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!






string-manipulation







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!






    string-manipulation







    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!






      string-manipulation







      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




      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

          votes

















          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





















              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: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: null,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              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%2f186649%2frearrangement-of-lists-of-strings%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              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






























                               

                              draft saved


                              draft discarded



















































                               


                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f186649%2frearrangement-of-lists-of-strings%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              Contact image not getting when fetch all contact list from iPhone by CNContact

                              Image
                              up vote 0 down vote favorite I know this question already asked but not getting solution. From this code I will get all the information from the contact but image not found when open vcf files on mac os, also not getting when share this file. I use this stackoverflow link here but It's not help full. var contacts = [CNContact]() let keys = [CNContactVCardSerialization.descriptorForRequiredKeys() ] as [Any] let request = CNContactFetchRequest(keysToFetch: keys as! [CNKeyDescriptor]) do { try self.contactStore.enumerateContacts(with: request) { (contact, stop) in // Array containing all unified contacts from everywhere contacts.append(contact) } } catch { print("unable to fetch contacts") } do { let data = try
                              Read more

                              count number of partitions of a set with n elements into k subsets

                              Image
                              up vote 6 down vote favorite 1 This program is for count number of partitions of a set with n elements into k subsets I am confusing here return k*countP(n-1, k) + countP(n-1, k-1); can some one explain what is happening here? why we are multiplying with k? NOTE->I know this is not the best way to calculate number of partitions that would be DP // A C++ program to count number of partitions // of a set with n elements into k subsets #include<iostream> using namespace std; // Returns count of different partitions of n // elements in k subsets int countP(int n, int k) { // Base cases if (n == 0 || k == 0 || k > n) return 0; if (k == 1 || k == n) return 1; // S(n+1, k) = k*S(n, k) + S(n, k-1) return k*countP(n-1, k) + countP(n-1, k-1); } // Driver program int ma
                              Read more

                              A CLEAN and SIMPLE way to add appendices to Table of Contents and bookmarks

                              Image
                              0 I found a lot of questions abount appendices and ToC. Many users want appendices to be grouped in an Appendix part, however some problems arise with ToC, hyperref, PDF viewer bookmarks, and so on. There are different solutions which require extra packages, command patching and other extra code, however none of them satisfies me. I almost found an easy way to accomplish a good result, where appendices are added to bookmarks in the right way and hyperref links point to the right page. However, the number of the "Appendix" part page is wrong (it's the number of appendix A). Is there any EASY way to fix that? This is a MWE: documentclass{book} usepackage[nottoc,notlot,notlof]{tocbibind} usepackage{hyperref} begin{document} frontmatter tableofcontents mainmatter part{First} chapter{
                              Read more