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
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
其他回答
下面是一个简单功能的简单解决方案:
.toJSON()方法
实现一个序列化器方法,而不是一个JSON可序列化类:
import json
class Object:
def toJSON(self):
return json.dumps(self, default=lambda o: o.__dict__,
sort_keys=True, indent=4)
所以你只需调用它来序列化:
me = Object()
me.name = "Onur"
me.age = 35
me.dog = Object()
me.dog.name = "Apollo"
print(me.toJSON())
将输出:
{
"age": 35,
"dog": {
"name": "Apollo"
},
"name": "Onur"
}
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))
为了给这场11年的大火再添一根柴,我想要一个满足以下条件的解决方案:
只允许使用json.dumps(obj)序列化类FileItem的实例 允许FileItem实例具有属性:FileItem .fname 允许FileItem实例提供给任何库,使用json.dumps(obj)序列化它 不需要将任何其他字段传递给json。转储(如自定义序列化器)
IE:
fileItem = FileItem('filename.ext')
assert json.dumps(fileItem) == '{"fname": "filename.ext"}'
assert fileItem.fname == 'filename.ext'
我的解决方案是:
obj的类是否继承自dict 将每个对象属性映射到底层字典
class FileItem(dict):
def __init__(self, fname):
self['fname'] = fname
#fname property
fname: str = property()
@fname.getter
def fname(self):
return self['fname']
@fname.setter
def fname(self, value: str):
self['fname'] = value
#Repeat for other properties
是的,如果你有很多属性,这有点冗长,但它是JSONSerializable,它的行为像一个对象,你可以把它给任何库,去json.dumps(obj)它。
对于更复杂的类,您可以考虑使用jsonpickle工具:
jsonpickle is a Python library for serialization and deserialization of complex Python objects to and from JSON. The standard Python libraries for encoding Python into JSON, such as the stdlib’s json, simplejson, and demjson, can only handle Python primitives that have a direct JSON equivalent (e.g. dicts, lists, strings, ints, etc.). jsonpickle builds on top of these libraries and allows more complex data structures to be serialized to JSON. jsonpickle is highly configurable and extendable–allowing the user to choose the JSON backend and add additional backends.
(链接到PyPi上的jsonpickle)
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