Timeline for Finding local maxima/minima with Numpy in a 1D numpy array
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 3 at 21:00 | comment | added | Nathan Ruffing | Элёржон Кимсанов, the result is kind of a sum of each element's value relative to its neighbors. With ascending or descending, 0 means higher + lower = 0. If both neighbors are lower, lower + lower = -2 so there's a maximum. If one neighbor is lower and the other equal, lower + equal = -1, so max of one side, but not a full local maximum. If there is another -1 before a 1, then the sequence drops again and you have a group that is the local maximum from the first -1 to the second -1. I like this answer the best. | |
| Aug 26, 2022 at 14:27 | comment | added | Элёржон Кимсанов | It ignores cases with repetitive elements. Eg [1,2,3,1,2,2,2,1,4,5]. How to fix it? | |
| Mar 29, 2021 at 10:46 | history | edited | desertnaut | CC BY-SA 4.0 |
deleted 35 characters in body
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| Aug 26, 2018 at 19:00 | review | Suggested edits | |||
| Aug 26, 2018 at 19:45 | |||||
| Aug 16, 2018 at 10:04 | comment | added | Brandon Rhodes | I think that this (good!) answer is the same as R. C.'s answer from 2012? He offers three one-line solutions, depending on whether the caller wants mins, maxes, or both, if I'm reading his solution correctly. | |
| Apr 24, 2018 at 11:06 | review | Late answers | |||
| Apr 24, 2018 at 11:07 | |||||
| Apr 24, 2018 at 10:48 | history | answered | Dave | CC BY-SA 3.0 |