Question :
I have a python object which includes some decimals. This is causing the json.dumps() to break.
I got the following solution from SO (e.g. Python JSON serialize a Decimal object) but
the recoomended solution still does not work. Python website – has the exact same answer.
Any suggestions how to make this work?
Thanks. Below is my code. It looks like the dumps() doesn’t even go into the specialized encoder.
clayton@mserver:~/python> cat test1.py
import json, decimal
class DecimalEncoder(json.JSONEncoder):
def _iterencode(self, o, markers=None):
print "here we go o is a == ", type(o)
if isinstance(o, decimal.Decimal):
print "woohoo! got a decimal"
return (str(o) for o in [o])
return super(DecimalEncoder, self)._iterencode(o, markers)
z = json.dumps( {'x': decimal.Decimal('5.5')}, cls=DecimalEncoder )
print z
clayton@mserver:~/python> python test1.py
Traceback (most recent call last):
File "test1.py", line 11, in <module>
z = json.dumps( {'x': decimal.Decimal('5.5')}, cls=DecimalEncoder )
File "/home/clayton/python/Python-2.7.3/lib/python2.7/json/__init__.py", line 238, in dumps
**kw).encode(obj)
File "/home/clayton/python/Python-2.7.3/lib/python2.7/json/encoder.py", line 201, in encode
chunks = self.iterencode(o, _one_shot=True)
File "/home/clayton/python/Python-2.7.3/lib/python2.7/json/encoder.py", line 264, in iterencode
return _iterencode(o, 0)
File "/home/clayton/python/Python-2.7.3/lib/python2.7/json/encoder.py", line 178, in default
raise TypeError(repr(o) + " is not JSON serializable")
TypeError: Decimal('5.5') is not JSON serializable
clayton@mserver:~/python>
Answer #1:
It is not (no longer) recommended you create a subclass; the json.dump() and json.dumps() functions take a default function:
def decimal_default(obj):
if isinstance(obj, decimal.Decimal):
return float(obj)
raise TypeError
json.dumps({'x': decimal.Decimal('5.5')}, default=decimal_default)
Demo:
>>> def decimal_default(obj):
... if isinstance(obj, decimal.Decimal):
... return float(obj)
... raise TypeError
...
>>> json.dumps({'x': decimal.Decimal('5.5')}, default=decimal_default)
'{"x": 5.5}'
The code you found only worked on Python 2.6 and overrides a private method that is no longer called in later versions.
Answer #2:
I can’t believe that no one here talked about using simplejson, which supports deserialization of Decimal out of the box.
import simplejson
from decimal import Decimal
simplejson.dumps({"salary": Decimal("5000000.00")})
'{"salary": 5000000.00}'
simplejson.dumps({"salary": Decimal("1.1")+Decimal("2.2")-Decimal("3.3")})
'{"salary": 0.0}'
Answer #3:
If you’re using Django. There is a great class for Decimal and date fields:
https://docs.djangoproject.com/en/1.10/topics/serialization/#djangojsonencoder
To use it:
import json
from django.core.serializers.json import DjangoJSONEncoder
json.dumps(value, cls=DjangoJSONEncoder)
