Re: Picking out pieces of a list

• To: mathgroup at smc.vnet.net
• Subject: [mg75973] Re: Picking out pieces of a list
• From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
• Date: Mon, 14 May 2007 03:41:57 -0400 (EDT)
• References: <f26nc2\$42l\$1@smc.vnet.net>

```Hi,

and

Graphics3D[{Opacity[.3],
EdgeForm[], {Line[#], Polygon[#]} &@pts[[All, 1]],
Polygon[#] &@pts[[All, 2]],
Polygon[Flatten[{#[[1]], Reverse[#[[2]]]}, 1] ] & /@
Partition[pts, 2, 1]}]

does not help ?

Regards
Jens

Hatto von Aquitanien wrote:
> Here's the motivation.  I want to draw slice out of a cone. (I don't know
> the proper mathematical term for this, but my meaning should be clear from
> the code.)  Note: See Feynman, Vol II, Page 1-8.
>
> (*A function that generates coordinates:*)
>
> R3[\[Theta]_, r_: 1.0, p_: {0, 0, 0}] :=
>  r {0, Cos@\[Theta], Sin@\[Theta]} + p
>
> (*Some "constant" values:*)
>
> r1 = 1.0;
> r2 = 0.8;
> \[CapitalDelta]x = 0.5;
>
> (*Points representing the centers of the circular faces:*)
>
> p1 = {\[CapitalDelta]x, 0, 0};
> p2 = {0, 0, 0};
>
> (*A list of point pairs:*)
>
> pts = {R3[#, r1, p1], R3[#, r2, p2]} & /@Range[-\[Pi], \[Pi], \[Pi]/20];
>
> (*Draw the circular faces, and the reference curve:*)
>
> Graphics3D[{
>   Opacity[.3]
>   , EdgeForm[]
>   , {Line[#], Polygon[#]} &@pts[[All, 1]]
>   , Polygon[#] &@pts[[All, 2]]
>   }]
>
> Now, I know I can use table, or some kind of brute force manipulation with
> (...)&/Range[Length@pts] to extract the points in the correct order to draw
> the polygons for the sides.  What I want to know is whether there is a way
> to use Part[] to get the four points in one statement.
>
> Here's a little scratch-pad code I created to explore the problem.
>
> c = CharacterRange["1", "9"]
> (lst = Table[
>     DisplayForm[
>      SubscriptBox["P", c[[i]] <> c[[j]] <> c[[k]]]]
>     , {i, 1, Length@c}
>     , {j, 1, 2}
>     , {k, 1, 2}]) // MatrixForm
>
> (*Get the points of the first ring*)
>
> lst[[All, 1]] // MatrixForm
>
> (*Get the points of the second ring*)
>
> lst[[All, 2]] // MatrixForm
>
> (*Here is /a/ solution to the problem:*)
>
> m[n_, lst_] := lst[[Mod[n + 1, Length@lst, 1]]]
>
> Join[lst[[#]][[{1, 2}]], m[#, lst][[{2, 1}]]] & /@ Range@len // MatrixForm
>
> Is there a "tighter" way to accomplish the same thing?
>

```

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