如果我有一个类：
class Person(object):
def __init__(self,name,**kwargs):
self.name = name
p = Person(name='joe',age=25) # age is ignored
额外的参数被忽略了.但如果我有一个namedtuple,我会得到`意外的关键字参数：
from collections import namedtuple
Person = namedtuple('Person','name')
p = Person(name='joe',age=25)
# Traceback (most recent call last):
# File "python",line 1,in <module>
# TypeError: __new__() got an unexpected keyword argument 'age'
我如何让namedtuple接受kwargs,这样我才能安全地传递额外的参数？
解决方法
解释器中的以下会话显示了解决问题的一种可能解决方案：
Python 3.5.0 (v3.5.0:374f501f4567,Sep 13 2015,02:27:37) [MSC v.1900 64 bit (AMD64)] on win32
Type "copyright","credits" or "license()" for more information.
>>> import collections
>>> class Person(collections.namedtuple('base','name')):
__slots__ = ()
def __new__(cls,*args,**kwargs):
for key in tuple(kwargs):
if key not in cls._fields:
del kwargs[key]
return super().__new__(cls,**kwargs)
>>> p = Person(name='joe',age=25)
>>> p
Person(name='joe')
>>>
替代方案：
由于您更倾向于使用更简单的解决方案,因此您可能会发现下一个程序更符合您的喜好：
#! /usr/bin/env python3
import collections
def main():
Person = namedtuple('Person','name')
p = Person(name='joe',age=25)
print(p)
def namedtuple(typename,field_names,verbose=False,rename=False):
base = collections.namedtuple('Base',verbose,rename)
return type(typename,(base,),{
'__slots__': (),'__new__': lambda cls,**kwargs: base.__new__(cls,**{
key: value for key,value in kwargs.items()
if key in base._fields})})
if __name__ == '__main__':
main()