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
扩展一下DS的答案，如果你需要对象是可变的(而namedtuple不是)，你可以使用记录类库而不是namedtuple:
import json
from recordclass import recordclass
data = '{"name": "John Smith", "hometown": {"name": "New York", "id": 123}}'
# Parse into a mutable object
x = json.loads(data, object_hook=lambda d: recordclass('X', d.keys())(*d.values()))
修改后的对象可以使用simplejson很容易地转换回json:
x.name = "John Doe"
new_json = simplejson.dumps(x)
其他回答
因此，我正在寻找一种不需要大量自定义反序列化代码就能解组任意类型(想想数据类的字典，或者数据类数组的字典的字典)的方法。
这是我的方法:
import json
from dataclasses import dataclass, make_dataclass
from dataclasses_json import DataClassJsonMixin, dataclass_json
@dataclass_json
@dataclass
class Person:
name: str
def unmarshal_json(data, t):
Unmarhsal = make_dataclass('Unmarhsal', [('res', t)],
bases=(DataClassJsonMixin,))
d = json.loads(data)
out = Unmarhsal.from_dict({"res": d})
return out.res
unmarshalled = unmarshal_json('{"1": {"name": "john"} }', dict[str, Person])
print(unmarshalled)
打印:{'1':Person(name='john')}
这不是代码高尔夫，但这里是我使用类型的最短技巧。SimpleNamespace作为JSON对象的容器。
与namedtuple解决方案相比，它是:
可能更快/更小，因为它没有为每个对象创建一个类 更短的 没有重命名选项，对于不是有效标识符的键可能有相同的限制(在幕后使用setattr)
例子:
from __future__ import print_function
import json
try:
from types import SimpleNamespace as Namespace
except ImportError:
# Python 2.x fallback
from argparse import Namespace
data = '{"name": "John Smith", "hometown": {"name": "New York", "id": 123}}'
x = json.loads(data, object_hook=lambda d: Namespace(**d))
print (x.name, x.hometown.name, x.hometown.id)
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
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)
使用json模块(Python 2.6新增)或几乎总是安装的simplejson模块。
