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

要添加另一个选项:您可以使用attrs包和asdict方法。

class ObjectEncoder(JSONEncoder):
    def default(self, o):
        return attr.asdict(o)

json.dumps(objects, cls=ObjectEncoder)

然后再转换回去

def from_json(o):
    if '_obj_name' in o:
        type_ = o['_obj_name']
        del o['_obj_name']
        return globals()[type_](**o)
    else:
        return o

data = JSONDecoder(object_hook=from_json).decode(data)

类看起来像这样

@attr.s
class Foo(object):
    x = attr.ib()
    _obj_name = attr.ib(init=False, default='Foo')

其他回答

基于Quinten Cabo的回答:

def sterilize(obj):
    """Make an object more ameniable to dumping as json
    """
    if type(obj) in (str, float, int, bool, type(None)):
        return obj
    elif isinstance(obj, dict):
        return {k: sterilize(v) for k, v in obj.items()}
    list_ret = []
    dict_ret = {}
    for a in dir(obj):
        if a == '__iter__' and callable(obj.__iter__):
            list_ret.extend([sterilize(v) for v in obj])
        elif a == '__dict__':
            dict_ret.update({k: sterilize(v) for k, v in obj.__dict__.items() if k not in ['__module__', '__dict__', '__weakref__', '__doc__']})
        elif a not in ['__doc__', '__module__']:
            aval = getattr(obj, a)
            if type(aval) in (str, float, int, bool, type(None)):
                dict_ret[a] = aval
            elif a != '__class__' and a != '__objclass__' and isinstance(aval, type):
                dict_ret[a] = sterilize(aval)
    if len(list_ret) == 0:
        if len(dict_ret) == 0:
            return repr(obj)
        return dict_ret
    else:
        if len(dict_ret) == 0:
            return list_ret
    return (list_ret, dict_ret)

区别在于

Works for any iterable instead of just list and tuple (it works for NumPy arrays, etc.) Works for dynamic types (ones that contain a __dict__). Includes native types float and None so they don't get converted to string. Classes that have __dict__ and members will mostly work (if the __dict__ and member names collide, you will only get one - likely the member) Classes that are lists and have members will look like a tuple of the list and a dictionary Python3 (that isinstance() call may be the only thing that needs changing)

为了在10年前的火灾中再添加一个日志,我还将为这个任务提供数据类向导,假设您使用的是Python 3.6+。这可以很好地用于数据类,这实际上是3.7+版本的python内置模块。

dataclass-wizard库将把对象(及其所有属性递归地)转换为dict,并使用fromdict使反向(反序列化)非常简单。另外,这里是PyPi链接:https://pypi.org/project/dataclass-wizard/。

import dataclass_wizard
import dataclasses

@dataclasses.dataclass
class A:
    hello: str
    a_field: int

obj = A('world', 123)
a_dict = dataclass_wizard.asdict(obj)
# {'hello': 'world', 'aField': 123}

或者如果你想要一个字符串:

a_str = jsons.dumps(dataclass_wizard.asdict(obj))

或者您的类是否从dataclass_wizard扩展。JSONWizard:

a_str = your_object.to_json()

最后,标准库还支持Union类型的数据类,这基本上意味着可以将dict反序列化为类C1或C2的对象。例如:

from dataclasses import dataclass

from dataclass_wizard import JSONWizard

@dataclass
class Outer(JSONWizard):

    class _(JSONWizard.Meta):
        tag_key = 'tag'
        auto_assign_tags = True

    my_string: str
    inner: 'A | B'  # alternate syntax: `inner: typing.Union['A', 'B']`

@dataclass
class A:
    my_field: int

@dataclass
class B:
    my_field: str


my_dict = {'myString': 'test', 'inner': {'tag': 'B', 'myField': 'test'}}
obj = Outer.from_dict(my_dict)

# True
assert repr(obj) == "Outer(my_string='test', inner=B(my_field='test'))"

obj.to_json()
# {"myString": "test", "inner": {"myField": "test", "tag": "B"}}

一个非常简单的一行程序解决方案

import json

json.dumps(your_object, default=lambda __o: __o.__dict__)

结束!

下面是一个测试。

import json
from dataclasses import dataclass


@dataclass
class Company:
    id: int
    name: str

@dataclass
class User:
    id: int
    name: str
    email: str
    company: Company


company = Company(id=1, name="Example Ltd")
user = User(id=1, name="John Doe", email="john@doe.net", company=company)


json.dumps(user, default=lambda __o: __o.__dict__)

输出:

{
  "id": 1, 
  "name": "John Doe", 
  "email": "john@doe.net", 
  "company": {
    "id": 1, 
    "name": "Example Ltd"
  }
}

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"
}

你们为什么要把事情搞得这么复杂?这里有一个简单的例子:

#!/usr/bin/env python3

import json
from dataclasses import dataclass

@dataclass
class Person:
    first: str
    last: str
    age: int

    @property
    def __json__(self):
        return {
            "name": f"{self.first} {self.last}",
            "age": self.age
        }

john = Person("John", "Doe", 42)
print(json.dumps(john, indent=4, default=lambda x: x.__json__))

这样你也可以序列化嵌套类,因为__json__返回一个python对象而不是字符串。不需要使用JSONEncoder,因为使用简单lambda的默认参数也可以很好地工作。

我使用@property代替了一个简单的函数,因为这样感觉更自然和现代。@dataclass也只是一个例子,它也适用于“普通”类。