Python **kwargs and self as an argument

Question

I don't have much experience in python but I am studying **kwargs.

Afer reading a lot I understood somethings about **kwargs but I have a small problem or I am not understanding something correct.

So this works:

def test_var_kwargs(farg, **kwargs):
    print "formal arg:", farg
    for key in kwargs:
        print "another keyword arg: %s: %s" % (key, kwargs[key])

test_var_kwargs(farg=1, myarg2="two", myarg3=3)

And prints:

formal arg: 1

another keyword arg: myarg2: two

another keyword arg: myarg3: 3

But if that function was an instance function then self would have to be included:

def test_var_kwargs(self, farg, **kwargs):
        print "formal arg:", farg
        for key in kwargs:
            print "another keyword arg: %s: %s" % (key, kwargs[key])
    
self.test_var_kwargs(farg=1, myarg2="two", myarg3=3)

But this produces an error:

TypeError: test_var_kwargs() takes exactly 2 arguments (1 given)

I understand that I have to pass self like:

self.test_var_kwargs(self, farg=1, myarg2="two", myarg3=3)

Why do I have to include self as an argument in the class instance's method?

Could you include the full code example of the **class function** example. I think you might be misunderstanding it. — Wessie, Jan 15 '13 at 16:15

Martijn Pieters · Accepted Answer · 2013-01-15T16:31:12.293

7

You cannot use farg as keyword argument in that case; it cannot be interpreted as both a positional and a keyword argument, Python interprets it as a keyword argument in this case.

Use

self.test_var_kwargs(self, 1, myarg2="two", myarg3=3)

instead.

Functions act as descriptors; when looked up on an instance, they get wrapped in a new object called a method, which automatically adds the instance as a first argument to the function when called. This wrapper essentially does this:

class Method(object):
    def __init__(self, function, instance, cls):
        self.func = function
        self.instance = instance
        self.cls = cls

    def __call__(self, *args, **kw):
        return self.func(self.instance, *args, **kw)

Your farg keyword argument is then lumped under the **kw catchall and passed on to the underlying function, with *args left empty.

But the test_var_kwargs method you defined has one positional argument (next to self, and thus an exception is raised.

edited Jan 15 '13 at 16:31

answered Jan 15 '13 at 16:15

Martijn Pieters

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but why have I to include self on the arguments? – Jimmy Kane Jan 15 '13 at 16:16
It doesn't appear to me that `farg` is interpreted as a positional and a keyword argument simultaneously. With proper class/object syntax, my machine produces the expected result. – Kyle Strand Jan 15 '13 at 16:24
You only need to include "self" for the definitions of instance methods. You NEVER have to include a reference to self when calling a method, that is done implicitly for you. – Zoran Pavlovic Jan 15 '13 at 16:25
@KyleStrand: That was a mis-wording on my part, updated with the expanded answer. – Martijn Pieters Jan 15 '13 at 16:31
Thanks. ouf at last. Thanks @ZoranPavlovic as well – Jimmy Kane Jan 15 '13 at 16:33

Python **kwargs and self as an argument

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