rearrangement of lists of strings











up vote
2
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
    2
    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
      2
      down vote

      favorite









      up vote
      2
      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 2 hours ago









      kglr

      172k8194399




      172k8194399










      asked 2 hours ago









      Suite401

      937312




      937312






















          2 Answers
          2






          active

          oldest

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          up vote
          4
          down vote













          You can use StringRiffle:



          StringRiffle /@ lis



          {"abc def ghi", "jkl mno"}







          share|improve this answer




























            up vote
            1
            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

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              4
              down vote













              You can use StringRiffle:



              StringRiffle /@ lis



              {"abc def ghi", "jkl mno"}







              share|improve this answer

























                up vote
                4
                down vote













                You can use StringRiffle:



                StringRiffle /@ lis



                {"abc def ghi", "jkl mno"}







                share|improve this answer























                  up vote
                  4
                  down vote










                  up vote
                  4
                  down vote









                  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 2 hours ago









                  kglr

                  172k8194399




                  172k8194399






















                      up vote
                      1
                      down vote













                      Use Riffle



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

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

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





                      share|improve this answer

























                        up vote
                        1
                        down vote













                        Use Riffle



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

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

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





                        share|improve this answer























                          up vote
                          1
                          down vote










                          up vote
                          1
                          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 2 hours ago









                          Bob Hanlon

                          57.5k23591




                          57.5k23591






























                               

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