我最喜欢Lost Koder的方法。当我试图序列化成员/方法不可序列化的更复杂的对象时，我遇到了问题。这是我的实现，工作在更多的对象:

class Serializer(object):
    @staticmethod
    def serialize(obj):
        def check(o):
            for k, v in o.__dict__.items():
                try:
                    _ = json.dumps(v)
                    o.__dict__[k] = v
                except TypeError:
                    o.__dict__[k] = str(v)
            return o
        return json.dumps(check(obj).__dict__, indent=2)

如果你正在使用Python3.5+，你可以使用jsons。(PyPi: https://pypi.org/project/jsons/)它将把你的对象(及其所有属性递归地)转换为字典。

import jsons

a_dict = jsons.dump(your_object)

或者如果你想要一个字符串:

a_str = jsons.dumps(your_object)

或者你的类实现了jsons。JsonSerializable:

a_dict = your_object.json

这是一个小库，它将一个对象及其所有子对象序列化为JSON，并将其解析回来:

https://github.com/tobiasholler/PyJSONSerialization/

我们经常在日志文件中转储JSON格式的复杂字典。虽然大多数字段携带重要信息，但我们不太关心内置的类对象(例如子进程)。Popen对象)。由于存在这些不可序列化的对象，对json.dumps()的调用会失败。

为了解决这个问题，我构建了一个小函数来转储对象的字符串表示形式，而不是转储对象本身。如果您正在处理的数据结构嵌套太多，您可以指定嵌套的最大级别/深度。

from time import time

def safe_serialize(obj , max_depth = 2):

    max_level = max_depth

    def _safe_serialize(obj , current_level = 0):

        nonlocal max_level

        # If it is a list
        if isinstance(obj , list):

            if current_level >= max_level:
                return "[...]"

            result = list()
            for element in obj:
                result.append(_safe_serialize(element , current_level + 1))
            return result

        # If it is a dict
        elif isinstance(obj , dict):

            if current_level >= max_level:
                return "{...}"

            result = dict()
            for key , value in obj.items():
                result[f"{_safe_serialize(key , current_level + 1)}"] = _safe_serialize(value , current_level + 1)
            return result

        # If it is an object of builtin class
        elif hasattr(obj , "__dict__"):
            if hasattr(obj , "__repr__"):
                result = f"{obj.__repr__()}_{int(time())}"
            else:
                try:
                    result = f"{obj.__class__.__name__}_object_{int(time())}"
                except:
                    result = f"object_{int(time())}"
            return result

        # If it is anything else
        else:
            return obj

    return _safe_serialize(obj)

由于字典也可以有不可序列化的键，转储它们的类名或对象表示将导致所有键都具有相同的名称，这将抛出错误，因为所有键都需要有唯一的名称，这就是为什么当前时间Since epoch被int(time())附加到对象名称。

可以使用以下具有不同级别/深度的嵌套字典来测试该函数

d = {
    "a" : {
        "a1" : {
            "a11" : {
                "a111" : "some_value" ,
                "a112" : "some_value" ,
            } ,
            "a12" : {
                "a121" : "some_value" ,
                "a122" : "some_value" ,
            } ,
        } ,
        "a2" : {
            "a21" : {
                "a211" : "some_value" ,
                "a212" : "some_value" ,
            } ,
            "a22" : {
                "a221" : "some_value" ,
                "a222" : "some_value" ,
            } ,
        } ,
    } ,
    "b" : {
        "b1" : {
            "b11" : {
                "b111" : "some_value" ,
                "b112" : "some_value" ,
            } ,
            "b12" : {
                "b121" : "some_value" ,
                "b122" : "some_value" ,
            } ,
        } ,
        "b2" : {
            "b21" : {
                "b211" : "some_value" ,
                "b212" : "some_value" ,
            } ,
            "b22" : {
                "b221" : "some_value" ,
                "b222" : "some_value" ,
            } ,
        } ,
    } ,
    "c" : subprocess.Popen("ls -l".split() , stdout = subprocess.PIPE , stderr = subprocess.PIPE) ,
}

执行以下命令将会得到-

print("LEVEL 3")
print(json.dumps(safe_serialize(d , 3) , indent = 4))

print("\n\n\nLEVEL 2")
print(json.dumps(safe_serialize(d , 2) , indent = 4))

print("\n\n\nLEVEL 1")
print(json.dumps(safe_serialize(d , 1) , indent = 4))

结果:

LEVEL 3
{
    "a": {
        "a1": {
            "a11": "{...}",
            "a12": "{...}"
        },
        "a2": {
            "a21": "{...}",
            "a22": "{...}"
        }
    },
    "b": {
        "b1": {
            "b11": "{...}",
            "b12": "{...}"
        },
        "b2": {
            "b21": "{...}",
            "b22": "{...}"
        }
    },
    "c": "<Popen: returncode: None args: ['ls', '-l']>"
}



LEVEL 2
{
    "a": {
        "a1": "{...}",
        "a2": "{...}"
    },
    "b": {
        "b1": "{...}",
        "b2": "{...}"
    },
    "c": "<Popen: returncode: None args: ['ls', '-l']>"
}



LEVEL 1
{
    "a": "{...}",
    "b": "{...}",
    "c": "<Popen: returncode: None args: ['ls', '-l']>"
}

[注意]:仅在不关心内置类对象的序列化时使用此选项。

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"
  }
}

我没有看到这里提到串行版本或backcompat，所以我将发布我的解决方案，我已经使用了一点。我可能还有很多东西要学习，特别是Java和Javascript可能比我更成熟，但我要这样做

https://gist.github.com/andy-d/b7878d0044a4242c0498ed6d67fd50fe