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How to get an exact representation of floats during `DataFrame.to_json`?

I observed the following behavior with DataFrame.to_json:

>>> df = pd.DataFrame([[eval(f'1.12345e-{i}') for i in range(8, 20)]])
>>> df
             0             1             2             3             4             5             6             7             8             9             10            11
0  1.123450e-08  1.123450e-09  1.123450e-10  1.123450e-11  1.123450e-12  1.123450e-13  1.123450e-14  1.123450e-15  1.123450e-16  1.123450e-17  1.123450e-18  1.123450e-19
>>> print(df.to_json(indent=2, orient='index'))
{
  "0":{
    "0":0.0000000112,
    "1":0.0000000011,
    "2":0.0000000001,
    "3":0.0,
    "4":0.0,
    "5":0.0,
    "6":0.0,
    "7":0.0,
    "8":1.12345e-16,
    "9":1.12345e-17,
    "10":1.12345e-18,
    "11":1.12345e-19
  }
}

So all numbers down to 1e-16 seem to be rounded to 10 decimal places (in agreement with the default value for double_precision) but all smaller values are represented exactly. Why is this the case and how can I turn off decimal rounding for the larger values too (i.e. using scientific notation instead)?


>>> pd.__version__
'1.3.1'

For reference, the standard library's json module doesn't do this:

>>> import json
>>> print(json.dumps([eval(f'1.12345e-{i}') for i in range(8, 20)], indent=2))
[
  1.12345e-08,
  1.12345e-09,
  1.12345e-10,
  1.12345e-11,
  1.12345e-12,
  1.12345e-13,
  1.12345e-14,
  1.12345e-15,
  1.12345e-16,
  1.12345e-17,
  1.12345e-18,
  1.12345e-19
]
over 4 years ago · Santiago Trujillo
3 Respuestas
Responde la pregunta

0

It refers to /pandas/io/json/_json.py codebase, by default the precision integer is upto 10, please see the below from the codebase..

def to_json(
    path_or_buf,
    obj,
    orient: Optional[str] = None,
    date_format: str = "epoch",
    double_precision: int = 10,
    force_ascii: bool = True,
    date_unit: str = "ms",
    default_handler: Optional[Callable[[Any], JSONSerializable]] = None,
    lines: bool = False,
    compression: Optional[str] = "infer",
    index: bool = True,
    indent: int = 0,

If you apply max precision then, you will be getting below..

>>> print(df.to_json(indent=2, orient='records', double_precision=15))
[
  {
    "0":0.0000000112345,
    "1":0.00000000112345,
    "2":0.000000000112345,
    "3":0.000000000011234,
    "4":0.000000000001123,
    "5":0.000000000000112,
    "6":0.000000000000011,
    "7":0.000000000000001,
    "8":1.12345e-16,
    "9":1.12345e-17,
    "10":1.12345e-18,
    "11":1.12345e-19,
    "12":1.12345e-20,
    "13":1.12345e-21,
    "14":1.12345e-22,
    "15":1.12345e-23,
    "16":1.12345e-24,
    "17":1.12345e-25,
    "18":1.12345e-26,
    "19":1.12345e-27,
    "20":1.12345e-28,
    "21":1.12345e-29,
    "22":1.12345e-30,
    "23":1.12345e-31,
    "24":1.12345e-32,
    "25":1.12345e-33,
    "26":1.12345e-34,
    "27":1.12345e-35,
    "28":1.12345e-36,
    "29":1.12345e-37,
    "30":1.12345e-38,
    "31":1.12345e-39
  }
]

Note: you will get value error if you use precision more than 15.

ValueError: Invalid value '20' for option 'double_precision', max is '15'

So, in a sense this is not same as json.dumps.

over 4 years ago · Santiago Trujillo Denunciar

0

I'm not sure on achieving this with pd.DataFrame.to_json, but we can use pd.DataFrame.to_dict, json, and pd.read_json to achieve a full precision json representation from a pandas dataframe.

json_df = json.dumps(df.to_dict('index'), indent=2)
>>> print(json_df)
{
  "0": {
    "0": 1.12345e-08,
    "1": 1.12345e-09,
    "2": 1.12345e-10,
    "3": 1.12345e-11,
    "4": 1.12345e-12,
    "5": 1.12345e-13,
    "6": 1.12345e-14,
    "7": 1.12345e-15,
    "8": 1.12345e-16,
    "9": 1.12345e-17,
    "10": 1.12345e-18,
    "11": 1.12345e-19
  }
}

To read it back in, we can then do:

>>> pd.read_json(json_df, orient='index')
             0             1             2   ...            9             10            11
0  1.123450e-08  1.123450e-09  1.123450e-10  ...  1.123450e-17  1.123450e-18  1.123450e-19

[1 rows x 12 columns]
over 4 years ago · Santiago Trujillo Denunciar

0

pd.DataFrame.to_json uses internal library pandas._libs.json and not a standard json module. This explains the difference in behavior. The former "normalizes" the numbers internally and does not expose API to control it. As a result, you have following options:

Either convert to dictionary and dump using standard json library (as mentioned earlier):

>>> print(json.dumps(df.to_dict(orient='records'), indent=2))

[
  {
    "0": 1.12345e-08,
    "1": 1.12345e-09,
    "2": 1.12345e-10,
    "3": 1.12345e-11,
    "4": 1.12345e-12,
    "5": 1.12345e-13,
    "6": 1.12345e-14,
    "7": 1.12345e-15,
    "8": 1.12345e-16,
    "9": 1.12345e-17,
    "10": 1.12345e-18,
    "11": 1.12345e-19
  }
]

This is perfectly legitimate solution.

You can use CSV format instead of JSON and specify desired floating point format:

>>> print(df.to_csv(float_format='%.10e', index=False))

0,1,2,3,4,5,6,7,8,9,10,11
1.1234500000e-08,1.1234500000e-09,1.1234500000e-10,1.1234500000e-11,1.1234500000e-12,1.1234500000e-13,1.1234500000e-14,1.1234500000e-15,1.1234500000e-16,1.1234500000e-17,1.1234500000e-18,1.1234500000e-19

Another option is to convert values to string before the "normalization" kicks in:

>>> print(df.astype(str).to_json(indent=2, orient='index'))

{
  "0":{
    "0":"1.12345e-08",
    "1":"1.12345e-09",
    "2":"1.12345e-10",
    "3":"1.12345e-11",
    "4":"1.12345e-12",
    "5":"1.12345e-13",
    "6":"1.12345e-14",
    "7":"1.12345e-15",
    "8":"1.12345e-16",
    "9":"1.12345e-17",
    "10":"1.12345e-18",
    "11":"1.12345e-19"
  }
}

The conversion to string will require special attention when reading back the JSON.

And finally, if you need the exact values, just use binary format such as parquet or pickle.

over 4 years ago · Santiago Trujillo Denunciar
Responde la pregunta
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