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
你们为什么要把事情搞得这么复杂?这里有一个简单的例子:
#!/usr/bin/env python3
import json
from dataclasses import dataclass
@dataclass
class Person:
first: str
last: str
age: int
@property
def __json__(self):
return {
"name": f"{self.first} {self.last}",
"age": self.age
}
john = Person("John", "Doe", 42)
print(json.dumps(john, indent=4, default=lambda x: x.__json__))
这样你也可以序列化嵌套类,因为__json__返回一个python对象而不是字符串。不需要使用JSONEncoder,因为使用简单lambda的默认参数也可以很好地工作。
我使用@property代替了一个简单的函数,因为这样感觉更自然和现代。@dataclass也只是一个例子,它也适用于“普通”类。
其他回答
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"
}
}
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
我没有看到这里提到串行版本或backcompat,所以我将发布我的解决方案,我已经使用了一点。我可能还有很多东西要学习,特别是Java和Javascript可能比我更成熟,但我要这样做
https://gist.github.com/andy-d/b7878d0044a4242c0498ed6d67fd50fe
如果你能够安装一个软件包,我建议你试试dill,它在我的项目中工作得很好。这个包的一个优点是它具有与pickle相同的接口,因此如果您已经在项目中使用了pickle,则可以简单地替换为dill并查看脚本是否运行,而无需更改任何代码。所以这是一个非常便宜的解决方案!
(完全反披露:我与莳萝项目没有任何关联,也从未参与过。)
安装包:
pip install dill
然后编辑你的代码导入莳萝而不是pickle:
# import pickle
import dill as pickle
运行脚本,看看它是否有效。(如果是的话,你可能想要清理你的代码,这样你就不再隐藏pickle模块的名字了!)
关于dill可以和不能序列化的数据类型的一些细节,来自项目页面:
dill can pickle the following standard types: none, type, bool, int, long, float, complex, str, unicode, tuple, list, dict, file, buffer, builtin, both old and new style classes, instances of old and new style classes, set, frozenset, array, functions, exceptions dill can also pickle more ‘exotic’ standard types: functions with yields, nested functions, lambdas, cell, method, unboundmethod, module, code, methodwrapper, dictproxy, methoddescriptor, getsetdescriptor, memberdescriptor, wrapperdescriptor, xrange, slice, notimplemented, ellipsis, quit dill cannot yet pickle these standard types: frame, generator, traceback
前几天我遇到了这个问题,并为Python对象实现了一个更通用的Encoder版本,可以处理嵌套对象和继承字段:
import json
import inspect
class ObjectEncoder(json.JSONEncoder):
def default(self, obj):
if hasattr(obj, "to_json"):
return self.default(obj.to_json())
elif hasattr(obj, "__dict__"):
d = dict(
(key, value)
for key, value in inspect.getmembers(obj)
if not key.startswith("__")
and not inspect.isabstract(value)
and not inspect.isbuiltin(value)
and not inspect.isfunction(value)
and not inspect.isgenerator(value)
and not inspect.isgeneratorfunction(value)
and not inspect.ismethod(value)
and not inspect.ismethoddescriptor(value)
and not inspect.isroutine(value)
)
return self.default(d)
return obj
例子:
class C(object):
c = "NO"
def to_json(self):
return {"c": "YES"}
class B(object):
b = "B"
i = "I"
def __init__(self, y):
self.y = y
def f(self):
print "f"
class A(B):
a = "A"
def __init__(self):
self.b = [{"ab": B("y")}]
self.c = C()
print json.dumps(A(), cls=ObjectEncoder, indent=2, sort_keys=True)
结果:
{
"a": "A",
"b": [
{
"ab": {
"b": "B",
"i": "I",
"y": "y"
}
}
],
"c": {
"c": "YES"
},
"i": "I"
}