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I am trying to pass each element of foo_list into a function expensive_call, and get a list of all the items whose output of expensive_call is Truthy. I am trying to do it with list comprehensions, is it possible? Something like:

Something like this:

 result_list = [y := expensive_call(x) for x in foo_list if y]

or....

 result_list = [y for x in foo_list if y := expensive_call(x)]

Note: This is not a solution because it call expensive call twice:

 result_list = [expensive_call(x) for x in foo_list if expensive_call(x)]

And before someone recommends none list comprehension, I know one can do:

result_list = []
for x in foo_list:
   result = expensive_call(x)
   result and result_list.append(result)
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  • 5
    "a list of all the items whose output of expensive_call is Truthy." Wait, doesn't that mean you want the result to contain original elements from the input? Isn't that just [x for x in foo_list if expensive_call(x)]? What's the difficulty? (That corresponds to the code in your non-comprehension version, too.) Commented Jan 18, 2022 at 18:37
  • 3
    What is wrong with your second version? Did you try it? Commented Jan 18, 2022 at 18:38
  • 3
    How about: [y for y in map(expensive_call,foo_list) if y] Commented Jan 18, 2022 at 18:40
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    In your loop version, avoid cutesy things like result and result_list.append(x), use if result: result_list.append(x), idiomatic Python values clarity and being explicit, not brevity Commented Jan 18, 2022 at 18:41
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    You might want to edit your loop variation, because it is different logic than the others. You clearly wanted the last line: result and result_list.append(result). Though as @juanpa.arrivillaga says, that's "cutesy" and could mislead many. Commented Jan 18, 2022 at 18:48

3 Answers 3

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9

Quoting from above:

result_list = [y for x in foo_list if y := expensive_call(x)]

It should work almost exactly as how you have it; just remember to parenthesize the assignment with the := operator, as shown below.

foo_list = [1, 2, 3]
def check(x): return x if x != 2 else None

result_list = [y for x in foo_list if (y := check(x))]
print(result_list)

Result:

[1, 3]
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4 Comments

Does it fail without the parentheses?
Yep, definitely seems to fail for me; at least when I test with a Python 3.9 environment.
Interesting. I wonder how it parses.
From the PEP on the walrus operator, it looks like there are edge cases where the () are required around the expression; tbh I'm not sure why it's needed here though.
2

As suggested by RufusVS, this works and has the expected complexity.

def expensive_call(x):
    return x == 5

foo_list = [1,2,3,4,5,6,6,5]
print([y for y in map(expensive_call,foo_list) if y])

results in

[True, True]

As there are two 5 that satisfy expensive_call

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Comments

1

You want to check if expensive_call(x) is Truthy, and if so append it to the list? Something like this should work. Doesn't need to use a walrus op:

[x for x in foo_list if expensive_call(x)]

If you want to store the output of expensive_call(x) and not x then your second approach needs parentheses around the walrus assignment due to how the assignment operator works. Refer to PEP 572 for more info.

[y for x in foo_list if (y := expensive_call(x))]
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4 Comments

The OP wants to save the result of the call, not the argument. AFAICT
@RufusVS if you refer to OP's non-comprehension they use result_list.append(x) not result_list.append(result) and they said they want to store the item if the output of the func is truthy not store the ouput of the func if it's truthy.
I did not even read the loop version he posted. Clearly it's not the same logic as his other attempts, but now I see where you were coming from.
Sorry for causing confusion, I corrected my code, thank you!

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