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
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"
}
其他回答
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
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))
我喜欢Onur的答案,但会扩展到包括一个可选的toJSON()方法,用于对象序列化自己:
def dumper(obj):
try:
return obj.toJSON()
except:
return obj.__dict__
print json.dumps(some_big_object, default=dumper, indent=2)
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"
}
只需要像这样添加to_json方法到你的类中:
def to_json(self):
return self.message # or how you want it to be serialized
然后将这段代码(来自这个答案)添加到所有内容的顶部:
from json import JSONEncoder
def _default(self, obj):
return getattr(obj.__class__, "to_json", _default.default)(obj)
_default.default = JSONEncoder().default
JSONEncoder.default = _default
这将会在导入json模块时monkey-patch,所以 JSONEncoder.default()自动检查特殊的to_json() 方法,并使用它对找到的对象进行编码。
就像Onur说的,但是这次你不需要更新项目中的每个json.dumps()。