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
下面是一个简单功能的简单解决方案:
.toJSON()方法
实现一个序列化器方法,而不是一个JSON可序列化类:
import json
class Object:
def toJSON(self):
return json.dumps(self, default=lambda o: o.__dict__,
sort_keys=True, indent=4)
所以你只需调用它来序列化:
me = Object()
me.name = "Onur"
me.age = 35
me.dog = Object()
me.dog.name = "Apollo"
print(me.toJSON())
将输出:
{
"age": 35,
"dog": {
"name": "Apollo"
},
"name": "Onur"
}
Kyle Delaney的评论是正确的,所以我尝试使用https://stackoverflow.com/a/15538391/1497139以及https://stackoverflow.com/a/10254820/1497139的改进版本
创建一个“JSONAble”mixin。
因此,要使一个类JSON可序列化使用“JSONAble”作为超类,并调用:
instance.toJSON()
or
instance.asJSON()
对于这两种方法。您还可以使用本文提供的其他方法扩展JSONAble类。
家庭和个人单元测试样本的测试示例结果如下:
toJSOn ():
{
"members": {
"Flintstone,Fred": {
"firstName": "Fred",
"lastName": "Flintstone"
},
"Flintstone,Wilma": {
"firstName": "Wilma",
"lastName": "Flintstone"
}
},
"name": "The Flintstones"
}
asJSOn ():
{'name': 'The Flintstones', 'members': {'Flintstone,Fred': {'firstName': 'Fred', 'lastName': 'Flintstone'}, 'Flintstone,Wilma': {'firstName': 'Wilma', 'lastName': 'Flintstone'}}}
使用家庭和个人样本进行单元测试
def testJsonAble(self):
family=Family("The Flintstones")
family.add(Person("Fred","Flintstone"))
family.add(Person("Wilma","Flintstone"))
json1=family.toJSON()
json2=family.asJSON()
print(json1)
print(json2)
class Family(JSONAble):
def __init__(self,name):
self.name=name
self.members={}
def add(self,person):
self.members[person.lastName+","+person.firstName]=person
class Person(JSONAble):
def __init__(self,firstName,lastName):
self.firstName=firstName;
self.lastName=lastName;
JSONAble .py定义JSONAble mixin
'''
Created on 2020-09-03
@author: wf
'''
import json
class JSONAble(object):
'''
mixin to allow classes to be JSON serializable see
https://stackoverflow.com/questions/3768895/how-to-make-a-class-json-serializable
'''
def __init__(self):
'''
Constructor
'''
def toJSON(self):
return json.dumps(self, default=lambda o: o.__dict__,
sort_keys=True, indent=4)
def getValue(self,v):
if (hasattr(v, "asJSON")):
return v.asJSON()
elif type(v) is dict:
return self.reprDict(v)
elif type(v) is list:
vlist=[]
for vitem in v:
vlist.append(self.getValue(vitem))
return vlist
else:
return v
def reprDict(self,srcDict):
'''
get my dict elements
'''
d = dict()
for a, v in srcDict.items():
d[a]=self.getValue(v)
return d
def asJSON(self):
'''
recursively return my dict elements
'''
return self.reprDict(self.__dict__)
您将发现这些方法现在集成在https://github.com/WolfgangFahl/pyLoDStorage项目中,该项目可在https://pypi.org/project/pylodstorage/上获得
我有了自己的解决办法。使用此方法,将任何文档(字典、列表、ObjectId等)传递给序列化。
def getSerializable(doc):
# check if it's a list
if isinstance(doc, list):
for i, val in enumerate(doc):
doc[i] = getSerializable(doc[i])
return doc
# check if it's a dict
if isinstance(doc, dict):
for key in doc.keys():
doc[key] = getSerializable(doc[key])
return doc
# Process ObjectId
if isinstance(doc, ObjectId):
doc = str(doc)
return doc
# Use any other custom serializting stuff here...
# For the rest of stuff
return doc
import json
class Foo(object):
def __init__(self):
self.bar = 'baz'
self._qux = 'flub'
def somemethod(self):
pass
def default(instance):
return {k: v
for k, v in vars(instance).items()
if not str(k).startswith('_')}
json_foo = json.dumps(Foo(), default=default)
assert '{"bar": "baz"}' == json_foo
print(json_foo)
如果你不介意为它安装一个包,你可以使用json-tricks:
pip install json-tricks
之后,你只需要从json_tricks导入dump(s)而不是json,它通常会工作:
from json_tricks import dumps
json_str = dumps(cls_instance, indent=4)
这将给
{
"__instance_type__": [
"module_name.test_class",
"MyTestCls"
],
"attributes": {
"attr": "val",
"dct_attr": {
"hello": 42
}
}
}
基本上就是这样!
这在一般情况下会很有效。有一些例外,例如,如果特殊的事情发生在__new__中,或者更多的元类魔法正在发生。
显然加载也可以(否则有什么意义):
from json_tricks import loads
json_str = loads(json_str)
这确实假设module_name.test_class。MyTestCls可以导入,并且没有以不兼容的方式进行更改。您将返回一个实例,而不是某个字典或其他东西,它应该是您转储的实例的相同副本。
如果你想自定义一些东西是如何(反)序列化的,你可以添加特殊的方法到你的类,像这样:
class CustomEncodeCls:
def __init__(self):
self.relevant = 42
self.irrelevant = 37
def __json_encode__(self):
# should return primitive, serializable types like dict, list, int, string, float...
return {'relevant': self.relevant}
def __json_decode__(self, **attrs):
# should initialize all properties; note that __init__ is not called implicitly
self.relevant = attrs['relevant']
self.irrelevant = 12
其中仅序列化部分属性参数,作为示例。
作为免费的奖励,你可以获得numpy数组、日期和时间、有序地图的(反)序列化,以及在json中包含注释的能力。
免责声明:我创建了json_tricks,因为我遇到了与您相同的问题。