如何持久化和加载数据类的所有属性

Question

我想保留作为数据类实例的对象的所有属性。然后我想从我保留的文件中加载回那个对象。

这是一个完成任务的例子：

from dataclasses import dataclass
import pickle


@dataclass
class Circle:
    radius: float
    centre: tuple

    def save(self, path: str):
        name = ".".join(("radius", "pkl"))
        with open("/".join((path, name)), "wb") as f:
            pickle.dump(self.radius, f)
        name = ".".join(("centre", "pkl"))
        with open("/".join((path, name)), "wb") as f:
            pickle.dump(self.centre, f)

    @classmethod
    def load(cls, path):
        my_model = {}
        name = "radius"
        file_name = ".".join((name, "pkl"))
        with open("\".join((path, file_name)), "rb") as f:
            my_model[name] = pickle.load(f)
        name = "centre"
        file_name = ".".join((name, "pkl"))
        with open("\".join((path, file_name)), "rb") as f:
            my_model[name] = pickle.load(f)
        return cls(**my_model)

>>> c = Circle(2, (0, 0))
>>> c.save(r".\Circle")
>>> c_loaded = Circle.load(r".\Circle")
>>> c_loaded == c
True

如您所见，我需要为每个属性重复相同的代码，更好的方法是什么？

Answer 1

在保存方法中使用self.__dict__。它包含作为字典的所有属性名称和值。 Load 是一个类方法，所以在那个阶段没有 __dict__ 。但是，cls.__annotations__ 包含属性名称和类型，仍然存储在字典中。

这是最终结果：

from dataclasses import dataclass
import pickle

@dataclass
class Circle:
    radius: float
    centre: tuple

    def save(self, path):
        for name, attribute in self.__dict__.items():
            name = ".".join((name, "pkl"))
            with open("/".join((path, name)), "wb") as f:
                pickle.dump(attribute, f)

    @classmethod
    def load(cls, path):
        my_model = {}
        for name in cls.__annotations__:
            file_name = ".".join((name, "pkl"))
            with open("/".join((path, file_name)), "rb") as f:
                my_model[name] = pickle.load(f)
        return cls(**my_model)

如何持久化和加载数据类的所有属性

How to persist and load all attributes of a dataclass

python

pickle

python-dataclasses