Disable scientific notation in python json.dumps o

Question

json.dumps outputs small float or decimal values using scientific notation, which is unacceptable to the json-rpc application this output is sent to.

>>> import json
>>> json.dumps({"x": 0.0000001})
'{"x": 1e-07}'

I want this output instead:

'{"x": 0.0000001}'

It would be ideal to avoid introducing additional dependencies.

Answer 1

found somewhere on the internet

import json
from json import encoder
encoder.FLOAT_REPR = lambda o: format(o, '.2f')

after this

>>> json.dumps({'gps': [51.5, 17.5]})
'{"gps": [51.5, 17.5]}'

Answer 2

One way to format

evil = {"x": 0.00000000001}

is to steal Decimal's "f" formatter. It's the only easy way I've found that avoids both cropping problems and exponents, but it's not space efficient.

class FancyFloat(float):
    def __repr__(self):
        return format(Decimal(self), "f")

To use it you can make an encoder that "decimalize"s the input

class JsonRpcEncoder(json.JSONEncoder):
    def decimalize(self, val):
        if isinstance(val, dict):
            return {k:self.decimalize(v) for k,v in val.items()}

        if isinstance(val, (list, tuple)):
            return type(val)(self.decimalize(v) for v in val)

        if isinstance(val, float):
            return FancyFloat(val)

        return val

    def encode(self, val):
        return super().encode(self.decimalize(val))

JsonRpcEncoder().encode(evil)
#>>> '{"x": 0.00000000000999999999999999939496969281939810930172340963650867706746794283390045166015625}'

or, of course, you could move the decimalization out into a function and call that before json.dumps.

That's how I would do it, even if it's a lame method.