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
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)
其他回答
你知道预期产量是多少吗?例如,这个可以吗?
>>> f = FileItem("/foo/bar")
>>> magic(f)
'{"fname": "/foo/bar"}'
在这种情况下,你只需调用json.dumps(f.__dict__)。
如果您想要更多自定义输出,那么您必须继承JSONEncoder并实现您自己的自定义序列化。
对于一个简单的例子,请参见下面。
>>> from json import JSONEncoder
>>> class MyEncoder(JSONEncoder):
def default(self, o):
return o.__dict__
>>> MyEncoder().encode(f)
'{"fname": "/foo/bar"}'
然后你把这个类作为cls kwarg传递给json.dumps()方法:
json.dumps(cls=MyEncoder)
如果还想解码,则必须向JSONDecoder类提供一个自定义object_hook。例如:
>>> def from_json(json_object):
if 'fname' in json_object:
return FileItem(json_object['fname'])
>>> f = JSONDecoder(object_hook = from_json).decode('{"fname": "/foo/bar"}')
>>> f
<__main__.FileItem object at 0x9337fac>
>>>
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。
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()。
我最喜欢Lost Koder的方法。当我试图序列化成员/方法不可序列化的更复杂的对象时,我遇到了问题。这是我的实现,工作在更多的对象:
class Serializer(object):
@staticmethod
def serialize(obj):
def check(o):
for k, v in o.__dict__.items():
try:
_ = json.dumps(v)
o.__dict__[k] = v
except TypeError:
o.__dict__[k] = str(v)
return o
return json.dumps(check(obj).__dict__, indent=2)
这是一个小库,它将一个对象及其所有子对象序列化为JSON,并将其解析回来:
https://github.com/tobiasholler/PyJSONSerialization/