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

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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












    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






      share|improve this question















      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






          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










          Your Answer




          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "387"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          convertImagesToLinks: false,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













           

          draft saved


          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f183538%2felegant-way-of-partitioning-in-two-starting-from-the-second-element%23new-answer', 'question_page');

          );

          Post as a guest

















          1 Answer
          1






          active

          oldest

          votes








          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

















           

          draft saved


          draft discarded















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f183538%2felegant-way-of-partitioning-in-two-starting-from-the-second-element%23new-answer', 'question_page');

          );

          Post as a guest
































































          Comments

          Post a Comment

          Popular posts from this blog

          What does second last employer means? [closed]

          Image
          Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
          Read more

          Installing NextGIS Connect into QGIS 3?

          Image
          Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugins nextgis share | improve this question edited 3 hours ago PolyGeo ♦ 51.9k 17 77 233 asked 4 hours ago IanGS 16 4 New contributor IanGS is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugin...
          Read more

          One-line joke

          A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
          Read more