Timeline for Sum Square Difference, which way is more Pythonic?

Current License: CC BY-SA 4.0

5 events

when toggle format	what		by	license	comment
Aug 16, 2019 at 13:04	comment	added	maxb		@AdamBarnes It is entirely possible to learn Python without ever touching `numpy`, but I'd definitely argue that the example I provided is in no way unreadable. Even to someone who has barely touched Python, I'd argue that `(nums2).sum()` is more readable than `sum(n2 for n in nums)`. However, I might be biased from having used `numpy` extensively. I definitely find Nikos Oikou's answer clear, concise and readable, but I wanted to present alternative approaches that might be usable. And in my own opinion, anyone learning Python should also put some focus on `numpy`.
Aug 16, 2019 at 12:52	comment	added	Adam Barnes		I -1'd this (even though I don't have the reputation to), because I would argue this is not answering the OP's question. He's not asking for the best computational way to do the sum of squares, he's asking for the most readable. `numpy` is very much unreadable to the initiated. Compare your examples to Nikos Oikou's, from the perspective of someone who knows Python, but has never used `numpy`.
Aug 16, 2019 at 7:36	comment	added	maxb		@NickMatteo Of course you could simplify it even further, I just wanted to present the way I would have done it. I prefer splitting up the logic, since generally you could have more complicated formulas, and combining them adds an extra layer of debugging for a future developer. I just googled the closed expressions for these sums, and wrote them into the code. That way, it will be easy for any future developer to verify that these formulas are correct.
Aug 16, 2019 at 6:51	comment	added	Nick Matteo		If you're going to do the math, you might as well do the subtraction as well, no? Then `sum_square_analytical` is just `return (3n4 + 2n*3 - 3n*2 - 2n)//12`. Or factor it, which seems faster: `return n(n+1)(n-1)(3n+2)//12`
Aug 15, 2019 at 7:51	history	answered	maxb	CC BY-SA 4.0