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?
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)
如果你正在寻找将JSON或任何复杂字典的类型安全反序列化到python类中,我强烈推荐python 3.7+的pydantic。它不仅有一个简洁的API(不需要编写“helper”样板),可以与Python数据类集成,而且具有复杂和嵌套数据结构的静态和运行时类型验证。
使用示例:
from pydantic import BaseModel
from datetime import datetime
class Item(BaseModel):
field1: str | int # union
field2: int | None = None # optional
field3: str = 'default' # default values
class User(BaseModel):
name: str | None = None
username: str
created: datetime # default type converters
items: list[Item] = [] # nested complex types
data = {
'name': 'Jane Doe',
'username': 'user1',
'created': '2020-12-31T23:59:00+10:00',
'items': [
{'field1': 1, 'field2': 2},
{'field1': 'b'},
{'field1': 'c', 'field3': 'override'}
]
}
user: User = User(**data)
要了解更多细节和特性,请查看文档中的pydantic的rational部分。
改进lovasoa非常好的答案。
如果你正在使用python 3.6+,你可以使用:
PIP安装棉花糖-enum和
PIP安装棉花糖数据类
它简单且类型安全。
你可以在string-json中转换你的类,反之亦然:
从对象到字符串Json:
from marshmallow_dataclass import dataclass
user = User("Danilo","50","RedBull",15,OrderStatus.CREATED)
user_json = User.Schema().dumps(user)
user_json_str = user_json.data
从String Json到Object:
json_str = '{"name":"Danilo", "orderId":"50", "productName":"RedBull", "quantity":15, "status":"Created"}'
user, err = User.Schema().loads(json_str)
print(user,flush=True)
类定义:
class OrderStatus(Enum):
CREATED = 'Created'
PENDING = 'Pending'
CONFIRMED = 'Confirmed'
FAILED = 'Failed'
@dataclass
class User:
def __init__(self, name, orderId, productName, quantity, status):
self.name = name
self.orderId = orderId
self.productName = productName
self.quantity = quantity
self.status = status
name: str
orderId: str
productName: str
quantity: int
status: OrderStatus
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')}
这是我的办法。
特性
支持类型提示
如果缺少键则引发错误。
跳过数据中的额外值
import typing
class User:
name: str
age: int
def __init__(self, data: dict):
for k, _ in typing.get_type_hints(self).items():
setattr(self, k, data[k])
data = {
"name": "Susan",
"age": 18
}
user = User(data)
print(user.name, user.age)
# Output: Susan 18
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)