Rearrangement of lists of strings

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up vote
5
down vote

favorite












I have a list:



lis = "abc","def","ghi","jkl","mno"


and wish to get:



res = "abc def ghi","jkl mno"


This:



Table[StringJoin[lis[[i]]], i, Length[lis]]


doesn't produce the desired " " between the original elements in lis. As always, thanks for suggestions!










share|improve this question



























    up vote
    5
    down vote

    favorite












    I have a list:



    lis = "abc","def","ghi","jkl","mno"


    and wish to get:



    res = "abc def ghi","jkl mno"


    This:



    Table[StringJoin[lis[[i]]], i, Length[lis]]


    doesn't produce the desired " " between the original elements in lis. As always, thanks for suggestions!










    share|improve this question

























      up vote
      5
      down vote

      favorite









      up vote
      5
      down vote

      favorite











      I have a list:



      lis = "abc","def","ghi","jkl","mno"


      and wish to get:



      res = "abc def ghi","jkl mno"


      This:



      Table[StringJoin[lis[[i]]], i, Length[lis]]


      doesn't produce the desired " " between the original elements in lis. As always, thanks for suggestions!










      share|improve this question















      I have a list:



      lis = "abc","def","ghi","jkl","mno"


      and wish to get:



      res = "abc def ghi","jkl mno"


      This:



      Table[StringJoin[lis[[i]]], i, Length[lis]]


      doesn't produce the desired " " between the original elements in lis. As always, thanks for suggestions!







      string-manipulation






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      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 25 at 9:27









      kglr

      174k9196402




      174k9196402










      asked Nov 25 at 2:43









      Suite401

      959312




      959312




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          14
          down vote



          accepted










          You can use StringRiffle:



          StringRiffle /@ lis



          "abc def ghi", "jkl mno"







          share|improve this answer



























            up vote
            6
            down vote













            Use Riffle



            lis = "abc", "def", "ghi", "jkl", "mno";

            StringJoin[Riffle[#, " "]] & /@ lis

            (* "abc def ghi", "jkl mno" *)





            share|improve this answer




















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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              14
              down vote



              accepted










              You can use StringRiffle:



              StringRiffle /@ lis



              "abc def ghi", "jkl mno"







              share|improve this answer
























                up vote
                14
                down vote



                accepted










                You can use StringRiffle:



                StringRiffle /@ lis



                "abc def ghi", "jkl mno"







                share|improve this answer






















                  up vote
                  14
                  down vote



                  accepted







                  up vote
                  14
                  down vote



                  accepted






                  You can use StringRiffle:



                  StringRiffle /@ lis



                  "abc def ghi", "jkl mno"







                  share|improve this answer












                  You can use StringRiffle:



                  StringRiffle /@ lis



                  "abc def ghi", "jkl mno"








                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 25 at 2:56









                  kglr

                  174k9196402




                  174k9196402




















                      up vote
                      6
                      down vote













                      Use Riffle



                      lis = "abc", "def", "ghi", "jkl", "mno";

                      StringJoin[Riffle[#, " "]] & /@ lis

                      (* "abc def ghi", "jkl mno" *)





                      share|improve this answer
























                        up vote
                        6
                        down vote













                        Use Riffle



                        lis = "abc", "def", "ghi", "jkl", "mno";

                        StringJoin[Riffle[#, " "]] & /@ lis

                        (* "abc def ghi", "jkl mno" *)





                        share|improve this answer






















                          up vote
                          6
                          down vote










                          up vote
                          6
                          down vote









                          Use Riffle



                          lis = "abc", "def", "ghi", "jkl", "mno";

                          StringJoin[Riffle[#, " "]] & /@ lis

                          (* "abc def ghi", "jkl mno" *)





                          share|improve this answer












                          Use Riffle



                          lis = "abc", "def", "ghi", "jkl", "mno";

                          StringJoin[Riffle[#, " "]] & /@ lis

                          (* "abc def ghi", "jkl mno" *)






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Nov 25 at 2:52









                          Bob Hanlon

                          57.9k23593




                          57.9k23593



























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