How do I add default parameters to functions when using type hinting?

If I have a function like this:

def foo(name, opts={}): pass

And I want to add type hints to the parameters, how do I do it? The way I assumed gives me a syntax error:

def foo(name: str, opts={}: dict) -> str: pass

The following doesn't throw a syntax error but it doesn't seem like the intuitive way to handle this case:

def foo(name: str, opts: dict={}) -> str: pass

I can't find anything in the typing documentation or on a Google search.

Edit: I didn't know how default arguments worked in Python, but for the sake of this question, I will keep the examples above. In general it's much better to do the following:

def foo(name: str, opts: dict=None) -> str: if not opts: opts={} pass

3 Answers

Your second way is correct.

def foo(opts: dict = {}): pass
print(foo.__annotations__)

this outputs

{'opts': <class 'dict'>}

It's true that's it's not listed in PEP 484, but type hints are an application of function annotations, which are documented in PEP 3107. The syntax section makes it clear that keyword arguments works with function annotations in this way.

I strongly advise against using mutable keyword arguments. More information here.

If you're using typing (introduced in Python 3.5) you can use typing.Optional, where Optional[X] is equivalent to Union[X, None]. It is used to signal that the explicit value of None is allowed . From typing.Optional:

def foo(arg: Optional[int] = None) -> None: ...

I recently saw this one-liner:

def foo(name: str, opts: dict=None) -> str: opts = {} if not opts else opts pass