我想将JSON数据转换为Python对象。

我从Facebook API收到JSON数据对象,我想将其存储在数据库中。

我的当前视图在Django (Python)(请求。POST包含JSON):

response = request.POST
user = FbApiUser(user_id = response['id'])
user.name = response['name']
user.username = response['username']
user.save()

这很好,但是如何处理复杂的JSON数据对象呢? 如果我能以某种方式将这个JSON对象转换为易于使用的Python对象,是不是会更好?


当前回答

我认为最简单的解决方法是

import orjson  # faster then json =)
from typing import NamedTuple

_j = '{"name":"Иван","age":37,"mother":{"name":"Ольга","age":58},"children":["Маша","Игорь","Таня"],"married": true,' \
     '"dog":null} '


class PersonNameAge(NamedTuple):
    name: str
    age: int


class UserInfo(NamedTuple):
    name: str
    age: int
    mother: PersonNameAge
    children: list
    married: bool
    dog: str


j = orjson.loads(_j)
u = UserInfo(**j)

print(u.name, u.age, u.mother, u.children, u.married, u.dog)

>>> Ivan 37 {'name': 'Olga', 'age': 58} ['Mary', 'Igor', 'Jane'] True None

其他回答

如果你使用的是Python 3.6或更新版本,你可以看看squema——一个用于静态类型数据结构的轻量级模块。它使您的代码易于阅读,同时提供简单的数据验证,转换和序列化,而无需额外的工作。你可以把它看作是命名元组和数据类的一种更复杂、更有见解的选择。下面是你如何使用它:

from uuid import UUID
from squema import Squema


class FbApiUser(Squema):
    id: UUID
    age: int
    name: str

    def save(self):
        pass


user = FbApiUser(**json.loads(response))
user.save()

在寻找解决方案时,我偶然发现了这个博客:https://blog.mosthege.net/2016/11/12/json-deserialization-of-nested-objects/

它使用与前面回答中相同的技术,但使用了装饰器。 我发现另一件有用的事情是,它在反序列化结束时返回一个类型化对象

class JsonConvert(object):
    class_mappings = {}

    @classmethod
    def class_mapper(cls, d):
        for keys, cls in clsself.mappings.items():
            if keys.issuperset(d.keys()):   # are all required arguments present?
                return cls(**d)
        else:
            # Raise exception instead of silently returning None
            raise ValueError('Unable to find a matching class for object: {!s}'.format(d))

    @classmethod
    def complex_handler(cls, Obj):
        if hasattr(Obj, '__dict__'):
            return Obj.__dict__
        else:
            raise TypeError('Object of type %s with value of %s is not JSON serializable' % (type(Obj), repr(Obj)))

    @classmethod
    def register(cls, claz):
        clsself.mappings[frozenset(tuple([attr for attr,val in cls().__dict__.items()]))] = cls
        return cls

    @classmethod
    def to_json(cls, obj):
        return json.dumps(obj.__dict__, default=cls.complex_handler, indent=4)

    @classmethod
    def from_json(cls, json_str):
        return json.loads(json_str, object_hook=cls.class_mapper)

用法:

@JsonConvert.register
class Employee(object):
    def __init__(self, Name:int=None, Age:int=None):
        self.Name = Name
        self.Age = Age
        return

@JsonConvert.register
class Company(object):
    def __init__(self, Name:str="", Employees:[Employee]=None):
        self.Name = Name
        self.Employees = [] if Employees is None else Employees
        return

company = Company("Contonso")
company.Employees.append(Employee("Werner", 38))
company.Employees.append(Employee("Mary"))

as_json = JsonConvert.to_json(company)
from_json = JsonConvert.from_json(as_json)
as_json_from_json = JsonConvert.to_json(from_json)

assert(as_json_from_json == as_json)

print(as_json_from_json)

你可以使用

x = Map(json.loads(response))
x.__class__ = MyClass

在哪里

class Map(dict):
    def __init__(self, *args, **kwargs):
        super(Map, self).__init__(*args, **kwargs)
        for arg in args:
            if isinstance(arg, dict):
                for k, v in arg.iteritems():
                    self[k] = v
                    if isinstance(v, dict):
                        self[k] = Map(v)

        if kwargs:
            # for python 3 use kwargs.items()
            for k, v in kwargs.iteritems():
                self[k] = v
                if isinstance(v, dict):
                    self[k] = Map(v)

    def __getattr__(self, attr):
        return self.get(attr)

    def __setattr__(self, key, value):
        self.__setitem__(key, value)

    def __setitem__(self, key, value):
        super(Map, self).__setitem__(key, value)
        self.__dict__.update({key: value})

    def __delattr__(self, item):
        self.__delitem__(item)

    def __delitem__(self, key):
        super(Map, self).__delitem__(key)
        del self.__dict__[key]

对于通用的、经得起未来考验的解决方案。

对于复杂的对象,可以使用JSON Pickle

Python库,用于将任意对象图序列化为JSON。 它几乎可以接受任何Python对象并将对象转换为JSON。 此外,它还可以将对象重新构造回Python。

数据类向导是一种现代的选项,可以类似地为您工作。它支持自动键大小写转换,如camelCase或TitleCase,这两者在API响应中都很常见。

当将实例转储到dict/JSON时,默认的键转换是camelCase,但这可以很容易地使用主数据类上提供的Meta配置来覆盖。

https://pypi.org/project/dataclass-wizard/

from dataclasses import dataclass

from dataclass_wizard import fromdict, asdict


@dataclass
class User:
    name: str
    age: int
    is_active: bool


data = {
    'name': 'John',
    'age': 30,
    'isActive': True,
}

user = fromdict(User, data)
assert user == User(name='John', age=30, is_active=True)

json_dict = asdict(user)
assert json_dict == {'name': 'John', 'age': 30, 'isActive': True}

设置元配置的例子,当序列化为dict/JSON时,将字段转换为lisp-case:

DumpMeta(key_transform='LISP').bind_to(User)