如何使一个Python类序列化?
class FileItem:
def __init__(self, fname):
self.fname = fname
尝试序列化为JSON:
>>> import json
>>> x = FileItem('/foo/bar')
>>> json.dumps(x)
TypeError: Object of type 'FileItem' is not JSON serializable
如何使一个Python类序列化?
class FileItem:
def __init__(self, fname):
self.fname = fname
尝试序列化为JSON:
>>> import json
>>> x = FileItem('/foo/bar')
>>> json.dumps(x)
TypeError: Object of type 'FileItem' is not JSON serializable
当前回答
前几天我遇到了这个问题,并为Python对象实现了一个更通用的Encoder版本,可以处理嵌套对象和继承字段:
import json
import inspect
class ObjectEncoder(json.JSONEncoder):
def default(self, obj):
if hasattr(obj, "to_json"):
return self.default(obj.to_json())
elif hasattr(obj, "__dict__"):
d = dict(
(key, value)
for key, value in inspect.getmembers(obj)
if not key.startswith("__")
and not inspect.isabstract(value)
and not inspect.isbuiltin(value)
and not inspect.isfunction(value)
and not inspect.isgenerator(value)
and not inspect.isgeneratorfunction(value)
and not inspect.ismethod(value)
and not inspect.ismethoddescriptor(value)
and not inspect.isroutine(value)
)
return self.default(d)
return obj
例子:
class C(object):
c = "NO"
def to_json(self):
return {"c": "YES"}
class B(object):
b = "B"
i = "I"
def __init__(self, y):
self.y = y
def f(self):
print "f"
class A(B):
a = "A"
def __init__(self):
self.b = [{"ab": B("y")}]
self.c = C()
print json.dumps(A(), cls=ObjectEncoder, indent=2, sort_keys=True)
结果:
{
"a": "A",
"b": [
{
"ab": {
"b": "B",
"i": "I",
"y": "y"
}
}
],
"c": {
"c": "YES"
},
"i": "I"
}
其他回答
只需要像这样添加to_json方法到你的类中:
def to_json(self):
return self.message # or how you want it to be serialized
然后将这段代码(来自这个答案)添加到所有内容的顶部:
from json import JSONEncoder
def _default(self, obj):
return getattr(obj.__class__, "to_json", _default.default)(obj)
_default.default = JSONEncoder().default
JSONEncoder.default = _default
这将会在导入json模块时monkey-patch,所以 JSONEncoder.default()自动检查特殊的to_json() 方法,并使用它对找到的对象进行编码。
就像Onur说的,但是这次你不需要更新项目中的每个json.dumps()。
这对我来说很有效:
class JsonSerializable(object):
def serialize(self):
return json.dumps(self.__dict__)
def __repr__(self):
return self.serialize()
@staticmethod
def dumper(obj):
if "serialize" in dir(obj):
return obj.serialize()
return obj.__dict__
然后
class FileItem(JsonSerializable):
...
and
log.debug(json.dumps(<my object>, default=JsonSerializable.dumper, indent=2))
正如在许多其他答案中提到的,您可以将函数传递给json。转储将不是默认支持的类型之一的对象转换为受支持的类型。令人惊讶的是,他们都没有提到最简单的情况,即使用内置函数vars将对象转换为包含其所有属性的dict:
json.dumps(obj, default=vars)
注意,这只涵盖了基本的情况,如果你需要对某些类型进行更具体的序列化(例如排除某些属性或没有__dict__属性的对象),你需要使用自定义函数或JSONEncoder,如其他答案中所述。
另一种选择是将JSON转储打包到它自己的类中:
import json
class FileItem:
def __init__(self, fname):
self.fname = fname
def __repr__(self):
return json.dumps(self.__dict__)
或者,更好的是,从JsonSerializable类继承FileItem类:
import json
class JsonSerializable(object):
def toJson(self):
return json.dumps(self.__dict__)
def __repr__(self):
return self.toJson()
class FileItem(JsonSerializable):
def __init__(self, fname):
self.fname = fname
测试:
>>> f = FileItem('/foo/bar')
>>> f.toJson()
'{"fname": "/foo/bar"}'
>>> f
'{"fname": "/foo/bar"}'
>>> str(f) # string coercion
'{"fname": "/foo/bar"}'
TLDR:复制-粘贴下面的选项1或选项2
真正的/完整的答案:让Pythons json模块与你的类一起工作
AKA,求解:json。dump ({"thing": YOUR_CLASS()})
解释:
Yes, a good reliable solution exists No, there is no python "official" solution By official solution, I mean there is no way (as of 2023) to add a method to your class (like toJSON in JavaScript) and/or no way to register your class with the built-in json module. When something like json.dumps([1,2, your_obj]) is executed, python doesn't check a lookup table or object method. I'm not sure why other answers don't explain this The closest official approach is probably andyhasit's answer which is to inherit from a dictionary. However, inheriting from a dictionary doesn't work very well for many custom classes like AdvancedDateTime, or pytorch tensors. The ideal workaround is this: Mutate json.dumps (affects everywhere, even pip modules that import json) Add def __json__(self) method to your class
选项1:让一个模块来做补丁
PIP安装json-fix (扩展+包装版FancyJohn的回答,谢谢@FancyJohn)
your_class_definition.py
import json_fix
class YOUR_CLASS:
def __json__(self):
# YOUR CUSTOM CODE HERE
# you probably just want to do:
# return self.__dict__
return "a built-in object that is naturally json-able"
这是它。
使用示例:
from your_class_definition import YOUR_CLASS
import json
json.dumps([1,2, YOUR_CLASS()], indent=0)
# '[\n1,\n2,\n"a built-in object that is naturally json-able"\n]'
生成json。dump适用于Numpy数组,Pandas DataFrames和其他第三方对象,请参阅模块(只有大约2行代码,但需要解释)。
它是如何工作的?嗯…
选项2:补丁json。把你自己
注意:这种方法是简化的,它在已知的edgcase上失败(例如:如果你的自定义类继承了dict或其他内置类),并且它错过了控制外部类的json行为(numpy数组,datetime, dataframes,张量等)。
some_file_thats_imported_before_your_class_definitions.py
# Step: 1
# create the patch
from json import JSONEncoder
def wrapped_default(self, obj):
return getattr(obj.__class__, "__json__", wrapped_default.default)(obj)
wrapped_default.default = JSONEncoder().default
# apply the patch
JSONEncoder.original_default = JSONEncoder.default
JSONEncoder.default = wrapped_default
your_class_definition.py
# Step 2
class YOUR_CLASS:
def __json__(self, **options):
# YOUR CUSTOM CODE HERE
# you probably just want to do:
# return self.__dict__
return "a built-in object that is natually json-able"
_
其他答案似乎都是“序列化自定义对象的最佳实践/方法”
在这里的文档中已经介绍过了(搜索“complex”可以找到编码复数的例子)