is there possible to pass more than one keyword argument to a function in python?
foo(self, **kwarg) # Want to pass one more keyword argument here 
- as Chris says below, your function would accept any number of keyword arguments without any changes.Yuji 'Tomita' Tomita– Yuji 'Tomita' Tomita2011-01-11 02:37:18 +00:00Commented Jan 11, 2011 at 2:37
- What are you trying to do that needs this?detly– detly2011-01-11 03:05:12 +00:00Commented Jan 11, 2011 at 3:05
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4 Answers
You only need one keyword-arguments parameter; it receives any number of keyword arguments.
def foo(**kwargs): return kwargs >>> foo(bar=1, baz=2) {'baz': 2, 'bar': 1} 3 Comments
user469652
I just want them in two separate dictionary.
Mark Ransom
@user469652, that's a very strange request.
C. K. Young
@user469652: How would you decide which dictionary should hold which keys?
I would write a function to do this for you
def partition_mapping(mapping, keys): """ Return two dicts. The first one has key, value pair for any key from the keys argument that is in mapping and the second one has the other keys from mapping """ # This could be modified to take two sequences of keys and map them into two dicts # optionally returning a third dict for the leftovers d1 = {} d2 = {} keys = set(keys) for key, value in mapping.iteritems(): if key in keys: d1[key] = value else: d2[key] = value return d1, d2 You can then use it like this
def func(**kwargs): kwargs1, kwargs2 = partition_mapping(kwargs, ("arg1", "arg2", "arg3")) This will get them into two separate dicts. It doesn't make any sense for python to provide this behavior as you have to manually specify which dict you want them to end up in. Another alternative is to just manually specify it in the function definition
def func(arg1=None, arg2=None, arg3=None, **kwargs): # do stuff Now you have one dict for the ones you don't specify and regular local variables for the ones you do want to name.
Comments
You cannot. But keyword arguments are dictionaries and when calling you can call as many keyword arguments as you like. They all will be captured in the single **kwarg. Can you explain a scenario where you would need more than one of those **kwarg in the function definition?
>>> def fun(a, **b): ... print b.keys() ... # b is dict here. You can do whatever you want. ... ... >>> fun(10,key1=1,key2=2,key3=3) ['key3', 'key2', 'key1'] 
2 Comments
user469652
I somehow has to make some values go into another dictionary. Otherwise, it won't work.
Senthil Kumaran
The arg you pass, 'b' is actually a dict. You can do any dict operation there. Copy the keys and values conditionally, etc.
Maybe this helps. Can you clarify how the kw arguments are divided into the two dicts?
>>> def f(kw1, kw2): ... print kw1 ... print kw2 ... >>> f(dict(a=1,b=2), dict(c=3,d=4)) {'a': 1, 'b': 2} {'c': 3, 'd': 4} 