Avoid calling a function twice

Question

I have the following situation in Python:

def func_0(x):
    # connects to a third party software
    val_1, val_2 = output_extracted_from_a_third_party_software
    return [val_1, val_2]

def func_1(x):
    return func_0()[0]

def func_2(x):
    return func_0()[1]

another_third_party_func(func_1, func_2) # this is a function that I cannot modify

Facts:

It is not possible for me to modify func_0 because it just extracts the output of a third party software (a computationally expensive process).
As a part of an algorithm, I have to pass func_1 and func_2 as 2 separate arguments into another third party library.

I'm looking for a more efficient way to define func_1 and func_2, so that I avoid calling func_0 twice.

Thank you in advance.

EDIT: Memorization doesn't work in this case because x has to be a numpy array.

func_1 and func_2 should take the values it needs as arguments, then you can do somethng liek val1, val2 = func_0(x) then simply call y1 = func_1(x, val1) and y2 = func_2(x, val2) — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Oct 5, 2022 at 17:22
@Moshmet it would probably good if you edited your question with more real code as people are getting hung up the fact that you are returning a list and that val1 and val2 are not defined — Joran Beasley
– Joran Beasley, Commented Oct 5, 2022 at 17:25
Your code is confusing: You're not calling func_0 from func_1 and func_2 but are rather treating it like an array, and you don't use the values of x passed to func_1 and func_2. †he key to your question seems to be if you can call func_0 just once, but it is unclear if you want to call it only once with the same value for the x parameter it takes or with two different x values. If you want to call it with the same x value, then you can use some sort of caching, and @JoranBeasley's answer is a great way to do this. — CryptoFool
– CryptoFool, Commented Oct 5, 2022 at 17:27

Joran Beasley · Accepted Answer · 2022-10-05 17:19:12Z

2

you can use memoization to solve this ... i think just the builtin LRU cache would work

@functools.lru_cache
def func_0(x):
    do something using x
    return [val_1, val_2]

of coarse this only works if x is the same in both calls ...

you can see it work as follows

import functools

@functools.lru_cache
def func_0(x):
    print("do something using x")
    return [x,x+1]

print(func_0(5)) # see print 'about do something using x'
print(func_0(6)) # see print 'about do something using x'
print(func_0(5)) # DO NOT see print 'about do something using x'

answered Oct 5, 2022 at 17:19

Joran Beasley

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2 Comments

Moshmet Over a year ago

I re-formulated the question, so that people don't get confused. Do you think it is clearer now?

Moshmet Over a year ago

Unfortunaltey it doesn't work because x is a numpy array (unhashable type).

Hassan's Cheesy Chips · Accepted Answer · 2022-10-05 17:19:41Z

-1

Could you be looking for Tuple unpacking?

answered Oct 5, 2022 at 17:19

Hassan's Cheesy Chips

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2 Comments

Joran Beasley Over a year ago

i believe you are trying to help him solve the wrong problem

CryptoFool Over a year ago

This isn't an answer to the question. This should be in a comment.

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