Is it possible to create a partial function without lambda or library function

Question

This is more curiosity than practical.

I have a library function that accepts a function and passes in a variable, but I want to also call that function with other arguments.

def libFunc(fn):
  fn("passes in a useless value")

def foo(a, b):
  print(f"i want to call this function using libFunc with args {a}, {b}")

# This works fine
num1, num2 = 5, 10
libFunc(lambda x, a=num1, b=num2 : foo(a, b))

# is there a way to avoid using a lambda function here, maybe using a named function?
# i.e. what is the equivalent named function of the above lambda function
def bar(uselessValue, a, b):
  return foo(a, b)

# my attempt but I'm not sure how to deal with the first argument being unknown
# without using partial functions from libraries
libFunc( bar(uselessValue=???, a=num1, b=num2) )

Am I missing something? Why not def bar(uselessValue, a=num1, b=num2): and call it with libFunc(bar)? Lambdas are nothing special, you can always replace them with named functions. — ggorlen
– ggorlen, Commented Jul 20, 2021 at 6:01
@ggorlen ah that works indeed. I was originally trying to reuse a function that was defined before num1 and num2 were defined. With or without the lambda I would have the repeated function definition. Thank you! — Danny
– Danny, Commented Jul 20, 2021 at 6:09

leaf_soba · Accepted Answer · 2021-07-20 06:33:16Z

2

I think ggorlen's answer is the best answer, but if you still want to know how to create a partial function without built-in library to do it, you can try this.

code:

def libFunc(fn):
  fn("passes in a useless value")

def foo(a, b):
  print(f"i want to call this function using libFunc with args {a}, {b}")

def partial(func, *args,**kwargs):
    def wrapper(uselessvalue):
        return func(*args,**kwargs)
    return wrapper

num1, num2 = 5, 10
libFunc(partial(foo,a=num1, b=num2))

result:

i want to call this function using libFunc with args 5, 10

answered Jul 20, 2021 at 6:33

leaf_soba

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