为什么我不能腌制 typing.NamedTuple 而我可以腌制 collections.namedtuple?

Why can't I pickle a typing.NamedTuple while I can pickle a collections.namedtuple?

为什么我不能腌制 typing.NamedTuple 而我可以腌制 collections.namedtuple?我怎样才能做 pickle a NamedTuple?

这段代码显示了我到目前为止所做的尝试:

from collections import namedtuple
from typing import NamedTuple

PersonTyping = NamedTuple('PersonTyping', [('firstname',str),('lastname',str)])
PersonCollections = namedtuple('PersonCollections', ['firstname','lastname'])

pt = PersonTyping("John","Smith")
pc = PersonCollections("John","Smith")


import pickle
import traceback

try:
    with open('personTyping.pkl', 'wb') as f:
        pickle.dump(pt, f)
except:
    traceback.print_exc()
try:
    with open('personCollections.pkl', 'wb') as f:
        pickle.dump(pc, f)
except:
    traceback.print_exc()

shell 上的输出:

$ python3 prova.py 
Traceback (most recent call last):
  File "prova.py", line 16, in <module>
    pickle.dump(pt, f)
_pickle.PicklingError: Can't pickle <class 'typing.PersonTyping'>: attribute lookup PersonTyping on typing failed
$ 

这是一个错误。我已经在上面开票了:http://bugs.python.org/issue25665

问题是 namedtuple 函数在创建 class 时通过从调用框架的全局变量中查找 __name__ 属性来设置其 __module__ 属性。在这种情况下,调用者是 typing.NamedTuple.

result.__module__ = _sys._getframe(1).f_globals.get('__name__', '__main__')

因此,在这种情况下最终将其设置为 'typing'

>>> type(pt)
<class 'typing.PersonTyping'>  # this should be __main__.PersonTyping
>>> type(pc)
<class '__main__.PersonCollections'>
>>> import typing
>>> typing.NamedTuple.__globals__['__name__']
'typing'

修复:

NamedTuple 函数应该自行设置它:

def NamedTuple(typename, fields):

    fields = [(n, t) for n, t in fields]
    cls = collections.namedtuple(typename, [n for n, t in fields])
    cls._field_types = dict(fields)
    try:
        cls.__module__ = sys._getframe(1).f_globals.get('__name__', '__main__')
    except (AttributeError, ValueError):
        pass
    return cls

现在您还可以:

PersonTyping = NamedTuple('PersonTyping', [('firstname',str),('lastname',str)])
PersonTyping.__module__ = __name__