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__接受额外的**kwargs

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

Answer 2

这样就可以了，只需更新self.__dict__

即可

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

示例：

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'

修改

要忽略附加内容，请不要使用@ wong2中的kwargs：

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

如何在Python中忽略额外的关键字参数？

2 个答案: