More KWARGS!

Object destructuring of function parameters for Python!

Motivation

Javascript has object destructuring, and it can be used for function parameters. This has a couple of benefts over Python's keyword arguments:

Extra caller parameters are ignored (eg f({a, b, c}))
Duplicate parameters are handled elegantly (eg f({a, a}))

The mo-kwargs library provides this functionality with the @override decorator, with additional benefits:

required parameters throw an error if missing, just like regular Python
all parameters, even ones not in the argument list, are passed in the optional kwargs parameter

The @override decorator adds a kwargs argument which can be passed a dict of call parameters; but unlike **kwargs, it will not raise duplicate key exceptions.

Provide default values

We decorate the login() function with @override. username is a required parameter, and password will default to None.

    @override
    def login(username, password=None):
        pass

Define some dicts for use with our kwargs parameter:

    creds = {"userame": "ekyle", "password": "password123"}
    alt_creds = {"username": "klahnakoski"}

The simplest case is when we use kwargs with no overrides

    login(kwargs=creds)
    # SAME AS
    login(**creds)
    # SAME AS
    login(username="ekyle", password="password123")

You may override any property in kwargs: In this case it is password

    login(password="123", kwargs=creds)
    # SAME AS
    login(username="ekyle", password="123")

There is no problem with overriding everything in kwargs:

    login(username="klahnakoski", password="asd213", kwargs=creds)
    # SAME AS
    login(username="klahnakoski", password="asd213")

You may continue to use **kwargs; which provides a way to overlay one parameter template (creds) with another (alt_creds)

    login(kwargs=creds, **alt_creds)
    # SAME AS
    login(username="klahnakoski", password="password123")

Handle too many parameters

Sometimes your method parameters come from a configuration file, or some other outside source which is outside your control. There may be more parameters than your method is willing to accept.

    creds = {"username": "ekyle", "password": "password123", "port":9000}
    def login(username, password=None):
         print(kwargs.get("port"))

Without mo-kwargs, passing the creds dictionary directly to login() would raise a key error

    >>> login(**creds)
    Traceback (most recent call last):
      File "<stdin>", line 1, in <module>
    TypeError: login() got an unexpected keyword argument 'port'

The traditional solution is to pass the parameters explicitly:

    login(username=creds.username, password=creds.password)

but that can get get tedious when done often, or the parameter list get long. mo-kwargs allows you to pass the whole dictionary to the kwargs parameter; only the parameters used by the method are used:

    @override
    def login(username, password=None):
        pass
     
    login(kwargs=creds)
    # SAME AS
    login(username=creds.username, password=creds.password)

Package all parameters

Your method can accept kwargs as a parameter. If it does, ensure it defaults to None so that it's not required.

    @override
    def login(username, password=None, kwargs=None):
        print(kwargs.get("username"))
        print(kwargs.get("port"))

kwargs will always be a dict, possibly empty, with the full set of parameters. This is different from using **kwargs which contains only the remainder of the keyword parameters.

    >>> creds = {"username": "ekyle", "password": "password123", "port":9000}
    >>> login(**creds)
    ekyle
    9000

Version Changes, Features

Version 8

March 2026

the kwargs parameter no longer incudes itself, and no longer includes first argument self, cls of instance and class methods.
the variable keyword argument (**) includes all parameters not listed in the method signature. (this was broken in previous versions).

Metadata

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