Unpickle namedtuple 具有向后兼容性（忽略其他属性）

Question

这是一个模拟运行较旧版本的 Python 程序针对较新版本编写的搁置数据库的场景。理想情况下，User 对象仍会被解析和读入； favouritePet 属性将被忽略。可以理解，它会抛出一个错误，抱怨元组不匹配。

是否有一种好的方法可以使这种情况与命名元组一起使用，或者更好地切换到存储字典或 class 如果需要这种灵活性？

import shelve
from collections import namedtuple

shelf = shelve.open("objectshelf", flag='n')

User = namedtuple("User", ("username", "password", "firstname", "surname", "favouritePet"))
shelf["namedtupleAndrew"] = User("andrew@example.com", "mypassword", "Andrew", "Smith", "cat")

# Redefine User to simulate a previous version of the User object that didn't record favouritePet;
# someone using an old version of the program against a database written to by a new version
User = namedtuple("User", ("username", "password", "firstname", "surname"))
# Throws error "takes 5 positional arguments but 6 were given"
namedTupleRead = shelf["namedtupleAndrew"]

print(namedTupleRead.username)

编辑：为了完整起见，这里使用了相同的想法 class:

import shelve

shelf = shelve.open("objectshelf", flag='n')

class User:

    def __init__(self, username, password, firstname, surname, favouritePet):
        self.username = username
        self.password = password
        self.firstname = firstname
        self.surname = surname
        self.favouritePet = favouritePet

shelf["objectAndrew"] = User("andrew@example.com", "mypassword", "Andrew", "Smith", "cat")

# Redefine User to simulate a previous version of the User object that didn't record favouritePet;
# someone using an old version of the program against a database written to by a new version
class User:

    def __init__(self, username, password, firstname, surname):
        self.username = username
        self.password = password
        self.firstname = firstname
        self.surname = surname

objectRead = shelf["objectAndrew"]

print(objectRead.username)
# favouritePet is still there; it's just a dictionary, after all.
print(objectRead.favouritePet)

Answer 1

我建议使用字典或自定义 class。

命名元组需要的参数与它具有的字段一样多，因此要直接使用命名元组进行此操作，您必须更改 class' __new__ 方法以使用 *args 和 **kwargs 而不是固定的参数列表。如果您查看 User class 的定义（通过添加 verbose=True 参数），您将看到 class 是如何定义的：

...
class User(tuple):
    'User(username, password, firstname, surname, favouritePet)'

    __slots__ = ()

    _fields = ('username', 'password', 'firstname', 'surname', 'favouritePet')

    def __new__(_cls, username, password, firstname, surname, favouritePet):
        'Create new instance of User(username, password, firstname, surname, favouritePet)'
        return _tuple.__new__(_cls, (username, password, firstname, surname, favouritePet))
...

__new__ 必须变成 __new__(_cls, *args, **kwargs)，然后正确解析 args 和 kwargs（您仍然希望能够使用 User('a', 'b', 'c', ...) 以及 User('a', password='b', firstname='c', ...) 但不是 User('a', username='A', ...) 以与 namedtuple 保持一致），然后将结果序列与 tuple.__new__ 一起使用。使用专用 class 而不是以这种方式修改 namedtuple 的行为可能更好。

使用自定义构造函数更改 User namedtuple 的 pickle 方式会更容易，例如：

from collections import namedtuple
import shelve
import copyreg

shelf = shelve.open("test")

User = namedtuple("User", ("username", "password", "firstname", "surname", "favouritePet"))

User.__reduce__ = lambda user: (construct_user, tuple(user))
# or: copyreg.pickle(User, lambda user: (construct_user, tuple(user)))

def construct_user(*args):
    print('creating new user:', args)       # for debugging
    return User(*args[:len(User._fields)])


user = User("andrew@example.com", "mypassword", "Andrew", "Smith", "cat")
print(user)
shelf["namedtupleAndrew"] = user

# redefine User
User = namedtuple("User", ("username", "password", "firstname", "surname"))

print(shelf["namedtupleAndrew"])

只要 construct_user 函数在所有兼容版本中可用，这就可以工作，但正如最初所说，我仍然建议使用不同的数据结构。

Unpickle namedtuple 具有向后兼容性（忽略其他属性）

Unpickle namedtuple with backwards compatibility (ignoring additional attributes)

python

python-3.x

namedtuple