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Taking the first element in a list of associations


Using ListPlot and lists of AssociationsGiven a list made up of subsets of an alphabet, how can I use associations to create a Boolean dictionary of the given listDataset vs an association of associationsFastest way to look up in a list of AssociationsConverting a list of associations into a single associationHow to extract the first element in nested listsJoinAcross with multiple associationsSort Associations by its Values (which are nested lists)






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








4












$begingroup$


I have the following list which is built up of associations, that is,



z = <


I am trying to to pick the first elements in each association such that it gives:



 >


to do that I do,



Map[First, z]


but this only returns the values, how can one fix this?










share|improve this question









$endgroup$




















    4












    $begingroup$


    I have the following list which is built up of associations, that is,



    z = <


    I am trying to to pick the first elements in each association such that it gives:



     >


    to do that I do,



    Map[First, z]


    but this only returns the values, how can one fix this?










    share|improve this question









    $endgroup$
















      4












      4








      4





      $begingroup$


      I have the following list which is built up of associations, that is,



      z = <


      I am trying to to pick the first elements in each association such that it gives:



       >


      to do that I do,



      Map[First, z]


      but this only returns the values, how can one fix this?










      share|improve this question









      $endgroup$




      I have the following list which is built up of associations, that is,



      z = <


      I am trying to to pick the first elements in each association such that it gives:



       >


      to do that I do,



      Map[First, z]


      but this only returns the values, how can one fix this?







      list-manipulation associations






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 10 hours ago









      WilliamWilliam

      1,1066 silver badges9 bronze badges




      1,1066 silver badges9 bronze badges























          2 Answers
          2






          active

          oldest

          votes


















          6













          $begingroup$

          You can use the fact that Part ([[…]]) preserves the keys when you supply a list of indices to extract:



          z[[All, 1]]
          (* < *)

          (* without the list *)
          z[[All, 1]]
          (* 1, 4, 6, 10 *)


          Alternatively you could use Take:



          Map[Take[#, 1] &, z]
          (* < *)


          Other possibilites:



          Map[#[[1]] &, z]
          (* < *)

          z[[All, ;; 1]]
          (* < *)


          In essence, anything that could in principle return multiple elements will preserve the keys (or, more generally, will preserve that level of the nested expression)






          share|improve this answer









          $endgroup$






















            2













            $begingroup$

            For completeness sake, a few more:



            Extract[z, All, 1]
            (* < *)

            Cases[z, KeyValuePattern[x_ -> y_] :> x -> y]
            (*"a", "b" -> 1, "g", "h" -> 4, "k", "l" -> 6, "s", "t" -> 10*)





            share|improve this answer









            $endgroup$

















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              2 Answers
              2






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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              6













              $begingroup$

              You can use the fact that Part ([[…]]) preserves the keys when you supply a list of indices to extract:



              z[[All, 1]]
              (* < *)

              (* without the list *)
              z[[All, 1]]
              (* 1, 4, 6, 10 *)


              Alternatively you could use Take:



              Map[Take[#, 1] &, z]
              (* < *)


              Other possibilites:



              Map[#[[1]] &, z]
              (* < *)

              z[[All, ;; 1]]
              (* < *)


              In essence, anything that could in principle return multiple elements will preserve the keys (or, more generally, will preserve that level of the nested expression)






              share|improve this answer









              $endgroup$



















                6













                $begingroup$

                You can use the fact that Part ([[…]]) preserves the keys when you supply a list of indices to extract:



                z[[All, 1]]
                (* < *)

                (* without the list *)
                z[[All, 1]]
                (* 1, 4, 6, 10 *)


                Alternatively you could use Take:



                Map[Take[#, 1] &, z]
                (* < *)


                Other possibilites:



                Map[#[[1]] &, z]
                (* < *)

                z[[All, ;; 1]]
                (* < *)


                In essence, anything that could in principle return multiple elements will preserve the keys (or, more generally, will preserve that level of the nested expression)






                share|improve this answer









                $endgroup$

















                  6














                  6










                  6







                  $begingroup$

                  You can use the fact that Part ([[…]]) preserves the keys when you supply a list of indices to extract:



                  z[[All, 1]]
                  (* < *)

                  (* without the list *)
                  z[[All, 1]]
                  (* 1, 4, 6, 10 *)


                  Alternatively you could use Take:



                  Map[Take[#, 1] &, z]
                  (* < *)


                  Other possibilites:



                  Map[#[[1]] &, z]
                  (* < *)

                  z[[All, ;; 1]]
                  (* < *)


                  In essence, anything that could in principle return multiple elements will preserve the keys (or, more generally, will preserve that level of the nested expression)






                  share|improve this answer









                  $endgroup$



                  You can use the fact that Part ([[…]]) preserves the keys when you supply a list of indices to extract:



                  z[[All, 1]]
                  (* < *)

                  (* without the list *)
                  z[[All, 1]]
                  (* 1, 4, 6, 10 *)


                  Alternatively you could use Take:



                  Map[Take[#, 1] &, z]
                  (* < *)


                  Other possibilites:



                  Map[#[[1]] &, z]
                  (* < *)

                  z[[All, ;; 1]]
                  (* < *)


                  In essence, anything that could in principle return multiple elements will preserve the keys (or, more generally, will preserve that level of the nested expression)







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 10 hours ago









                  Lukas LangLukas Lang

                  9,0841 gold badge11 silver badges34 bronze badges




                  9,0841 gold badge11 silver badges34 bronze badges


























                      2













                      $begingroup$

                      For completeness sake, a few more:



                      Extract[z, All, 1]
                      (* < *)

                      Cases[z, KeyValuePattern[x_ -> y_] :> x -> y]
                      (*"a", "b" -> 1, "g", "h" -> 4, "k", "l" -> 6, "s", "t" -> 10*)





                      share|improve this answer









                      $endgroup$



















                        2













                        $begingroup$

                        For completeness sake, a few more:



                        Extract[z, All, 1]
                        (* < *)

                        Cases[z, KeyValuePattern[x_ -> y_] :> x -> y]
                        (*"a", "b" -> 1, "g", "h" -> 4, "k", "l" -> 6, "s", "t" -> 10*)





                        share|improve this answer









                        $endgroup$

















                          2














                          2










                          2







                          $begingroup$

                          For completeness sake, a few more:



                          Extract[z, All, 1]
                          (* < *)

                          Cases[z, KeyValuePattern[x_ -> y_] :> x -> y]
                          (*"a", "b" -> 1, "g", "h" -> 4, "k", "l" -> 6, "s", "t" -> 10*)





                          share|improve this answer









                          $endgroup$



                          For completeness sake, a few more:



                          Extract[z, All, 1]
                          (* < *)

                          Cases[z, KeyValuePattern[x_ -> y_] :> x -> y]
                          (*"a", "b" -> 1, "g", "h" -> 4, "k", "l" -> 6, "s", "t" -> 10*)






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 5 hours ago









                          sakrasakra

                          3,29314 silver badges30 bronze badges




                          3,29314 silver badges30 bronze badges






























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