如何正确腌制一个namedtuple实例

问题描述

我正在学习如何使用泡菜.我创建了一个namedtuple对象，并将其附加到列表中，并尝试使该列表腌制.但是，出现以下错误:

I'm learning how to use pickle. I've created a namedtuple object, appended it to a list, and tried to pickle that list. However, I get the following error:

pickle.PicklingError: Can't pickle <class '__main__.P'>: it's not found as __main__.P

我发现，如果我在运行代码时未将其包装在函数中，则它可以完美运行.当包裹在函数中时，是否需要额外的步骤来腌制一个对象?

I found that if I ran the code without wrapping it inside a function, it works perfectly. Is there an extra step required to pickle an object when wrapped inside a function?

这是我的代码:

from collections import namedtuple
import pickle

def pickle_test():
    P = namedtuple("P", "one two three four")
    my_list = []
    abe = P("abraham", "lincoln", "vampire", "hunter")
    my_list.append(abe)
    f = open('abe.pickle', 'w')
    pickle.dump(abe, f)
    f.close()

pickle_test()

推荐答案

创建函数的命名元组外部:

from collections import namedtuple
import pickle

P = namedtuple("P", "one two three four")

def pickle_test():
    my_list = []
    abe = P("abraham", "lincoln", "vampire", "hunter")
    my_list.append(abe)
    f = open('abe.pickle', 'w')
    pickle.dump(abe, f)
    f.close()

pickle_test()

现在pickle可以找到它；现在它是一个全局模块.拆线时，所有pickle模块要做的就是再次定位__main__.P.在您的版本中，P是pickle_test()函数的 local ，并且不是自省的或可导入的.

Now pickle can find it; it is a module global now. When unpickling, all the pickle module has to do is locate __main__.P again. In your version, P is a local, to the pickle_test() function, and that is not introspectable or importable.

请记住，namedtuple()是类工厂，这一点很重要；您给它参数，它返回一个类对象供您创建实例. pickle仅存储实例中包含的 data ，外加对原始类的字符串引用以再次重建实例.

It is important to remember that namedtuple() is a class factory; you give it parameters and it returns a class object for you to create instances from. pickle only stores the data contained in the instances, plus a string reference to the original class to reconstruct the instances again.

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