如何找到有问题的（不可腌制的）对象？

Question

尝试 pickle 对象时，我收到以下错误消息：

TypeError: cannot serialize '_io.TextIOWrapper' object

有问题的对象 class 继承了另一个 class (pytorch_lightning.LightningModule)，所以我假设有问题的对象属于超级 class。我怎样才能找到它？错误跟踪只会导致我的 pickling 命令，即

with open(save_path, "wb") as p:
    pickle.dump(self, p)

Answer 1

根据pickle documentation you can influence on how an object is pickled by defining __getstate__函数。

Classes can further influence how their instances are pickled; if the class defines the method __getstate__(), it is called and the returned object is pickled as the contents for the instance, instead of the contents of the instance’s dictionary. If the __getstate__() method is absent, the instance’s __dict__ is pickled as usual.

尝试执行以下操作：

def __getstate__(self):
    for variable_name, value in vars(self).items():
        try:
            pickle.dumps(value)
        expect:
            print('{variable_name} with value {value} is not pickable')

当您尝试 pickle 对象时，它应该打印出对象的 un-picklable 对象。

替代方式：

If the __getstate__() method is absent, the instance’s __dict__ is pickled as usual.

以相同的方式检查对象 __dict__，尝试对每个变量进行 pickle 以找出有问题的变量。

如何找到有问题的（不可腌制的）对象？

How to locate problematic (un-picklable) object?

python

pickle

替代方式：