:
me = Object()
me.name = "Onur"
me.age = 35
me.dog = Object()
me.dog.name = "Apollo"
print(me.toJSON())
will output:
{
"age": 35,
"dog": {
"name": "Apollo"
},
"name": "Onur"
}
Do you have an idea about the expected output? For example, will this do?
>>> f = FileItem("/foo/bar")
>>> magic(f)
'{"fname": "/foo/bar"}'
In that case you can merely call json.dumps(f.__dict__)
.
If you want more customized output then you will have to subclass JSONEncoder
and implement your own custom serialization.
For a trivial example, see below.
>>> from json import JSONEncoder
>>> class MyEncoder(JSONEncoder):
def default(self, o):
return o.__dict__
>>> MyEncoder().encode(f)
'{"fname": "/foo/bar"}'
Then you pass this class into the json.dumps()
method as cls
kwarg:
json.dumps(cls=MyEncoder)
If you also want to decode then you'll have to supply a custom object_hook
to the JSONDecoder
class. For example:
>>> def from_json(json_object):
if 'fname' in json_object:
return FileItem(json_object['fname'])
>>> f = JSONDecoder(object_hook = from_json).decode('{"fname": "/foo/bar"}')
>>> f
<__main__.FileItem object at 0x9337fac>
>>>
For more complex classes you could consider the tool jsonpickle:
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.
(link to jsonpickle on PyPi)
Most of the answers involve changing the call to json.dumps(), which is not always possible or desirable (it may happen inside a framework component for example).
If you want to be able to call json.dumps(obj) as is, then a simple solution is inheriting from dict:
class FileItem(dict):
def __init__(self, fname):
dict.__init__(self, fname=fname)
f = FileItem('tasks.txt')
json.dumps(f) #No need to change anything here
This works if your class is just basic data representation, for trickier things you can always set keys explicitly.
I like Onur's answer but would expand to include an optional toJSON()
method for objects to serialize themselves:
def dumper(obj):
try:
return obj.toJSON()
except:
return obj.__dict__
print json.dumps(some_big_object, default=dumper, indent=2)