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
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)
其他回答
这个函数使用递归迭代遍历字典的每个部分,然后调用非内置类型类的repr()方法。
def sterilize(obj):
object_type = type(obj)
if isinstance(obj, dict):
return {k: sterilize(v) for k, v in obj.items()}
elif object_type in (list, tuple):
return [sterilize(v) for v in obj]
elif object_type in (str, int, bool, float):
return obj
else:
return obj.__repr__()
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"
}
为了给这场11年的大火再添一根柴,我想要一个满足以下条件的解决方案:
只允许使用json.dumps(obj)序列化类FileItem的实例 允许FileItem实例具有属性:FileItem .fname 允许FileItem实例提供给任何库,使用json.dumps(obj)序列化它 不需要将任何其他字段传递给json。转储(如自定义序列化器)
IE:
fileItem = FileItem('filename.ext')
assert json.dumps(fileItem) == '{"fname": "filename.ext"}'
assert fileItem.fname == 'filename.ext'
我的解决方案是:
obj的类是否继承自dict 将每个对象属性映射到底层字典
class FileItem(dict):
def __init__(self, fname):
self['fname'] = fname
#fname property
fname: str = property()
@fname.getter
def fname(self):
return self['fname']
@fname.setter
def fname(self, value: str):
self['fname'] = value
#Repeat for other properties
是的,如果你有很多属性,这有点冗长,但它是JSONSerializable,它的行为像一个对象,你可以把它给任何库,去json.dumps(obj)它。
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"}'
解决这个问题有很多方法。'ObjDict' (pip install object)是另一个。重点是提供像javascript一样的对象,它也可以像字典一样最好地处理从JSON加载的数据,但还有其他功能也很有用。这为原始问题提供了另一种解决方案。