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

def load_model_from_dict(self, data: dict):
    for key, value in data.items():
        self.__dict__[key] = value
    return self

它帮助返回你自己的模型,从字典中不可预见的变量。

其他回答

Dacite也可能是您的解决方案,它支持以下功能:

嵌套结构 (基本)类型检查 可选字段(即typing.Optional) 工会 向前引用 集合 自定义类型钩子

https://pypi.org/project/dacite/

from dataclasses import dataclass
from dacite import from_dict


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


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

user = from_dict(data_class=User, data=data)

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

Python3.x

以我的知识,我能找到的最好的方法是。 注意,这段代码也处理set()。 这种方法是通用的,只需要类的扩展(在第二个例子中)。 请注意,我只是对文件执行此操作,但是很容易根据自己的喜好修改行为。

然而,这是一个编解码器。

再做一点工作,就可以用其他方式构造类。 我假设有一个默认构造函数来实例它,然后更新类dict。

import json
import collections


class JsonClassSerializable(json.JSONEncoder):

    REGISTERED_CLASS = {}

    def register(ctype):
        JsonClassSerializable.REGISTERED_CLASS[ctype.__name__] = ctype

    def default(self, obj):
        if isinstance(obj, collections.Set):
            return dict(_set_object=list(obj))
        if isinstance(obj, JsonClassSerializable):
            jclass = {}
            jclass["name"] = type(obj).__name__
            jclass["dict"] = obj.__dict__
            return dict(_class_object=jclass)
        else:
            return json.JSONEncoder.default(self, obj)

    def json_to_class(self, dct):
        if '_set_object' in dct:
            return set(dct['_set_object'])
        elif '_class_object' in dct:
            cclass = dct['_class_object']
            cclass_name = cclass["name"]
            if cclass_name not in self.REGISTERED_CLASS:
                raise RuntimeError(
                    "Class {} not registered in JSON Parser"
                    .format(cclass["name"])
                )
            instance = self.REGISTERED_CLASS[cclass_name]()
            instance.__dict__ = cclass["dict"]
            return instance
        return dct

    def encode_(self, file):
        with open(file, 'w') as outfile:
            json.dump(
                self.__dict__, outfile,
                cls=JsonClassSerializable,
                indent=4,
                sort_keys=True
            )

    def decode_(self, file):
        try:
            with open(file, 'r') as infile:
                self.__dict__ = json.load(
                    infile,
                    object_hook=self.json_to_class
                )
        except FileNotFoundError:
            print("Persistence load failed "
                  "'{}' do not exists".format(file)
                  )


class C(JsonClassSerializable):

    def __init__(self):
        self.mill = "s"


JsonClassSerializable.register(C)


class B(JsonClassSerializable):

    def __init__(self):
        self.a = 1230
        self.c = C()


JsonClassSerializable.register(B)


class A(JsonClassSerializable):

    def __init__(self):
        self.a = 1
        self.b = {1, 2}
        self.c = B()

JsonClassSerializable.register(A)

A().encode_("test")
b = A()
b.decode_("test")
print(b.a)
print(b.b)
print(b.c.a)

Edit

通过更多的研究,我发现了一种不需要SUPERCLASS寄存器方法调用的泛化方法,使用元类

import json
import collections

REGISTERED_CLASS = {}

class MetaSerializable(type):

    def __call__(cls, *args, **kwargs):
        if cls.__name__ not in REGISTERED_CLASS:
            REGISTERED_CLASS[cls.__name__] = cls
        return super(MetaSerializable, cls).__call__(*args, **kwargs)


class JsonClassSerializable(json.JSONEncoder, metaclass=MetaSerializable):

    def default(self, obj):
        if isinstance(obj, collections.Set):
            return dict(_set_object=list(obj))
        if isinstance(obj, JsonClassSerializable):
            jclass = {}
            jclass["name"] = type(obj).__name__
            jclass["dict"] = obj.__dict__
            return dict(_class_object=jclass)
        else:
            return json.JSONEncoder.default(self, obj)

    def json_to_class(self, dct):
        if '_set_object' in dct:
            return set(dct['_set_object'])
        elif '_class_object' in dct:
            cclass = dct['_class_object']
            cclass_name = cclass["name"]
            if cclass_name not in REGISTERED_CLASS:
                raise RuntimeError(
                    "Class {} not registered in JSON Parser"
                    .format(cclass["name"])
                )
            instance = REGISTERED_CLASS[cclass_name]()
            instance.__dict__ = cclass["dict"]
            return instance
        return dct

    def encode_(self, file):
        with open(file, 'w') as outfile:
            json.dump(
                self.__dict__, outfile,
                cls=JsonClassSerializable,
                indent=4,
                sort_keys=True
            )

    def decode_(self, file):
        try:
            with open(file, 'r') as infile:
                self.__dict__ = json.load(
                    infile,
                    object_hook=self.json_to_class
                )
        except FileNotFoundError:
            print("Persistence load failed "
                  "'{}' do not exists".format(file)
                  )


class C(JsonClassSerializable):

    def __init__(self):
        self.mill = "s"


class B(JsonClassSerializable):

    def __init__(self):
        self.a = 1230
        self.c = C()


class A(JsonClassSerializable):

    def __init__(self):
        self.a = 1
        self.b = {1, 2}
        self.c = B()


A().encode_("test")
b = A()
b.decode_("test")
print(b.a)
# 1
print(b.b)
# {1, 2}
print(b.c.a)
# 1230
print(b.c.c.mill)
# s

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。

更新

在Python3中,你可以使用SimpleNamespace和object_hook在一行中完成:

import json
from types import SimpleNamespace

data = '{"name": "John Smith", "hometown": {"name": "New York", "id": 123}}'

# Parse JSON into an object with attributes corresponding to dict keys.
x = json.loads(data, object_hook=lambda d: SimpleNamespace(**d))
print(x.name, x.hometown.name, x.hometown.id)

旧答案(Python2)

在Python2中,你可以使用namedtuple和object_hook在一行中完成(但对于嵌套对象非常慢):

import json
from collections import namedtuple

data = '{"name": "John Smith", "hometown": {"name": "New York", "id": 123}}'

# Parse JSON into an object with attributes corresponding to dict keys.
x = json.loads(data, object_hook=lambda d: namedtuple('X', d.keys())(*d.values()))
print x.name, x.hometown.name, x.hometown.id

或者,为了便于重用:

def _json_object_hook(d): return namedtuple('X', d.keys())(*d.values())
def json2obj(data): return json.loads(data, object_hook=_json_object_hook)

x = json2obj(data)

如果希望它处理不是很好的属性名称的键,请检查namedtuple的rename参数。

这里有一个快速而肮脏的json pickle替代方案

import json

class User:
    def __init__(self, name, username):
        self.name = name
        self.username = username

    def to_json(self):
        return json.dumps(self.__dict__)

    @classmethod
    def from_json(cls, json_str):
        json_dict = json.loads(json_str)
        return cls(**json_dict)

# example usage
User("tbrown", "Tom Brown").to_json()
User.from_json(User("tbrown", "Tom Brown").to_json()).to_json()