Timeline for How to find rows that have maximum value?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:56 | history | edited | CommunityBot | replaced http://mathematica.stackexchange.com/ with https://mathematica.stackexchange.com/ | |
| Jul 15, 2014 at 6:00 | comment | added | Mr.Wizard | @MichaelE2 That's for pointing out that new function. I'll probably keep using my own (or rather my version of Timo's original) out of habit, but it makes a good companion. AccurateTiming uses AbsoluteTiming rather than Timing so should reveal differences when used for paralleled functions. | |
| Jul 15, 2014 at 5:44 | comment | added | Michael E2 | This is the earliest ref. to timeAvg I've found on Mma.SE. I thought you might like to compare it with a new function in V10, Needs["GeneralUtilities"]; AccurateTiming[Range[10^4]]`. | |
| Sep 18, 2013 at 19:45 | comment | added | Michael E2 | That's remarkbly faster! | |
| Sep 18, 2013 at 18:53 | comment | added | Mr.Wizard | @MichaelE2 Please see update. | |
| Sep 18, 2013 at 18:53 | history | edited | Mr.Wizard | CC BY-SA 3.0 | added 632 characters in body |
| Sep 18, 2013 at 18:46 | comment | added | Mr.Wizard | @MichaelE2 Thanks. Time for more optimization. :-) | |
| Sep 18, 2013 at 18:38 | comment | added | Michael E2 | The compiled version of Position is even faster (I think it is optimized to check only level 1). Try sticking pos = Compile[{{list, _Real, 1}, {pat, _Real}}, Position[list, pat]] into the last two. I get a speedup of more than a factor of 2 in V9.0.1 | |
| Sep 18, 2013 at 18:29 | history | edited | Mr.Wizard | CC BY-SA 3.0 | added 1249 characters in body |
| May 12, 2012 at 9:55 | vote | accept | Prashant Bhate | ||
| Jan 29, 2012 at 16:56 | comment | added | Mr.Wizard | @Istvan perhaps, but I was typing these (and updating the question) as I thought of them. I didn't compare performance (or significantly consider performance, really) until after I posted the four methods. After testing I could have changed my answer to keep only the Pick version but by that time someone else posted a method using Pick. Also, I think it is interesting to show different approaches, even if for a specific problem some of them are contrived or awkward because they may not be on another problem. | |
| Jan 29, 2012 at 9:45 | comment | added | István Zachar | @Spartacus: Overkill... | |
| Jan 28, 2012 at 23:30 | comment | added | Mr.Wizard | @J.M. I was attempting to keep this general. | |
| Jan 28, 2012 at 23:26 | comment | added | J. M.'s missing motivation | In this case, the #[[3]] & is replaceable with Last[], e.g. Last[SplitBy[SortBy[data, Last], Last]]. | |
| Jan 28, 2012 at 23:25 | history | edited | Mr.Wizard | CC BY-SA 3.0 | added 71 characters in body |
| Jan 28, 2012 at 23:17 | history | answered | Mr.Wizard | CC BY-SA 3.0 |