如何使一个Python类序列化?
class FileItem:
def __init__(self, fname):
self.fname = fname
尝试序列化为JSON:
>>> import json
>>> x = FileItem('/foo/bar')
>>> json.dumps(x)
TypeError: Object of type 'FileItem' is not JSON serializable
如何使一个Python类序列化?
class FileItem:
def __init__(self, fname):
self.fname = fname
尝试序列化为JSON:
>>> import json
>>> x = FileItem('/foo/bar')
>>> json.dumps(x)
TypeError: Object of type 'FileItem' is not JSON serializable
当前回答
这对我来说很有效:
class JsonSerializable(object):
def serialize(self):
return json.dumps(self.__dict__)
def __repr__(self):
return self.serialize()
@staticmethod
def dumper(obj):
if "serialize" in dir(obj):
return obj.serialize()
return obj.__dict__
然后
class FileItem(JsonSerializable):
...
and
log.debug(json.dumps(<my object>, default=JsonSerializable.dumper, indent=2))
其他回答
基于Quinten Cabo的回答:
def sterilize(obj):
"""Make an object more ameniable to dumping as json
"""
if type(obj) in (str, float, int, bool, type(None)):
return obj
elif isinstance(obj, dict):
return {k: sterilize(v) for k, v in obj.items()}
list_ret = []
dict_ret = {}
for a in dir(obj):
if a == '__iter__' and callable(obj.__iter__):
list_ret.extend([sterilize(v) for v in obj])
elif a == '__dict__':
dict_ret.update({k: sterilize(v) for k, v in obj.__dict__.items() if k not in ['__module__', '__dict__', '__weakref__', '__doc__']})
elif a not in ['__doc__', '__module__']:
aval = getattr(obj, a)
if type(aval) in (str, float, int, bool, type(None)):
dict_ret[a] = aval
elif a != '__class__' and a != '__objclass__' and isinstance(aval, type):
dict_ret[a] = sterilize(aval)
if len(list_ret) == 0:
if len(dict_ret) == 0:
return repr(obj)
return dict_ret
else:
if len(dict_ret) == 0:
return list_ret
return (list_ret, dict_ret)
区别在于
Works for any iterable instead of just list and tuple (it works for NumPy arrays, etc.) Works for dynamic types (ones that contain a __dict__). Includes native types float and None so they don't get converted to string. Classes that have __dict__ and members will mostly work (if the __dict__ and member names collide, you will only get one - likely the member) Classes that are lists and have members will look like a tuple of the list and a dictionary Python3 (that isinstance() call may be the only thing that needs changing)
Json在它可以打印的对象方面受到限制,而jsonpickle(你可能需要一个PIP安装jsonpickle)在它不能缩进文本方面受到限制。如果你想检查一个你不能改变类的对象的内容,我仍然找不到比:
import json
import jsonpickle
...
print json.dumps(json.loads(jsonpickle.encode(object)), indent=2)
注意:他们仍然不能打印对象方法。
一个非常简单的一行程序解决方案
import json
json.dumps(your_object, default=lambda __o: __o.__dict__)
结束!
下面是一个测试。
import json
from dataclasses import dataclass
@dataclass
class Company:
id: int
name: str
@dataclass
class User:
id: int
name: str
email: str
company: Company
company = Company(id=1, name="Example Ltd")
user = User(id=1, name="John Doe", email="john@doe.net", company=company)
json.dumps(user, default=lambda __o: __o.__dict__)
输出:
{
"id": 1,
"name": "John Doe",
"email": "john@doe.net",
"company": {
"id": 1,
"name": "Example Ltd"
}
}
Jsonweb似乎是我的最佳解决方案。参见http://www.jsonweb.info/en/latest/
from jsonweb.encode import to_object, dumper
@to_object()
class DataModel(object):
def __init__(self, id, value):
self.id = id
self.value = value
>>> data = DataModel(5, "foo")
>>> dumper(data)
'{"__type__": "DataModel", "id": 5, "value": "foo"}'
Kyle Delaney的评论是正确的,所以我尝试使用https://stackoverflow.com/a/15538391/1497139以及https://stackoverflow.com/a/10254820/1497139的改进版本
创建一个“JSONAble”mixin。
因此,要使一个类JSON可序列化使用“JSONAble”作为超类,并调用:
instance.toJSON()
or
instance.asJSON()
对于这两种方法。您还可以使用本文提供的其他方法扩展JSONAble类。
家庭和个人单元测试样本的测试示例结果如下:
toJSOn ():
{
"members": {
"Flintstone,Fred": {
"firstName": "Fred",
"lastName": "Flintstone"
},
"Flintstone,Wilma": {
"firstName": "Wilma",
"lastName": "Flintstone"
}
},
"name": "The Flintstones"
}
asJSOn ():
{'name': 'The Flintstones', 'members': {'Flintstone,Fred': {'firstName': 'Fred', 'lastName': 'Flintstone'}, 'Flintstone,Wilma': {'firstName': 'Wilma', 'lastName': 'Flintstone'}}}
使用家庭和个人样本进行单元测试
def testJsonAble(self):
family=Family("The Flintstones")
family.add(Person("Fred","Flintstone"))
family.add(Person("Wilma","Flintstone"))
json1=family.toJSON()
json2=family.asJSON()
print(json1)
print(json2)
class Family(JSONAble):
def __init__(self,name):
self.name=name
self.members={}
def add(self,person):
self.members[person.lastName+","+person.firstName]=person
class Person(JSONAble):
def __init__(self,firstName,lastName):
self.firstName=firstName;
self.lastName=lastName;
JSONAble .py定义JSONAble mixin
'''
Created on 2020-09-03
@author: wf
'''
import json
class JSONAble(object):
'''
mixin to allow classes to be JSON serializable see
https://stackoverflow.com/questions/3768895/how-to-make-a-class-json-serializable
'''
def __init__(self):
'''
Constructor
'''
def toJSON(self):
return json.dumps(self, default=lambda o: o.__dict__,
sort_keys=True, indent=4)
def getValue(self,v):
if (hasattr(v, "asJSON")):
return v.asJSON()
elif type(v) is dict:
return self.reprDict(v)
elif type(v) is list:
vlist=[]
for vitem in v:
vlist.append(self.getValue(vitem))
return vlist
else:
return v
def reprDict(self,srcDict):
'''
get my dict elements
'''
d = dict()
for a, v in srcDict.items():
d[a]=self.getValue(v)
return d
def asJSON(self):
'''
recursively return my dict elements
'''
return self.reprDict(self.__dict__)
您将发现这些方法现在集成在https://github.com/WolfgangFahl/pyLoDStorage项目中,该项目可在https://pypi.org/project/pylodstorage/上获得