Re: arrange lists side by side

*To*: mathgroup at smc.vnet.net*Subject*: [mg56681] Re: [mg56538] arrange lists side by side*From*: John Kiehl <john.kiehl at soundtrackny.com>*Date*: Tue, 3 May 2005 05:26:33 -0400 (EDT)*Reply-to*: John Kiehl <john.kiehl at soundtrackny.com>*Sender*: owner-wri-mathgroup at wolfram.com

Somehow, my post from yesterday didn't include the actual line of Mathematica code. So here it is again. Your problem immediately made me think of using MapIndexed[ ] which uses a second parameter (#2) to refer to the position of each element that's being Mapped onto. I use that position information to refer to the appropriate member of the shorter list. ( There's an inelegant detail with subtracting 1 inside the Mod[ ] and adding 1 outside the Mod[ ], as well as needing Sequence[] to get the bracketing as desired, but I've grown use to these details with MapIndexed[ ]. ) So, using your a and b as defined: In[] = MapIndexed[{#, Sequence @@ a[[ Mod[#2-1,Length[a]] +1]] }&, b ] john kiehl On Thursday, April 28, 2005 2:40 AM, marloo3 at mail15.com wrote: >Hi >is there a way to spread out a small list over a bigger list recurrently like >this: >a={28, 30, 17}; >b={1, 2, 9, 4, 5, 7, 3, 8}; >to give the output: >{{1,28},{2,30},{9,17},{4,28},{5,30},{7,17},{3,28},{8,30}}; >the number of items in "b" do not neccesary multiples of the number of items >in "a" >mark > >okay this is my approach > >a={28,30,17}; >b={1,2,9,4,5,7,3,8}; >aa={};aa=Table[Join[aa,a],{i,Length[b]/Length[a]}] > >Out[]= >{{28,30,17},{28,30,17}} > >frc=Mod[Length[b],Length[a]]; >gg=Flatten[Join[aa,Part[a,{1,frc}]]]; >Transpose[Join[{b},{gg}]] > >Out[]= >{{1,28},{2,30},{9,17},{4,28},{5,30},{7,17},{3,28},{8,30}} > >---------------------------------------------------------------------- > > >