Cache variable in list comprehension

Question

Lets say I have an expensive operation expensive(x: int) -> int and the following list comprehension:

# expensive(x: int) -> int
# check(x: int) -> bool
[expensive(i) for i in range(LARGE_NUMBER) if check(expensive(i))]

If I want to avoid running expensive(i) twice for each i, is there any way to save it's value with list comprehension?

[Related](https://stackoverflow.com/questions/57484396/vectorizing-a-pure-function-with-numpy-assuming-many-duplicates). — hilberts_drinking_problem, Feb 06 '20 at 02:02
If the function is pure, you could wrap it with `functools.lru_cache`. — hilberts_drinking_problem, Feb 06 '20 at 02:02
Or `[e for e in map(expensive, range(LARGE_NUMBER)) if check(e)]`. — hilberts_drinking_problem, Feb 06 '20 at 02:03

score 5 · Accepted Answer · answered Feb 06 '20 at 02:04

5

Using the walrus:

[cache for i in range(LARGE_NUMBER) if check(cache := expensive(i))]

answered Feb 06 '20 at 02:04

Kelly Bundy

1

Note this feature is only available as of Python 3.8 – Tyberius Feb 06 '20 at 02:14

Alain T. · Answer 2 · 2020-02-07T13:15:16.627

0

You could simulate a nested comprehension:

 [val for i in range(LARGE_NUMBER) for val in [expensive(i)] if check(val)]

edited Feb 07 '20 at 13:15

answered Feb 06 '20 at 22:10

Alain T.

2 Answers2