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Timeline for Fluctuating ranges

Current License: CC BY-SA 3.0

9 events
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Jun 17, 2020 at 9:04 history edited CommunityBot
Commonmark migration
Oct 7, 2017 at 16:18 comment added Jonathan Frech n[0:1] is equivalent to n[:1].
Mar 9, 2016 at 9:13 history edited user45941 CC BY-SA 3.0
added 12 characters in body
Mar 9, 2016 at 9:04 comment added Erwan i think you can replace [1,-1][y+1<x] by 2*(y>x)-1 (also i don't understand why you use y<=x and not simply y<x )
Mar 5, 2016 at 23:43 comment added user45941 I was very tired while writing this :) Answer first, improve later.
Mar 5, 2016 at 23:43 history edited user45941 CC BY-SA 3.0
added 69 characters in body
Mar 5, 2016 at 22:46 comment added Denker why not go with lambda n:n[0:1]+sum([range(x,y,[1,-1][y+1<x])[1:]+[y]for(x,y)in zip(n,n[1:])],[])? saves some bytes.
Mar 5, 2016 at 20:21 comment added Neil Surely the +1 is unnecessary?
Mar 5, 2016 at 17:16 history answered user45941 CC BY-SA 3.0