Shortening redundant calculation

Question

Is there a faster or shorter way of writing the following code to aviod calculating f(x) twice without defining another variable?

# could be any function or condition
if f(x) == 0:
    print(f(x))

# with additional variable
y = f(x)
if y == 0:
    print(y)

If you need this frequently, define a `print_if_0()` function. — Barmar, Sep 10 '21 at 15:58
@Barmar the condition is arbitrary, I am only concerned with shortening the syntax — Christian, Sep 10 '21 at 16:12

score 4 · Accepted Answer · answered Sep 10 '21 at 15:56

4

The walrus operator would save you an extra line (but not the assignment of y).

if (y := f(x)) == 0:
    print(y)

answered Sep 10 '21 at 15:56

timgeb

This is not actually shorter if you put it all on one line, because the `():` are longer than the newline that's saved. – Barmar Sep 10 '21 at 16:29
@Barmar But if you put it all in one line someone's gonna PEP8 you. – timgeb Sep 10 '21 at 16:30
1

True, but the OP seems to be looking for a one-liner, and he also asked how to do it "without defining another variable". I'm not sure why he accepted this, since it doesn't seem to address either of his wishes. – Barmar Sep 10 '21 at 16:33
@Barmar sometimes OPs don't know what they should actually want before they see it, eh. – timgeb Sep 10 '21 at 16:34

Barmar · Answer 2 · 2021-09-10T16:32:23.003

0

Define a print_if_0() function, and use that where you need it.

def print_if_0(n):
    if n == 0:
        print(n)

print_if_0(f(x))

You can generalize it to any test:

def print_if(f, val):
    if f(val):
        print(val)

print_if(lambda x: x == 0, f(x))

edited Sep 10 '21 at 16:32

answered Sep 10 '21 at 16:00

Barmar

score 0 · Answer 3 · answered Sep 10 '21 at 16:04

0

Use lru_cache to cache the results of f(x).

from functools import lru_cache

@lru_cache
def f(x):
    ...

answered Sep 10 '21 at 16:04

gold_cy

2

This will only work if the function is idempotent. – Barmar Sep 10 '21 at 16:14
@Barmar do you mean "pure"? afaik "idempotent" means f(f(x)) = f(x). – mkrieger1 Sep 10 '21 at 16:22
@mkrieger1 That's the mathematical meaning. In CS, it means that invoking the same operation multiple times has the same result. – Barmar Sep 10 '21 at 16:25
@mkrieger1 See https://stackoverflow.com/questions/1077412/what-is-an-idempotent-operation – Barmar Sep 10 '21 at 16:27
@Barmar okay, thanks for the reminder. tbh I didn't quite understand what you meant. Nevermind. – mkrieger1 Sep 10 '21 at 16:52

3 Answers3