Typehinting lambda function as function argument

Question

I have a function that takes in a lambda:

def my_function(some_lambda):
  # do stuff
  some_other_variable = some_lambda(some_variable)

my_function(lambda x: x + 2)

I would like to typehint the lambda function passed.

I've tried

def my_function(some_lambda: lambda) -> None:
# SyntaxError: invalid syntax
from typing import Lambda
# ImportError: cannot import name 'Lambda'

My IDE complains about similar things on 2.7 straddled typehints, eg

def my_function(some_lambda: lambda) -> None:
  # type: (lambda) -> None
# formal parameter name expected

This question is similar to: How can I specify the function type in my type hints?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — InSync
– InSync, Commented Jul 27, 2024 at 13:37

wonton · Accepted Answer · 2017-09-07 21:16:12Z

11

This is obvious when you think about it, but it took a while to register in the head. A lambda is a function. There is no function type but there is a Callable type in the typing package. The solution to this problem is

from typing import Callable
def my_function(some_lambda: Callable) -> None:

Python 2 version:

from typing import Callable
def my_function(some_lambda):
  # type: (Callable) -> None

answered Sep 7, 2017 at 21:16

wonton

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2 Comments

chepner Over a year ago

You can also further restrict the function type: e.g., Callable[[int], int] is the type of a one-argument function that takes an int and returns an int.

chepner Over a year ago

There is a function type (types.FunctionType), but it's not something you usually want to use, because it's more of an implementation detail (specifically, it's the type of things defined by def statements and lambda expressions; other types exist for callable things like built-in functions, methods, etc.)

Lain · Accepted Answer · 2024-07-09 19:24:04Z

0

Callable is the answer to your question.

from typing import Callable


some_lambda: Callable[[int], int] = lambda x: x + 2


def my_function(func: Callable[[int], int]) -> None:
    # do some stuff
        

my_function(some_lambda)

edited Jul 9, 2024 at 19:24

Lain

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answered Aug 18, 2020 at 16:14

Vlad Bezden

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