Re: Re: arrange lists side by side

*To*: mathgroup at smc.vnet.net*Subject*: [mg56623] Re: [mg56591] Re: arrange lists side by side*From*: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>*Date*: Sat, 30 Apr 2005 01:28:05 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Sorry, this is the right way to tell the joke (see below): >-----Original Message----- >From: Wolf, Hartmut To: mathgroup at smc.vnet.net >Sent: Friday, April 29, 2005 11:19 AM >Subject: [mg56623] RE: [mg56591] Re: arrange lists side by side > > >>-----Original Message----- >>From: Carl K. Woll [mailto:carlw at u.washington.edu] To: mathgroup at smc.vnet.net >>Sent: Friday, April 29, 2005 9:22 AM >>Subject: [mg56623] [mg56591] Re: arrange lists side by side >> >><marloo3 at mail15.com> wrote in message >news:d4q1qj$oh8$1 at smc.vnet.net... >>> 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}} >>> >>> >>---------------------------------------------------------------------- >>> >>> >> >>Undoubtedly you will get a lot of responses. Here is one more. >> >>Transpose[ PadRight[ {{}}, {2, Length[b]}, {b,a} ] ] >> >>Carl Woll >> >> >> > >my first idea, quite similar, was this: > >In[5]:= Transpose[{b, PadRight[{}, Length[b], a]}] >Out[5]= >{{1, 28}, {2, 30}, {9, 17}, {4, 28}, {5, 30}, {7, 17}, {3, >28}, {8, 30}} > > >This however is a bit obfuscated: > >In[15]:= First@ListCorrelate[b, a, {1, Length[a]}, a, List, List] >Out[15]= >{{1, 28}, {2, 30}, {9, 17}, {4, 28}, {5, 30}, {7, 17}, {3, >28}, {8, 30}} > Perhaps this was more appropriate: ListCorrelate[b, a, {1, Length[a]}, a, List, Sequence] >or this > >In[30]:= >Transpose@Prepend[Partition[a, Length[b], Length[b], {1, 1}, a], b] >Out[30]= >{{1, 28}, {2, 30}, {9, 17}, {4, 28}, {5, 30}, {7, 17}, {3, >28}, {8, 30}} > > >-- >Hartmut >