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
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)
其他回答
既然没有人给出了和我一样的答案，我就把它贴在这里。
这是一个健壮的类，可以轻松地在JSON str和dict之间来回转换，我已经从我的答案复制到另一个问题:
import json
class PyJSON(object):
def __init__(self, d):
if type(d) is str:
d = json.loads(d)
self.from_dict(d)
def from_dict(self, d):
self.__dict__ = {}
for key, value in d.items():
if type(value) is dict:
value = PyJSON(value)
self.__dict__[key] = value
def to_dict(self):
d = {}
for key, value in self.__dict__.items():
if type(value) is PyJSON:
value = value.to_dict()
d[key] = value
return d
def __repr__(self):
return str(self.to_dict())
def __setitem__(self, key, value):
self.__dict__[key] = value
def __getitem__(self, key):
return self.__dict__[key]
json_str = """... JSON string ..."""
py_json = PyJSON(json_str)
JSON到python对象
下面的代码递归地使用对象键创建动态属性。
JSON对象- fb_data.json:
{
"name": "John Smith",
"hometown": {
"name": "New York",
"id": 123
},
"list": [
"a",
"b",
"c",
1,
{
"key": 1
}
],
"object": {
"key": {
"key": 1
}
}
}
在转换中我们有三种情况:
列表 Dicts(新对象) Bool, int, float和STR
import json
class AppConfiguration(object):
def __init__(self, data=None):
if data is None:
with open("fb_data.json") as fh:
data = json.loads(fh.read())
else:
data = dict(data)
for key, val in data.items():
setattr(self, key, self.compute_attr_value(val))
def compute_attr_value(self, value):
if isinstance(value, list):
return [self.compute_attr_value(x) for x in value]
elif isinstance(value, dict):
return AppConfiguration(value)
else:
return value
if __name__ == "__main__":
instance = AppConfiguration()
print(instance.name)
print(instance.hometown.name)
print(instance.hometown.id)
print(instance.list[4].key)
print(instance.object.key.key)
键值对是属性-对象。
输出:
John Smith
New York
123
1
1
将JSON作为代码粘贴
支持TypeScript、Python、Go、Ruby、c#、Java、Swift、Rust、Kotlin、c++、Flow、Objective-C、JavaScript、Elm、JSON Schema。
从JSON、JSON Schema和TypeScript中交互式地生成类型和(反)序列化代码 将JSON/JSON Schema/TypeScript作为代码粘贴
quicktype从示例JSON数据中推断类型，然后输出强类型模型和序列化器，以便用所需的编程语言处理这些数据。
输出:
# Generated by https://quicktype.io
#
# To change quicktype's target language, run command:
#
# "Set quicktype target language"
from typing import List, Union
class Hometown:
name: str
id: int
def __init__(self, name: str, id: int) -> None:
self.name = name
self.id = id
class Key:
key: int
def __init__(self, key: int) -> None:
self.key = key
class Object:
key: Key
def __init__(self, key: Key) -> None:
self.key = key
class FbData:
name: str
hometown: Hometown
list: List[Union[Key, int, str]]
object: Object
def __init__(self, name: str, hometown: Hometown, list: List[Union[Key, int, str]], object: Object) -> None:
self.name = name
self.hometown = hometown
self.list = list
self.object = object
这个扩展可以在Visual Studio代码市场中免费获得。
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。
使用json模块(Python 2.6新增)或几乎总是安装的simplejson模块。
