Python JSON serialize a Decimal object

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

Note that this will convert the decimal to its string representation (e.g.; "1.2300") to a. not lose significant digits and b. prevent rounding errors.

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.

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)