How to ignore extra keyword arguments in Python?

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'

Is there any methods to ignore the parameter variations of dict d?

You pass kwargs, but you don't handle them. Use **kwargs or something in the init signature — Tim
– Tim, Commented Mar 8, 2016 at 8:14
What do you mean by "deal with" the extra keyword args? Ignore them? Use them to set attributes of the object? Something else? — mhawke
– mhawke, Commented Mar 8, 2016 at 8:20

wong2 · Accepted Answer · 2016-03-08 08:14:27Z

13

You can make Test.__init__ accept an extra **kwargs

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

answered Mar 8, 2016 at 8:14

wong2

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6 Comments

Tim Over a year ago

That is half the answer. How to go from here?

wong2 Over a year ago

@TimCastelijns I think that's enough, we can just ignore kwargs

Tim Over a year ago

Does it automatically assign to self from the keys in the dict?

mhawke Over a year ago

@TimCastelijns: in all fairness the OP hasn't really explained what is meant by "deal with" the extra kwargs. At this point ignoring them is perfectly valid. Using them to set attributes on the object might also be valid.

Tim Over a year ago

@mhawke that's true. However in cases where it's not clear exactly what OP is asking, it's better to not answer at all until it's cleared up. Right now we can only guess

|

Netwave · Accepted Answer · 2016-03-08 09:19:01Z

10

This will do the trick, simply update the self.__dict__

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

Example:

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'

EDIT:

To ignore the extras just dont use the kwargs as in @wong2 answer:

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

edited Mar 8, 2016 at 9:19

answered Mar 8, 2016 at 8:22

Netwave

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4 Comments

vaultah Over a year ago

Why not def __init__(self, **kwargs): self.__dict__.update(kwargs)?

Netwave Over a year ago

@vaultah, he may want to force the id and the name to be forced int the params, so you really neead an id and a name

Wilence Over a year ago

I want to ignore extra keyword args.

wong2 Over a year ago

@Wilence see my answer

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