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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









          $endgroup$




















            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












              $begingroup$

              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









              $endgroup$













                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





















                        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 35 mins ago









                            Jason B.Jason B.

                            48.9k389196




                            48.9k389196





















                                1












                                $begingroup$

                                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












                                  $begingroup$

                                  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












                                    1








                                    1





                                    $begingroup$

                                    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 25 mins ago









                                    Henrik SchumacherHenrik Schumacher

                                    59.5k582166




                                    59.5k582166



























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