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How to reverse every other sublist of a list?



The 2019 Stack Overflow Developer Survey Results Are InHow can the lowest non-zero value in a nested list be found and its position in the sublist returned?Comparing elements of the $n^textth$ sublist in a ragged list with the $n^textth$ member of a sequenceSublist inside simple listHow to split list into disjoint lists?Selecting every last element of a nested listAdd elements of one list to sublists of another listEliminating elements from sublists under a global conditionRemove the right sublistList manipulation - adding last element of sublist to each sublistHow to partition a 1D list according to specified lengths?










4












$begingroup$


I have a list with sublists list = 1, 2, 3, 4, 5, 6, 7, 8. I would like to reverse every other sublist, starting from the second one. The result would look like result = 1, 2, 4, 3, 5, 6, 8, 7. What would be a simple way to do this?










share|improve this question









$endgroup$
















    4












    $begingroup$


    I have a list with sublists list = 1, 2, 3, 4, 5, 6, 7, 8. I would like to reverse every other sublist, starting from the second one. The result would look like result = 1, 2, 4, 3, 5, 6, 8, 7. What would be a simple way to do this?










    share|improve this question









    $endgroup$














      4












      4








      4





      $begingroup$


      I have a list with sublists list = 1, 2, 3, 4, 5, 6, 7, 8. I would like to reverse every other sublist, starting from the second one. The result would look like result = 1, 2, 4, 3, 5, 6, 8, 7. What would be a simple way to do this?










      share|improve this question









      $endgroup$




      I have a list with sublists list = 1, 2, 3, 4, 5, 6, 7, 8. I would like to reverse every other sublist, starting from the second one. The result would look like result = 1, 2, 4, 3, 5, 6, 8, 7. What would be a simple way to do this?







      list-manipulation






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 1 hour ago









      nanjunnanjun

      43229




      43229




















          3 Answers
          3






          active

          oldest

          votes


















          5












          $begingroup$

          Use MapAt, which accepts the same syntax as Part:



          MapAt[Reverse, list, 2 ;; ;; 2]
          (* 1, 2, 4, 3, 5, 6, 8, 7 *)





          share|improve this answer









          $endgroup$




















            2












            $begingroup$

            Using Part and Span might not seem overly elegant but it is fast:



            list = RandomReal[-1, 1, 100000, 10];

            a = MapAt[Reverse, list, 2 ;; ;; 2]; // RepeatedTiming // First
            b = ReplacePart[list, i_?EvenQ :> Reverse@list[[i]]]; // RepeatedTiming // First
            c = Module[result = list,
            result[[2 ;; ;; 2]] =
            list[[2 ;; ;; 2, -1 ;; 1 ;; -1]];
            result
            ]; // RepeatedTiming // First
            a == b == c



            0.11



            0.317



            0.0036



            True







            share|improve this answer









            $endgroup$




















              1












              $begingroup$

              Another method is to use ReplacePart:



              ReplacePart[1, 2, 3, 4, 5, 6, 7, 8,
              i_ ? EvenQ :> Reverse@list[[i]]
              ]
              (* 1,2,4,3,5,6,8,7 *)





              share|improve this answer









              $endgroup$













                Your Answer





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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                5












                $begingroup$

                Use MapAt, which accepts the same syntax as Part:



                MapAt[Reverse, list, 2 ;; ;; 2]
                (* 1, 2, 4, 3, 5, 6, 8, 7 *)





                share|improve this answer









                $endgroup$

















                  5












                  $begingroup$

                  Use MapAt, which accepts the same syntax as Part:



                  MapAt[Reverse, list, 2 ;; ;; 2]
                  (* 1, 2, 4, 3, 5, 6, 8, 7 *)





                  share|improve this answer









                  $endgroup$















                    5












                    5








                    5





                    $begingroup$

                    Use MapAt, which accepts the same syntax as Part:



                    MapAt[Reverse, list, 2 ;; ;; 2]
                    (* 1, 2, 4, 3, 5, 6, 8, 7 *)





                    share|improve this answer









                    $endgroup$



                    Use MapAt, which accepts the same syntax as Part:



                    MapAt[Reverse, list, 2 ;; ;; 2]
                    (* 1, 2, 4, 3, 5, 6, 8, 7 *)






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 1 hour ago









                    marchmarch

                    17.5k22769




                    17.5k22769





















                        2












                        $begingroup$

                        Using Part and Span might not seem overly elegant but it is fast:



                        list = RandomReal[-1, 1, 100000, 10];

                        a = MapAt[Reverse, list, 2 ;; ;; 2]; // RepeatedTiming // First
                        b = ReplacePart[list, i_?EvenQ :> Reverse@list[[i]]]; // RepeatedTiming // First
                        c = Module[result = list,
                        result[[2 ;; ;; 2]] =
                        list[[2 ;; ;; 2, -1 ;; 1 ;; -1]];
                        result
                        ]; // RepeatedTiming // First
                        a == b == c



                        0.11



                        0.317



                        0.0036



                        True







                        share|improve this answer









                        $endgroup$

















                          2












                          $begingroup$

                          Using Part and Span might not seem overly elegant but it is fast:



                          list = RandomReal[-1, 1, 100000, 10];

                          a = MapAt[Reverse, list, 2 ;; ;; 2]; // RepeatedTiming // First
                          b = ReplacePart[list, i_?EvenQ :> Reverse@list[[i]]]; // RepeatedTiming // First
                          c = Module[result = list,
                          result[[2 ;; ;; 2]] =
                          list[[2 ;; ;; 2, -1 ;; 1 ;; -1]];
                          result
                          ]; // RepeatedTiming // First
                          a == b == c



                          0.11



                          0.317



                          0.0036



                          True







                          share|improve this answer









                          $endgroup$















                            2












                            2








                            2





                            $begingroup$

                            Using Part and Span might not seem overly elegant but it is fast:



                            list = RandomReal[-1, 1, 100000, 10];

                            a = MapAt[Reverse, list, 2 ;; ;; 2]; // RepeatedTiming // First
                            b = ReplacePart[list, i_?EvenQ :> Reverse@list[[i]]]; // RepeatedTiming // First
                            c = Module[result = list,
                            result[[2 ;; ;; 2]] =
                            list[[2 ;; ;; 2, -1 ;; 1 ;; -1]];
                            result
                            ]; // RepeatedTiming // First
                            a == b == c



                            0.11



                            0.317



                            0.0036



                            True







                            share|improve this answer









                            $endgroup$



                            Using Part and Span might not seem overly elegant but it is fast:



                            list = RandomReal[-1, 1, 100000, 10];

                            a = MapAt[Reverse, list, 2 ;; ;; 2]; // RepeatedTiming // First
                            b = ReplacePart[list, i_?EvenQ :> Reverse@list[[i]]]; // RepeatedTiming // First
                            c = Module[result = list,
                            result[[2 ;; ;; 2]] =
                            list[[2 ;; ;; 2, -1 ;; 1 ;; -1]];
                            result
                            ]; // RepeatedTiming // First
                            a == b == c



                            0.11



                            0.317



                            0.0036



                            True








                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered 30 mins ago









                            Henrik SchumacherHenrik Schumacher

                            59.6k582166




                            59.6k582166





















                                1












                                $begingroup$

                                Another method is to use ReplacePart:



                                ReplacePart[1, 2, 3, 4, 5, 6, 7, 8,
                                i_ ? EvenQ :> Reverse@list[[i]]
                                ]
                                (* 1,2,4,3,5,6,8,7 *)





                                share|improve this answer









                                $endgroup$

















                                  1












                                  $begingroup$

                                  Another method is to use ReplacePart:



                                  ReplacePart[1, 2, 3, 4, 5, 6, 7, 8,
                                  i_ ? EvenQ :> Reverse@list[[i]]
                                  ]
                                  (* 1,2,4,3,5,6,8,7 *)





                                  share|improve this answer









                                  $endgroup$















                                    1












                                    1








                                    1





                                    $begingroup$

                                    Another method is to use ReplacePart:



                                    ReplacePart[1, 2, 3, 4, 5, 6, 7, 8,
                                    i_ ? EvenQ :> Reverse@list[[i]]
                                    ]
                                    (* 1,2,4,3,5,6,8,7 *)





                                    share|improve this answer









                                    $endgroup$



                                    Another method is to use ReplacePart:



                                    ReplacePart[1, 2, 3, 4, 5, 6, 7, 8,
                                    i_ ? EvenQ :> Reverse@list[[i]]
                                    ]
                                    (* 1,2,4,3,5,6,8,7 *)






                                    share|improve this answer












                                    share|improve this answer



                                    share|improve this answer










                                    answered 40 mins ago









                                    Jason B.Jason B.

                                    48.9k389196




                                    48.9k389196



























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