How do I define a Python function with optional arguments?

General Discussions
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1

I want to define a function in Python where some arguments are optional and can be passed in any combination.

I’m aware that Python doesn’t support function overloading, so I’m looking for clean ways to handle Python optional arguments without using placeholders or sending empty strings. What’s the recommended approach?

2

I’ve run into this when writing utility functions where some parameters might or might not be needed.

The cleanest way I handle python optional arguments is by setting default values to None and checking inside the function.

For example:

def process_data(a, b, c=None, d=None, e=None):
    if c:
        # do something with c
    if d:
        # etc.

It’s flexible, readable, and you can pass just the arguments you need without worrying about order, just use keyword arguments when calling.

3

When I have a lot of optional parameters that aren’t always used, I like to go with kwargs. It’s super dynamic and clean. Here’s how I typically do it:

def my_function(a, b, **kwargs):
    if 'd' in kwargs:
        do_something(kwargs['d'])
Then I call it like this:

my_function(1, 2, d='optional', e='also optional')

It gives me total freedom and avoids the mess of too many positional parameters.

4

Since Python 3, I’ve started using keyword-only arguments to make optional values explicit and prevent confusion. You just use a * in the function signature:

def generate_report(a, b, *, format='pdf', include_summary=False):
    # your logic

Now, format and include_summary can only be passed by name, which improves readability and avoids bugs. It’s a great middle ground between defaults and flexibility!