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。
这是我的3美分… 这演示了一个树状python对象的显式json序列化。 注意:如果你真的想要这样的代码,你可以使用twisted FilePath类。
import json, sys, os
class File:
def __init__(self, path):
self.path = path
def isdir(self):
return os.path.isdir(self.path)
def isfile(self):
return os.path.isfile(self.path)
def children(self):
return [File(os.path.join(self.path, f))
for f in os.listdir(self.path)]
def getsize(self):
return os.path.getsize(self.path)
def getModificationTime(self):
return os.path.getmtime(self.path)
def _default(o):
d = {}
d['path'] = o.path
d['isFile'] = o.isfile()
d['isDir'] = o.isdir()
d['mtime'] = int(o.getModificationTime())
d['size'] = o.getsize() if o.isfile() else 0
if o.isdir(): d['children'] = o.children()
return d
folder = os.path.abspath('.')
json.dump(File(folder), sys.stdout, default=_default)
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/
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()。