Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2007
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

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?
> 


  • Prev by Date: Re: question 1 (Integrate and simplification)
  • Next by Date: Re: Package Help in Mathematica 6
  • Previous by thread: Picking out pieces of a list
  • Next by thread: Re: Picking out pieces of a list