Python JSON序列化Decimal对象

I have a Decimal('3.9') as part of an object, and wish to encode this to a JSON string which should look like {'x': 3.9}. I don’t care about precision on the client side, so a float is fine.

Is there a good way to serialize this? JSONDecoder doesn’t accept Decimal objects, and converting to a float beforehand yields {'x': 3.8999999999999999} which is wrong, and will be a big waste of bandwidth.

How about subclassing json.JSONEncoder?

class DecimalEncoder(json.JSONEncoder):
    def default(self, o):
        if isinstance(o, decimal.Decimal):
            # wanted a simple yield str(o) in the next line,
            # but that would mean a yield on the line with super(...),
            # which wouldn't work (see my comment below), so...
            return (str(o) for o in [o])
        return super(DecimalEncoder, self).default(o)

Then use it like so:

json.dumps({'x': decimal.Decimal('5.5')}, cls=DecimalEncoder)

Simplejson 2.1 and higher has native support for Decimal type:

>>> json.dumps(Decimal('3.9'), use_decimal=True)
'3.9'

Note that use_decimal is True by default:

def dumps(obj, skipkeys=False, ensure_ascii=True, check_circular=True,
    allow_nan=True, cls=None, indent=None, separators=None,
    encoding='utf-8', default=None, use_decimal=True,
    namedtuple_as_object=True, tuple_as_array=True,
    bigint_as_string=False, sort_keys=False, item_sort_key=None,
    for_json=False, ignore_nan=False, **kw):

So:

>>> json.dumps(Decimal('3.9'))
'3.9'

Hopefully, this feature will be included in standard library.

I would like to let everyone know that I tried Michał Marczyk’s answer on my web server that was running Python 2.6.5 and it worked fine. However, I upgraded to Python 2.7 and it stopped working. I tried to think of some sort of way to encode Decimal objects and this is what I came up with:

import decimal

class DecimalEncoder(json.JSONEncoder):
    def default(self, o):
        if isinstance(o, decimal.Decimal):
            return float(o)
        return super(DecimalEncoder, self).default(o)

This should hopefully help anyone who is having problems with Python 2.7. I tested it and it seems to work fine. If anyone notices any bugs in my solution or comes up with a better way, please let me know.

In my Flask app, Which uses python 2.7.11, flask alchemy(with ‘db.decimal’ types), and Flask Marshmallow ( for ‘instant’ serializer and deserializer), i had this error, every time i did a GET or POST. The serializer and deserializer, failed to convert Decimal types into any JSON identifiable format.

I did a “pip install simplejson”, then Just by adding

import simplejson as json

the serializer and deserializer starts to purr again. I did nothing else… DEciamls are displayed as ‘234.00’ float format.

I tried switching from simplejson to builtin json for GAE 2.7, and had issues with the decimal. If default returned str(o) there were quotes (because _iterencode calls _iterencode on the results of default), and float(o) would remove trailing 0.

If default returns an object of a class that inherits from float (or anything that calls repr without additional formatting) and has a custom __repr__ method, it seems to work like I want it to.

import json
from decimal import Decimal

class fakefloat(float):
    def __init__(self, value):
        self._value = value
    def __repr__(self):
        return str(self._value)

def defaultencode(o):
    if isinstance(o, Decimal):
        # Subclass float with custom repr?
        return fakefloat(o)
    raise TypeError(repr(o) + " is not JSON serializable")

json.dumps([10.20, "10.20", Decimal('10.20')], default=defaultencode)
'[10.2, "10.20", 10.20]'

The native option is missing so I’ll add it for the next guy/gall that looks for it.

Starting on Django 1.7.x there is a built-in DjangoJSONEncoder that you can get it from django.core.serializers.json.

import json
from django.core.serializers.json import DjangoJSONEncoder
from django.forms.models import model_to_dict

model_instance = YourModel.object.first()
model_dict = model_to_dict(model_instance)

json.dumps(model_dict, cls=DjangoJSONEncoder)

Presto!

My $.02!

I extend a bunch of the JSON encoder since I am serializing tons of data for my web server. Here’s some nice code. Note that it’s easily extendable to pretty much any data format you feel like and will reproduce 3.9 as "thing": 3.9

JSONEncoder_olddefault = json.JSONEncoder.default
def JSONEncoder_newdefault(self, o):
    if isinstance(o, UUID): return str(o)
    if isinstance(o, datetime): return str(o)
    if isinstance(o, time.struct_time): return datetime.fromtimestamp(time.mktime(o))
    if isinstance(o, decimal.Decimal): return str(o)
    return JSONEncoder_olddefault(self, o)
json.JSONEncoder.default = JSONEncoder_newdefault

Makes my life so much easier…

3.9 can not be exactly represented in IEEE floats, it will always come as 3.8999999999999999, e.g. try print repr(3.9), you can read more about it here:

http://en.wikipedia.org/wiki/Floating_point
http://docs.sun.com/source/806-3568/ncg_goldberg.html

So if you don’t want float, only option you have to send it as string, and to allow automatic conversion of decimal objects to JSON, do something like this:

import decimal
from django.utils import simplejson

def json_encode_decimal(obj):
    if isinstance(obj, decimal.Decimal):
        return str(obj)
    raise TypeError(repr(obj) + " is not JSON serializable")

d = decimal.Decimal('3.5')
print simplejson.dumps([d], default=json_encode_decimal)

For Django users:

Recently came across TypeError: Decimal('2337.00') is not JSON serializable while JSON encoding i.e. json.dumps(data)

Solution:

# converts Decimal, Datetime, UUIDs to str for Encoding
from django.core.serializers.json import DjangoJSONEncoder  

json.dumps(response.data, cls=DjangoJSONEncoder)

But, now the Decimal value will be a string, now we can explicitly set the decimal/float value parser when decoding data, using parse_float option in json.loads:

import decimal 

data = json.loads(data, parse_float=decimal.Decimal) # default is float(num_str)

From the JSON Standard Document, as linked in json.org:

JSON is agnostic about the semantics of numbers. In any programming language, there can be a variety of number types of various capacities and complements, fixed or floating, binary or decimal. That can make interchange between different programming languages difficult. JSON instead offers only the representation of numbers that humans use: a sequence of digits. All programming languages know how to make sense of digit sequences even if they disagree on internal representations. That is enough to allow interchange.

So it’s actually accurate to represent Decimals as numbers (rather than strings) in JSON. Bellow lies a possible solution to the problem.

Define a custom JSON encoder:

import json


class CustomJsonEncoder(json.JSONEncoder):

    def default(self, obj):
        if isinstance(obj, Decimal):
            return float(obj)
        return super(CustomJsonEncoder, self).default(obj)

Then use it when serializing your data:

json.dumps(data, cls=CustomJsonEncoder)

As noted from comments on the other answers, older versions of python might mess up the representation when converting to float, but that’s not the case anymore.

To get the decimal back in Python:

Decimal(str(value))

This solution is hinted in Python 3.0 documentation on decimals:

To create a Decimal from a float, first convert it to a string.

This is what I have, extracted from our class

class CommonJSONEncoder(json.JSONEncoder):

    """
    Common JSON Encoder
    json.dumps(myString, cls=CommonJSONEncoder)
    """

    def default(self, obj):

        if isinstance(obj, decimal.Decimal):
            return {'type{decimal}': str(obj)}

class CommonJSONDecoder(json.JSONDecoder):

    """
    Common JSON Encoder
    json.loads(myString, cls=CommonJSONEncoder)
    """

    @classmethod
    def object_hook(cls, obj):
        for key in obj:
            if isinstance(key, six.string_types):
                if 'type{decimal}' == key:
                    try:
                        return decimal.Decimal(obj[key])
                    except:
                        pass

    def __init__(self, **kwargs):
        kwargs['object_hook'] = self.object_hook
        super(CommonJSONDecoder, self).__init__(**kwargs)

Which passes unittest:

def test_encode_and_decode_decimal(self):
    obj = Decimal('1.11')
    result = json.dumps(obj, cls=CommonJSONEncoder)
    self.assertTrue('type{decimal}' in result)
    new_obj = json.loads(result, cls=CommonJSONDecoder)
    self.assertEqual(new_obj, obj)

    obj = {'test': Decimal('1.11')}
    result = json.dumps(obj, cls=CommonJSONEncoder)
    self.assertTrue('type{decimal}' in result)
    new_obj = json.loads(result, cls=CommonJSONDecoder)
    self.assertEqual(new_obj, obj)

    obj = {'test': {'abc': Decimal('1.11')}}
    result = json.dumps(obj, cls=CommonJSONEncoder)
    self.assertTrue('type{decimal}' in result)
    new_obj = json.loads(result, cls=CommonJSONDecoder)
    self.assertEqual(new_obj, obj)

You can create a custom JSON encoder as per your requirement.

import json
from datetime import datetime, date
from time import time, struct_time, mktime
import decimal

class CustomJSONEncoder(json.JSONEncoder):
    def default(self, o):
        if isinstance(o, datetime):
            return str(o)
        if isinstance(o, date):
            return str(o)
        if isinstance(o, decimal.Decimal):
            return float(o)
        if isinstance(o, struct_time):
            return datetime.fromtimestamp(mktime(o))
        # Any other serializer if needed
        return super(CustomJSONEncoder, self).default(o)

The Decoder can be called like this,

import json
from decimal import Decimal
json.dumps({'x': Decimal('3.9')}, cls=CustomJSONEncoder)

and the output will be:

>>'{"x": 3.9}'

For those who don’t want to use a third-party library… An issue with Elias Zamaria’s answer is that it converts to float, which can run into problems. For example:

>>> json.dumps({'x': Decimal('0.0000001')}, cls=DecimalEncoder)
'{"x": 1e-07}'
>>> json.dumps({'x': Decimal('100000000000.01734')}, cls=DecimalEncoder)
'{"x": 100000000000.01733}'

The JSONEncoder.encode() method lets you return the literal json content, unlike JSONEncoder.default(), which has you return a json compatible type (like float) that then gets encoded in the normal way. The problem with encode() is that it (normally) only works at the top level. But it’s still usable, with a little extra work (python 3.x):

import json
from collections.abc import Mapping, Iterable
from decimal import Decimal

class DecimalEncoder(json.JSONEncoder):
    def encode(self, obj):
        if isinstance(obj, Mapping):
            return '{' + ', '.join(f'{self.encode(k)}: {self.encode(v)}' for (k, v) in obj.items()) + '}'
        if isinstance(obj, Iterable) and (not isinstance(obj, str)):
            return '[' + ', '.join(map(self.encode, obj)) + ']'
        if isinstance(obj, Decimal):
            return f'{obj.normalize():f}'  # using normalize() gets rid of trailing 0s, using ':f' prevents scientific notation
        return super().encode(obj)

Which gives you:

>>> json.dumps({'x': Decimal('0.0000001')}, cls=DecimalEncoder)
'{"x": 0.0000001}'
>>> json.dumps({'x': Decimal('100000000000.01734')}, cls=DecimalEncoder)
'{"x": 100000000000.01734}'

Based on stdOrgnlDave answer I have defined this wrapper that it can be called with optional kinds so the encoder will work only for certain kinds inside your projects. I believe the work should be done inside your code and not to use this “default” encoder since “it is better explicit than implicit”, but I understand using this will save some of your time. :-)

import time
import json
import decimal
from uuid import UUID
from datetime import datetime

def JSONEncoder_newdefault(kind=['uuid', 'datetime', 'time', 'decimal']):
    '''
    JSON Encoder newdfeault is a wrapper capable of encoding several kinds
    Use it anywhere on your code to make the full system to work with this defaults:
        JSONEncoder_newdefault()  # for everything
        JSONEncoder_newdefault(['decimal'])  # only for Decimal
    '''
    JSONEncoder_olddefault = json.JSONEncoder.default

    def JSONEncoder_wrapped(self, o):
        '''
        json.JSONEncoder.default = JSONEncoder_newdefault
        '''
        if ('uuid' in kind) and isinstance(o, uuid.UUID):
            return str(o)
        if ('datetime' in kind) and isinstance(o, datetime):
            return str(o)
        if ('time' in kind) and isinstance(o, time.struct_time):
            return datetime.fromtimestamp(time.mktime(o))
        if ('decimal' in kind) and isinstance(o, decimal.Decimal):
            return str(o)
        return JSONEncoder_olddefault(self, o)
    json.JSONEncoder.default = JSONEncoder_wrapped

# Example
if __name__ == '__main__':
    JSONEncoder_newdefault()

If you want to pass a dictionary containing decimals to the requests library (using the json keyword argument), you simply need to install simplejson:

$ pip3 install simplejson    
$ python3
>>> import requests
>>> from decimal import Decimal
>>> # This won't error out:
>>> requests.post('https://www.google.com', json={'foo': Decimal('1.23')})

The reason of the problem is that requests uses simplejson only if it is present, and falls back to the built-in json if it is not installed.

this can be done by adding

    elif isinstance(o, decimal.Decimal):
        yield str(o)

in \Lib\json\encoder.py:JSONEncoder._iterencode, but I was hoping for a better solution