python - 更改已保存 pickle 的导入库

Question

前段时间我将一个类函数保存到 pickle 文件中。

<library.module.Class at 0x1c926b2e520>

该类导入了两个同时更改了名称的库。

是否可以编辑 pickle 文件，以便我可以更新这 2 个新导入而无需重新生成 pickle？

谢谢!问候

编辑:

这是我加载 pickle 的方式:

import pickle
model_path = os.getenv('MODELS_FOLDER') + 'model_20210130.pkl'
model = pickle.load(open(model, 'rb'))

这是pickle 类的内容。我要更新的两个库已精确定位。

**import socceraction.spadl.config** as spadlconfig
**from socceraction.spadl.base import** SPADLSchema

class ExpectedThreat:
    """An implementation of the model.

    """

    def __init__(self):
...
     

    def __solve(
        self) -> None:
        

    def fit(self, actions: DataFrame[SPADLSchema]) -> 'ExpectedThreat':
        """Fits the xT model with the given actions."""

        

    def predict(
        self, actions: DataFrame[SPADLSchema], use_interpolation: bool = False
    ) -> np.ndarray:
        """Predicts the model values for the given actions.

        

Answer 1

我认为你做不到。

Pickled 对象是序列化的内容，为了被修改，必须适当反序列化。

为什么不能直接加载、更改和覆盖它？

您仍然可以用新函数覆盖属性等函数:

# import new libraries
import new_socceraction.spadl.config as new_spadlconfig
from new_socceraction.spadl.base import new_SPADLSchema

import pickle

model_path = os.getenv('MODELS_FOLDER') + 'model_20210130.pkl'

with open(model_path, 'rb') as file:
    model = pickle.load(file)

# define new methods with different libraries 

def fit(self, actions: DataFrame[new_SPADLSchema]) -> 'ExpectedThreat':
    """Fits the xT model with the given actions."""
    pass # your new implementation
   
def predict(
        self, actions: DataFrame[new_SPADLSchema], use_interpolation: bool = False
    ) -> np.ndarray:
    """Predicts the model values for the given actions."""
    pass # your new implementation 


# overwrite new methods 
model.fit = fit
model.predict = predict 

# save it again
with open(model_path, 'wb') as file:
    pickle.dump(file)
...