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。