Question

与特定类关联的typename有什么用？例如，

Point = namedtuple('P', ['x', 'y'])

您通常会在哪里使用typename＆＃39; P＆＃39;？

谢谢！

Answer 1

为了理智，namedtuple的第一个参数应该与你赋给它的变量名相同：

>>> from collections import namedtuple
>>> Point = namedtuple('P','x y')
>>> pp = Point(1,2)
>>> type(pp)
<class '__main__.P'>

isinstance并不太关心这个，虽然只是'P'是不知道的：

>>> isinstance(pp,Point)
True
>>> isinstance(pp,P)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'P' is not defined

但是pickle是一个关心找到与typename匹配的类名的模块：

>>> import pickle
>>> ppp = pickle.dumps(pp)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "c:\python26\lib\pickle.py", line 1366, in dumps
    Pickler(file, protocol).dump(obj)
  File "c:\python26\lib\pickle.py", line 224, in dump
    self.save(obj)
  File "c:\python26\lib\pickle.py", line 331, in save
    self.save_reduce(obj=obj, *rv)
  File "c:\python26\lib\pickle.py", line 401, in save_reduce
    save(args)
  File "c:\python26\lib\pickle.py", line 286, in save
    f(self, obj) # Call unbound method with explicit self
  File "c:\python26\lib\pickle.py", line 562, in save_tuple
    save(element)
  File "c:\python26\lib\pickle.py", line 286, in save
    f(self, obj) # Call unbound method with explicit self
  File "c:\python26\lib\pickle.py", line 748, in save_global
    (obj, module, name))
pickle.PicklingError: Can't pickle <class '__main__.P'>: it's not found as __main__.P

如果我将namedtuple定义为'Point'，那么pickle很高兴：

>>> Point = namedtuple('Point','x y')
>>> pp = Point(1,2)
>>> ppp = pickle.dumps(pp)
>>>

不幸的是，由您来管理这种一致性。由于赋值是一个语句而不是Python中的运算符，因此namedtuple无法知道您为其分配输出的内容，因此您必须将正确的类名传递给namedtuple，并将结果类分配给相同的变量名。

在python中使用类类型名称

1 个答案: