Assignments in python list comprehension

Question

I am looking for a way to do assignments in a list comprehension. I would like to rewrite something like the following piece of code into a list comprehension.

I have this "costly" function:

import time
def f(x):
    time.sleep(1)
    return x + 1

And this loop:

l = []
for value in [1, 2, 3]:
    x = f(value)
    l.append(x + x)

I would like to rewrite this to a list comprehension:

l = [
     f(value) + f(fvalue)
     for value in [1, 2, 3]
]

But since calling f(value) is costly, I would like to minimize the number of calls (this following snippet doesn't run):

l = [
     (x = f(value))
     x + x
     for value in [1, 2, 3]
]

I've read about the assignment expression (:=) (https://www.python.org/dev/peps/pep-0572/#changing-the-scope-rules-for-comprehensions) but I can't seem to figure it out.

For the given example, does l = [x*10 for x in [1, 2, 3]] work? — arshovon
– arshovon, Commented Jan 16, 2022 at 13:12
Does this answer your question? How can I do assignments in a list comprehension? — Hector Haffenden
– Hector Haffenden, Commented Jan 16, 2022 at 13:14
Does not work! x, y = f(value) requires f to return an iterable of 2 values, while your example only return a scalar. — Serge Ballesta
– Serge Ballesta, Commented Jan 16, 2022 at 13:48
After a second look,I think that you wanted ... return x, x+1 — Serge Ballesta
– Serge Ballesta, Commented Jan 16, 2022 at 13:54

Lootcifer · Accepted Answer · 2022-01-16 13:42:22Z

2

My approach would be to nest multiple list comprehension, like

l_new = [x * x * y for x, y in [f(value) for value in [1, 2, 3]]]

So f() should only be called once for each value.

answered Jan 16, 2022 at 13:42

Lootcifer

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1 Comment

Steven Over a year ago

Thank you for your answer. The problem for me would be that we introduce an extra loop which makes it a bit unreadable to me

Ferenc Wágner · Accepted Answer · 2022-09-17 15:56:09Z

0

Here is another way to fake it:

[x+x for value in [1, 2, 3] for x in [f(value)]]

answered Sep 17, 2022 at 15:56

Ferenc Wágner

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Comments

liveware · Accepted Answer · 2022-01-16 16:15:27Z

-1

This can be done with a walrus operator. Note that you need Python 3.8 or later.

l = [x:=10 * value for value in [1, 2, 3]]

With one fast and slow function.

import time

def slow_function(a):
    time.sleep(1)
    return a + 1

def fast_function(b):
    return b + b  


l = [
     fast_function(x := slow_function(value))
     for value in [1, 2, 3]
]

The walrus operator can be skipped in both these examples, but perhaps the real problem would require it.

edited Jan 16, 2022 at 16:15

answered Jan 16, 2022 at 12:56

liveware

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8 Comments

Wizard.Ritvik Over a year ago

I think you don’t even need walrus assignment here, right?

Steven Over a year ago

Yeah for the original example I made this is true. I updated the post to demonstrate my problem more clearly.

liveware Over a year ago

In that case you could write one fast and one slow function, like this.

Steven Over a year ago

What would be the use of the walrus operator? Since x is not (re-)used

liveware Over a year ago

You could write it as e.g. l = [(x := f(value), x + x)[1] for value in [1, 2, 3]], if you're willing to create a tuple and take the second element of it inside the list comprehension.

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