Python 到 JSON 序列化在十进制上失败
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Python to JSON Serialization fails on Decimal
提问by Dinesh
I have a python object which includes some decimals. This is causing the json.dumps() to break.
我有一个包含一些小数的 python 对象。这导致 json.dumps() 中断。
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.
我从 SO 得到以下解决方案（例如Python JSON serialize a Decimal object），但推荐的解决方案仍然不起作用。Python 网站 - 有完全相同的答案。
Any suggestions how to make this work?
任何建议如何使这项工作？
Thanks. Below is my code. It looks like the dumps() doesn't even go into the specialized encoder.
谢谢。下面是我的代码。看起来 dumps() 甚至没有进入专门的编码器。
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>
采纳答案by Martijn Pieters
It is not (no longer) recommended you create a subclass; the json.dump()and json.dumps()functions take a defaultfunction:
不（不再）建议您创建子类；的json.dump()和json.dumps()函数采用一个default函数：
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:
演示：
>>> 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.
您发现的代码仅适用于 Python 2.6，并覆盖了在更高版本中不再调用的私有方法。
回答by FearlessFuture
I can't believe that no one here talked about using simplejson, which supports deserialization of Decimal out of the box.
我不敢相信这里没有人谈论使用 simplejson，它支持开箱即用的 Decimal 反序列化。
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}'
回答by Fran?ois Constant
If you're using Django. There is a great class for Decimal and date fields:
如果您使用的是 Django。Decimal 和 date 字段有一个很好的类：
https://docs.djangoproject.com/en/1.10/topics/serialization/#djangojsonencoder
https://docs.djangoproject.com/en/1.10/topics/serialization/#djangojsonencoder
To use it:
要使用它：
import json
from django.core.serializers.json import DjangoJSONEncoder
json.dumps(value, cls=DjangoJSONEncoder)
