Re: Picking out pieces of a list

*To*: mathgroup at smc.vnet.net*Subject*: [mg76076] Re: Picking out pieces of a list*From*: Hatto von Aquitanien <abbot at AugiaDives.hre>*Date*: Wed, 16 May 2007 05:19:09 -0400 (EDT)*References*: <f26nc2$42l$1@smc.vnet.net> <f29c20$po4$1@smc.vnet.net>

CKWong wrote: > I don't know if this qualifies as "tighter", but it doesn't require > the function m. > > Table[ Flatten[{lst[[i]], Reverse[lst[[Mod[i + 1, Length@lst, 1]]]]}, > 1], {i, Length@lst}] //TableForm Well, I left my question intentionally somewhat vague in order to see how other people might approach the general problem. So I can't complain that your reply didn't provide the answer I was looking for. I consolidated my code along the lines you suggested and got this: Join[lst[[#]], Reverse@lst[[Mod[# + 1, Length@lst, 1]]]] & /@ Range@Length@lst // MatrixForm There is nothing wrong with that approach. What I was fishing for is a way to grab all four points for each Polygon in one pts[[(*what goes here?*)]]. Here's the drawing code using the modifications based on your suggestions. R3[\[Theta]_, r_: 1.0, p_: {0, 0, 0}] := r {0, Cos@\[Theta], Sin@\[Theta]} + p r1 = 1.0; r2 = 0.8; \[CapitalDelta]x = 0.5; p1 = {\[CapitalDelta]x, 0, 0}; p2 = {0, 0, 0}; pts = {R3[#, r1, p1], R3[#, r2, p2]} & /@ Range[-\[Pi], \[Pi], \[Pi]/20]; Graphics3D[{Opacity[.3] , EdgeForm[] , {Line[#], Polygon[#]} &@pts[[All, 1]] , Polygon[#] &@pts[[All, 2]] , Polygon@ Join[pts[[#]], Reverse@pts[[Mod[# + 1, Length@pts, 1]]]] & /@ Range@Length@pts}] -- http://www.dailymotion.com/video/x1ek5w_wtc7-the-smoking-gun-of-911-updated http://911research.wtc7.net http://vehme.blogspot.com Virtus Tutissima Cassis