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

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。

其他回答

class SimpleClass:
    def __init__(self, **kwargs):
        for k, v in kwargs.items():
            if type(v) is dict:
                setattr(self, k, SimpleClass(**v))
            else:
                setattr(self, k, v)


json_dict = {'name': 'jane doe', 'username': 'jane', 'test': {'foo': 1}}

class_instance = SimpleClass(**json_dict)

print(class_instance.name, class_instance.test.foo)
print(vars(class_instance))

使用python 3.7,我发现下面的代码非常简单有效。在本例中,将JSON从文件加载到字典中:

class Characteristic:
    def __init__(self, characteristicName, characteristicUUID):
        self.characteristicName = characteristicName
        self.characteristicUUID = characteristicUUID


class Service:
    def __init__(self, serviceName, serviceUUID, characteristics):
        self.serviceName = serviceName
        self.serviceUUID = serviceUUID
        self.characteristics = characteristics

class Definitions:
    def __init__(self, services):
        self.services = []
        for service in services:
            self.services.append(Service(**service))


def main():
    parser = argparse.ArgumentParser(
        prog="BLEStructureGenerator",
        description="Taking in a JSON input file which lists all of the services, "
                    "characteristics and encoded properties. The encoding takes in "
                    "another optional template services and/or characteristics "
                    "file where the JSON file contents are applied to the templates.",
        epilog="Copyright Brown & Watson International"
    )

    parser.add_argument('definitionfile',
                        type=argparse.FileType('r', encoding='UTF-8'),
                        help="JSON file which contains the list of characteristics and "
                             "services in the required format")
    parser.add_argument('-s', '--services',
                        type=argparse.FileType('r', encoding='UTF-8'),
                        help="Services template file to be used for each service in the "
                             "JSON file list")
    parser.add_argument('-c', '--characteristics',
                        type=argparse.FileType('r', encoding='UTF-8'),
                        help="Characteristics template file to be used for each service in the "
                             "JSON file list")

    args = parser.parse_args()
    definition_dict = json.load(args.definitionfile)
    definitions = Definitions(**definition_dict)

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。

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)

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。