Your approach seems good. You can also do: (i.10) timex '+/i.10000'
Regarding the domain of timing code, I'd be surprised, and so curious, if the standard deviation wasn't a "fixed percentage" of average time, with a bias for quickest functions having the highest standard dev as a percentage. And also the timing on first iteration may be likely to be higher than remaining iterations. Situations that deviate from these rules, might be a sign of something "wrong". On Tuesday, January 11, 2022, 02:20:24 p.m. EST, Devon McCormick <[email protected]> wrote: Hi, I sometimes time expressions with a left argument of something like 10 or 100, e.g. "(100) 6!:2 'singlePair testSP0'". This gives me an average time for the multiple iterations. It would be nice to be able to get the standard deviation of the multiple invocations as well. I can do this myself with something like tms=. 6!:2 &>100$<'singlePair testSP0' then doing statistics on "tms" but I'm wondering if there's a way to do this more simply. tms=. 6!:2 &>100$<'singlePair testSP0' (mean,stddev) tms 0.260937 0.00313815 Thanks, Devon -- Devon McCormick, CFA Quantitative Consultant ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
