How to reverse every other sublist of a list?












5












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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?










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    5












    $begingroup$


    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









    $endgroup$















      5












      5








      5


      2



      $begingroup$


      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






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      asked 2 days ago









      nanjunnanjun

      44429




      44429






















          4 Answers
          4






          active

          oldest

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          13












          $begingroup$

          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









          $endgroup$





















            9












            $begingroup$

            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




              $begingroup$
              A bit cleaner, s/b comparable in speed: Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
              $endgroup$
              – ciao
              yesterday



















            4












            $begingroup$

            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









            $endgroup$





















              3












              $begingroup$

              This will work



              list = {{1, 2}, {3, 4}, {5, 6}, {7, 8}};

              Table[If[EvenQ[i], Reverse[list[[i]]], list[[i]]], {i, 1, Length[list]}]

              (*{{1, 2}, {4, 3}, {5, 6}, {8, 7}}*)
              ```





              share|improve this answer








              New contributor




              cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






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                Your Answer





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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                13












                $begingroup$

                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









                $endgroup$


















                  13












                  $begingroup$

                  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









                  $endgroup$
















                    13












                    13








                    13





                    $begingroup$

                    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 2 days ago









                    marchmarch

                    17.6k22770




                    17.6k22770























                        9












                        $begingroup$

                        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




                          $begingroup$
                          A bit cleaner, s/b comparable in speed: Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
                          $endgroup$
                          – ciao
                          yesterday
















                        9












                        $begingroup$

                        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




                          $begingroup$
                          A bit cleaner, s/b comparable in speed: Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
                          $endgroup$
                          – ciao
                          yesterday














                        9












                        9








                        9





                        $begingroup$

                        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








                        edited yesterday

























                        answered 2 days ago









                        Henrik SchumacherHenrik Schumacher

                        59.9k582167




                        59.9k582167








                        • 2




                          $begingroup$
                          A bit cleaner, s/b comparable in speed: Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
                          $endgroup$
                          – ciao
                          yesterday














                        • 2




                          $begingroup$
                          A bit cleaner, s/b comparable in speed: Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
                          $endgroup$
                          – ciao
                          yesterday








                        2




                        2




                        $begingroup$
                        A bit cleaner, s/b comparable in speed: Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
                        $endgroup$
                        – ciao
                        yesterday




                        $begingroup$
                        A bit cleaner, s/b comparable in speed: Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
                        $endgroup$
                        – ciao
                        yesterday











                        4












                        $begingroup$

                        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









                        $endgroup$


















                          4












                          $begingroup$

                          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









                          $endgroup$
















                            4












                            4








                            4





                            $begingroup$

                            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 2 days ago









                            Jason B.Jason B.

                            48.9k389197




                            48.9k389197























                                3












                                $begingroup$

                                This will work



                                list = {{1, 2}, {3, 4}, {5, 6}, {7, 8}};

                                Table[If[EvenQ[i], Reverse[list[[i]]], list[[i]]], {i, 1, Length[list]}]

                                (*{{1, 2}, {4, 3}, {5, 6}, {8, 7}}*)
                                ```





                                share|improve this answer








                                New contributor




                                cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.






                                $endgroup$


















                                  3












                                  $begingroup$

                                  This will work



                                  list = {{1, 2}, {3, 4}, {5, 6}, {7, 8}};

                                  Table[If[EvenQ[i], Reverse[list[[i]]], list[[i]]], {i, 1, Length[list]}]

                                  (*{{1, 2}, {4, 3}, {5, 6}, {8, 7}}*)
                                  ```





                                  share|improve this answer








                                  New contributor




                                  cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.






                                  $endgroup$
















                                    3












                                    3








                                    3





                                    $begingroup$

                                    This will work



                                    list = {{1, 2}, {3, 4}, {5, 6}, {7, 8}};

                                    Table[If[EvenQ[i], Reverse[list[[i]]], list[[i]]], {i, 1, Length[list]}]

                                    (*{{1, 2}, {4, 3}, {5, 6}, {8, 7}}*)
                                    ```





                                    share|improve this answer








                                    New contributor




                                    cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.






                                    $endgroup$



                                    This will work



                                    list = {{1, 2}, {3, 4}, {5, 6}, {7, 8}};

                                    Table[If[EvenQ[i], Reverse[list[[i]]], list[[i]]], {i, 1, Length[list]}]

                                    (*{{1, 2}, {4, 3}, {5, 6}, {8, 7}}*)
                                    ```






                                    share|improve this answer








                                    New contributor




                                    cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.









                                    share|improve this answer



                                    share|improve this answer






                                    New contributor




                                    cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.









                                    answered 2 days ago









                                    cphyscphys

                                    815




                                    815




                                    New contributor




                                    cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.





                                    New contributor





                                    cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.






                                    cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.






























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