How can I rewrite this code so that it gives me the same result in less than one second?

Asked Mar 05 '20 at 10:18

Active Mar 05 '20 at 10:38

Viewed 36 times

For[i = 1; list = {}, i <= 23456, i++, 
 l = AppendTo[list, i/Pi]]; Total@N@list

edited Mar 05 '20 at 10:38

xzczd

asked Mar 05 '20 at 10:18

User9495

2

Sum[N[i/Pi], {i, 23456}] takes 0.00096 seconds on my machine – მამუკა ჯიბლაძე Mar 05 '20 at 10:25
1

We can all learn from Gauß: N[# (# - 1)/(2 Pi)] &[23456] does this in 4 flops. – Henrik Schumacher Mar 05 '20 at 10:29
@HenrikSchumacher Strictly speaking this is different: rather than rounding the sum, OP rounds each summand before summing up. – მამუკა ჯიბლაძე Mar 05 '20 at 10:38
So even worse... ^^ – Henrik Schumacher Mar 05 '20 at 10:39
Well maybe - at any rate, results differ by 7466.28 – მამუკა ჯიბლაძე Mar 05 '20 at 10:41
(Sorry, "rounding" was a wrong word to use here) – მამუკა ჯიბლაძე Mar 05 '20 at 10:42
1

First and foremost would be to avoid AppendTo. For reasons behind that, check the documentation. – Daniel Lichtblau Mar 05 '20 at 16:07

0 Answers0