`math.sumprod` equivalent code is more complicated than needed

[Akuli]

Documentation

Docs say that math.sumprod(p, q) is roughly equivalent to:

sum(itertools.starmap(operator.mul, zip(p, q, strict=True)))

This is not useful for people who don't already know about itertools.starmap and operator.mul. IMO that is a shame, because math.sumprod is not that complicated to understand. It would be better to describe it as:

sum(x*y for x, y in zip(p, q, strict=True))

Linked PRs

• gh-130203: Further simplify math.sumprod documentation #130206

[picnixz]

I agree that it's better to avoid itertools and operator. Do you want to send a PR?

[Akuli]

It seems like this was already simplified in gh-126407. On the main branch, it is:

sum(map(operator.mul, p, q, strict=True))

I still feel like plain old * would be better than operator.mul. I will make a PR.

[picnixz]

sum(map(operator.mul, p, q, strict=True))

May be still fine IMO. The reason is that we're using the strict keyword of map which is nice to show. Otherwise we have the strict in zip.