Python arguments as a dictionary

Question

How can I get argument names and their values passed to a method as a dictionary?

I want to specify the optional and required parameters for a GET request as part of a HTTP API in order to build the URL. I'm not sure of the best way to make this pythonic.

Solution 1

Use a single argument prefixed with **.

>>> def foo(**args):
...     print(args)
...
>>> foo(a=1, b=2)
{'a': 1, 'b': 2}

Solution 2

For non-keyworded arguments, use a single *, and for keyworded arguments, use a **.

For example:

def test(*args, **kwargs):
    print args
    print kwargs

>>test(1, 2, a=3, b=4)
(1, 2)
{'a': 3, 'b': 4}

Non-keyworded arguments would be unpacked to a tuple and keyworded arguments would be unpacked to a dictionary. Unpacking Argument Lists

Comments

[Rob Wouters]

Post some code to illustrate, this could mean several things (or perhaps just two).

[Devin M]

Something like this? stackoverflow.com/a/196997/590177

[Marcin]

possible duplicate of Can you list the keyword arguments a Python function receives?

[user1066101]

"I’m wanting to wanting to specify the optional and required parameters for a GET request"? Can you expand on this? It makes very little sense.

[outis]

possible duplicate of Get a list/tuple/dict of the arguments passed to a function?

[jdi]

And just to add, a single * is used to accept an unnamed amount of non keyword args as a list: *args

[Makoto]

The result would be: (1, 2) {'a': 3, 'b': 4}.

[Fred Foo]

@Marcin: Well, maybe I give away code too easily, but this is the kind of construct that I personally always find hard to search for when learning a language.

[Vanuan]

What if I still want the possible keyword arguments (and defaults) to be provided in the function's definition?