Elegant way of partitioning in two starting from the second element?

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I wrote a code for partitioning a list. I want to group them into two groups following the same pattern: 1. element added, then 2. and 3. are omitted, then 4. and 5. added and goes like this. Partitioning is easy, then I have to add the first element manually but the problem is when I partition the list, it omits the last element because then the list contains odd number of elements and I have to add it to the list manually also. I thought maybe someone can figure out a more elegant way. My list is as follow:



 vars= c0r, c0i, c1r, c1i, c2r, c2i, c3r, c3i, c4r, c4i, c5r, c5i, c6r, 
 c6i, c7r, c7i, c8r, c8i, c9r, c9i, c10r, c10i, c11r, c11i, c12r, 
 c12i, c13r, c13i, c14r, c14i, c15r, c15i


and here is my code;



 varr1 = Partition[vars[[2 ;;]], 2]

 c0i, c1r, c1i, c2r, c2i, c3r, c3i, c4r, c4i, c5r, c5i, 
 c6r, c6i, c7r, c7i, c8r, c8i, c9r, c9i, c10r, c10i, 
 c11r, c11i, c12r, c12i, c13r, c13i, c14r, c14i, c15r 

 Join[vars[[1]], Flatten[varr1[[2 ;; ;; 2]]], vars[[-1]]]

 c0r, c1i, c2r, c3i, c4r, c5i, c6r, c7i, c8r, c9i, c10r, c11i, c12r, 
 c13i, c14r, c15i


and this is the list I want.






list-manipulation







share|improve this question



























    up vote
    4
    down vote

    favorite












    I wrote a code for partitioning a list. I want to group them into two groups following the same pattern: 1. element added, then 2. and 3. are omitted, then 4. and 5. added and goes like this. Partitioning is easy, then I have to add the first element manually but the problem is when I partition the list, it omits the last element because then the list contains odd number of elements and I have to add it to the list manually also. I thought maybe someone can figure out a more elegant way. My list is as follow:



     vars= c0r, c0i, c1r, c1i, c2r, c2i, c3r, c3i, c4r, c4i, c5r, c5i, c6r, 
     c6i, c7r, c7i, c8r, c8i, c9r, c9i, c10r, c10i, c11r, c11i, c12r, 
     c12i, c13r, c13i, c14r, c14i, c15r, c15i


    and here is my code;



     varr1 = Partition[vars[[2 ;;]], 2]

     c0i, c1r, c1i, c2r, c2i, c3r, c3i, c4r, c4i, c5r, c5i, 
     c6r, c6i, c7r, c7i, c8r, c8i, c9r, c9i, c10r, c10i, 
     c11r, c11i, c12r, c12i, c13r, c13i, c14r, c14i, c15r 

     Join[vars[[1]], Flatten[varr1[[2 ;; ;; 2]]], vars[[-1]]]

     c0r, c1i, c2r, c3i, c4r, c5i, c6r, c7i, c8r, c9i, c10r, c11i, c12r, 
     c13i, c14r, c15i


    and this is the list I want.






    list-manipulation







    share|improve this question

























      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      I wrote a code for partitioning a list. I want to group them into two groups following the same pattern: 1. element added, then 2. and 3. are omitted, then 4. and 5. added and goes like this. Partitioning is easy, then I have to add the first element manually but the problem is when I partition the list, it omits the last element because then the list contains odd number of elements and I have to add it to the list manually also. I thought maybe someone can figure out a more elegant way. My list is as follow:



       vars= c0r, c0i, c1r, c1i, c2r, c2i, c3r, c3i, c4r, c4i, c5r, c5i, c6r, 
       c6i, c7r, c7i, c8r, c8i, c9r, c9i, c10r, c10i, c11r, c11i, c12r, 
       c12i, c13r, c13i, c14r, c14i, c15r, c15i


      and here is my code;



       varr1 = Partition[vars[[2 ;;]], 2]

       c0i, c1r, c1i, c2r, c2i, c3r, c3i, c4r, c4i, c5r, c5i, 
       c6r, c6i, c7r, c7i, c8r, c8i, c9r, c9i, c10r, c10i, 
       c11r, c11i, c12r, c12i, c13r, c13i, c14r, c14i, c15r 

       Join[vars[[1]], Flatten[varr1[[2 ;; ;; 2]]], vars[[-1]]]

       c0r, c1i, c2r, c3i, c4r, c5i, c6r, c7i, c8r, c9i, c10r, c11i, c12r, 
       c13i, c14r, c15i


      and this is the list I want.






      list-manipulation







      share|improve this question















      I wrote a code for partitioning a list. I want to group them into two groups following the same pattern: 1. element added, then 2. and 3. are omitted, then 4. and 5. added and goes like this. Partitioning is easy, then I have to add the first element manually but the problem is when I partition the list, it omits the last element because then the list contains odd number of elements and I have to add it to the list manually also. I thought maybe someone can figure out a more elegant way. My list is as follow:



       vars= c0r, c0i, c1r, c1i, c2r, c2i, c3r, c3i, c4r, c4i, c5r, c5i, c6r, 
       c6i, c7r, c7i, c8r, c8i, c9r, c9i, c10r, c10i, c11r, c11i, c12r, 
       c12i, c13r, c13i, c14r, c14i, c15r, c15i


      and here is my code;



       varr1 = Partition[vars[[2 ;;]], 2]

       c0i, c1r, c1i, c2r, c2i, c3r, c3i, c4r, c4i, c5r, c5i, 
       c6r, c6i, c7r, c7i, c8r, c8i, c9r, c9i, c10r, c10i, 
       c11r, c11i, c12r, c12i, c13r, c13i, c14r, c14i, c15r 

       Join[vars[[1]], Flatten[varr1[[2 ;; ;; 2]]], vars[[-1]]]

       c0r, c1i, c2r, c3i, c4r, c5i, c6r, c7i, c8r, c9i, c10r, c11i, c12r, 
       c13i, c14r, c15i


      and this is the list I want.






      list-manipulation




      list-manipulation






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      share|improve this question













      share|improve this question




      share|improve this question








      edited 2 hours ago

























      asked 2 hours ago









      Buddha_the_Scientist

      6715




      6715




















          1 Answer
          1






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          up vote
          3
          down vote



          accepted










          You can use the five-argument form of Partition:



          Flatten[Partition[vars, 2, 4, -1, 1, ]]



          c0r, c1i, c2r, c3i, c4r, c5i, c6r, c7i, c8r, c9i, c10r, c11i, c12r,
          c13i, c14r, c15i




          You can also use



          Pick[#, Mod[Range @ Length @ #, 4], 1 | 0]& @ vars 
          Flatten[Partition[vars, 4][[All, 1, -1]]]
          Flatten[List @@@ ListConvolve[1, 0, 0, 1, vars][[ ;; ;; 4]]]





          share|improve this answer






















          • You find the more clever way of picking them as first and third.
            – Buddha_the_Scientist
            1 hour ago










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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          3
          down vote



          accepted










          You can use the five-argument form of Partition:



          Flatten[Partition[vars, 2, 4, -1, 1, ]]



          c0r, c1i, c2r, c3i, c4r, c5i, c6r, c7i, c8r, c9i, c10r, c11i, c12r,
          c13i, c14r, c15i




          You can also use



          Pick[#, Mod[Range @ Length @ #, 4], 1 | 0]& @ vars 
          Flatten[Partition[vars, 4][[All, 1, -1]]]
          Flatten[List @@@ ListConvolve[1, 0, 0, 1, vars][[ ;; ;; 4]]]





          share|improve this answer






















          • You find the more clever way of picking them as first and third.
            – Buddha_the_Scientist
            1 hour ago














          up vote
          3
          down vote



          accepted










          You can use the five-argument form of Partition:



          Flatten[Partition[vars, 2, 4, -1, 1, ]]



          c0r, c1i, c2r, c3i, c4r, c5i, c6r, c7i, c8r, c9i, c10r, c11i, c12r,
          c13i, c14r, c15i




          You can also use



          Pick[#, Mod[Range @ Length @ #, 4], 1 | 0]& @ vars 
          Flatten[Partition[vars, 4][[All, 1, -1]]]
          Flatten[List @@@ ListConvolve[1, 0, 0, 1, vars][[ ;; ;; 4]]]





          share|improve this answer






















          • You find the more clever way of picking them as first and third.
            – Buddha_the_Scientist
            1 hour ago












          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          You can use the five-argument form of Partition:



          Flatten[Partition[vars, 2, 4, -1, 1, ]]



          c0r, c1i, c2r, c3i, c4r, c5i, c6r, c7i, c8r, c9i, c10r, c11i, c12r,
          c13i, c14r, c15i




          You can also use



          Pick[#, Mod[Range @ Length @ #, 4], 1 | 0]& @ vars 
          Flatten[Partition[vars, 4][[All, 1, -1]]]
          Flatten[List @@@ ListConvolve[1, 0, 0, 1, vars][[ ;; ;; 4]]]





          share|improve this answer














          You can use the five-argument form of Partition:



          Flatten[Partition[vars, 2, 4, -1, 1, ]]



          c0r, c1i, c2r, c3i, c4r, c5i, c6r, c7i, c8r, c9i, c10r, c11i, c12r,
          c13i, c14r, c15i




          You can also use



          Pick[#, Mod[Range @ Length @ #, 4], 1 | 0]& @ vars 
          Flatten[Partition[vars, 4][[All, 1, -1]]]
          Flatten[List @@@ ListConvolve[1, 0, 0, 1, vars][[ ;; ;; 4]]]






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 1 hour ago

























          answered 1 hour ago









          kglr

          164k8188388




          164k8188388











          • You find the more clever way of picking them as first and third.
            – Buddha_the_Scientist
            1 hour ago
















          • You find the more clever way of picking them as first and third.
            – Buddha_the_Scientist
            1 hour ago















          You find the more clever way of picking them as first and third.
          – Buddha_the_Scientist
          1 hour ago




          You find the more clever way of picking them as first and third.
          – Buddha_the_Scientist
          1 hour ago

















           

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