如何忽略Python中的额外关键字参数？

Question

class Test(object): 
    def __init__(self, id, name): 
     self.id = id 
     self.name = name 

def foo(d): 
    return Test(**d) 

# this works well 
d = {'id': 'aaa', 'name': 'aaa-name'｝ 
foo(d) 
# this would not work well, because of parameter variations, a external para  add into dict d. 
d = {'id': 'aaa', 'name': 'aaa-name', 'test1': 'test1'｝ 

# traceback 
TypeError: __init__() got an unexpected keyword argument 'test1' 

是否有任何方法可以忽略dict d的参数变化？

Answer 1

您可以Test.__init__接受额外**kwargs

class Test(object): 
    def __init__(self, id, name, **kwargs): 
     self.id = id 
     self.name = name 

Answer 2

这将这样的伎俩，只需更新self.__dict__

class Test(object): 
    def __init__(self, id, name, **kwargs): 
     self.id = id 
     self.name = name 
     self.__dict__.update(kwargs) 

例子：

In[2]: class Test(object): 
    def __init__(self, id, name, **kwargs): 
     self.id = id 
     self.name = name 
     self.__dict__.update(kwargs) 
In[3]: d = {'id': 'aaa', 'name': 'aaa-name', 'test1': 'test1'} 
In[5]: t = Test(**d) 
In[6]: t.id 
Out[5]: 'aaa' 
In[7]: t.name 
Out[6]: 'aaa-name' 
In[8]: t.test1 
Out[7]: 'test1' 

编辑：

要忽略演员只是不使用kwargs在@ wong2答案：

class Test(object): 
    def __init__(self, id, name, **kwargs): 
     self.id = id 
     self.name = name