Terse Method to Swap Lowest for Highest?Efficient method for Inserting arrays into arraysSwap elements in list without copyBetter method to swap the values of two 2-D arraysHow to get this list with a terse methodBuilt-in (or Terse) Method to Combine and Transpose DatasetsAre there more readable and terse method can get this listefficiently method for generating a sequenceSimple method to sort versionsFunction for SortBySwap Elements of a continuous List, possible?

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Terse Method to Swap Lowest for Highest?


Efficient method for Inserting arrays into arraysSwap elements in list without copyBetter method to swap the values of two 2-D arraysHow to get this list with a terse methodBuilt-in (or Terse) Method to Combine and Transpose DatasetsAre there more readable and terse method can get this listefficiently method for generating a sequenceSimple method to sort versionsFunction for SortBySwap Elements of a continuous List, possible?













10












$begingroup$


I have built a solution to swap the lowest values with the highest values in a list.



With



SeedRandom[987]
test = RandomSample@*Join @@ Range @@@ 6, 10, 56, 60, 1, 5, -5, -1



-1, 2, 7, 8, 60, 57, 58, 10, 9, 4, -5, -3, 3, 59, 1, 5, -4, 6, -2, 56



Then



swapPositions =
PermutationReplace[
Ordering@Ordering@test,
With[len = Length@test,
Cycles@
Transpose@Range @@ 1, Floor[len/2], Reverse@*Range @@ Ceiling[len/2] + 1, len
]
];

Sort[test][[swapPositions]]



56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1



The largest half of the numbers have had their positions swapped with lowest half of the numbers.



However, it feels too verbose and I think Sort might be expensive in this case. Is there a built-in function or more terse method to achieve this. Of course with no loss in speed. The actual case is for list of length 100000 and more.










share|improve this question











$endgroup$
















    10












    $begingroup$


    I have built a solution to swap the lowest values with the highest values in a list.



    With



    SeedRandom[987]
    test = RandomSample@*Join @@ Range @@@ 6, 10, 56, 60, 1, 5, -5, -1



    -1, 2, 7, 8, 60, 57, 58, 10, 9, 4, -5, -3, 3, 59, 1, 5, -4, 6, -2, 56



    Then



    swapPositions =
    PermutationReplace[
    Ordering@Ordering@test,
    With[len = Length@test,
    Cycles@
    Transpose@Range @@ 1, Floor[len/2], Reverse@*Range @@ Ceiling[len/2] + 1, len
    ]
    ];

    Sort[test][[swapPositions]]



    56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1



    The largest half of the numbers have had their positions swapped with lowest half of the numbers.



    However, it feels too verbose and I think Sort might be expensive in this case. Is there a built-in function or more terse method to achieve this. Of course with no loss in speed. The actual case is for list of length 100000 and more.










    share|improve this question











    $endgroup$














      10












      10








      10





      $begingroup$


      I have built a solution to swap the lowest values with the highest values in a list.



      With



      SeedRandom[987]
      test = RandomSample@*Join @@ Range @@@ 6, 10, 56, 60, 1, 5, -5, -1



      -1, 2, 7, 8, 60, 57, 58, 10, 9, 4, -5, -3, 3, 59, 1, 5, -4, 6, -2, 56



      Then



      swapPositions =
      PermutationReplace[
      Ordering@Ordering@test,
      With[len = Length@test,
      Cycles@
      Transpose@Range @@ 1, Floor[len/2], Reverse@*Range @@ Ceiling[len/2] + 1, len
      ]
      ];

      Sort[test][[swapPositions]]



      56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1



      The largest half of the numbers have had their positions swapped with lowest half of the numbers.



      However, it feels too verbose and I think Sort might be expensive in this case. Is there a built-in function or more terse method to achieve this. Of course with no loss in speed. The actual case is for list of length 100000 and more.










      share|improve this question











      $endgroup$




      I have built a solution to swap the lowest values with the highest values in a list.



      With



      SeedRandom[987]
      test = RandomSample@*Join @@ Range @@@ 6, 10, 56, 60, 1, 5, -5, -1



      -1, 2, 7, 8, 60, 57, 58, 10, 9, 4, -5, -3, 3, 59, 1, 5, -4, 6, -2, 56



      Then



      swapPositions =
      PermutationReplace[
      Ordering@Ordering@test,
      With[len = Length@test,
      Cycles@
      Transpose@Range @@ 1, Floor[len/2], Reverse@*Range @@ Ceiling[len/2] + 1, len
      ]
      ];

      Sort[test][[swapPositions]]



      56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1



      The largest half of the numbers have had their positions swapped with lowest half of the numbers.



      However, it feels too verbose and I think Sort might be expensive in this case. Is there a built-in function or more terse method to achieve this. Of course with no loss in speed. The actual case is for list of length 100000 and more.







      list-manipulation performance-tuning sorting permutation






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 23 at 2:15









      J. M. is slightly pensive

      98.5k10308466




      98.5k10308466










      asked Mar 22 at 20:57









      EdmundEdmund

      26.7k330103




      26.7k330103




















          2 Answers
          2






          active

          oldest

          votes


















          13












          $begingroup$

          How about:



          Module[tmp = test,
          With[ord=Ordering[tmp],
          tmp[[ord]] = Reverse @ tmp[[ord]]];
          tmp
          ]



          56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1







          share|improve this answer









          $endgroup$








          • 1




            $begingroup$
            That is so obvious I want to cry. Thanks (+1).
            $endgroup$
            – Edmund
            Mar 22 at 21:15


















          7












          $begingroup$

          This is equivalent to Carl's procedure, except that it uses one less scratch list:



          With[ord = Ordering[test],
          test[[PermutationProduct[Reverse[ord], InversePermutation[ord]]]]]
          56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1


          Recall that list[[perm]] = list is equivalent to list = list[[InversePermutation[perm]]], where perm is a permutation list. (The situation is equivalent to list.pmat being the same as Transpose[pmat].list if pmat is a permutation matrix.) You can then use PermutationProduct[] to compose successive permutations.



          (This was supposed to be a comment, but it got too long.)






          share|improve this answer











          $endgroup$












          • $begingroup$
            This solution doesn't copy the list so may be faster than Carl's. (+1).
            $endgroup$
            – Edmund
            Mar 23 at 3:38










          • $begingroup$
            FWIW, I consistently get 56, -2, 6, -4, 5, 1, 59, 3, -3, -5, 4, 9, 10, 58, 57, 60, 8, 7, 2, -1 from this.
            $endgroup$
            – Rabbit
            2 days ago











          • $begingroup$
            @Rabbit, what version number of Mathematica is giving that result?
            $endgroup$
            – J. M. is slightly pensive
            2 days ago










          • $begingroup$
            11.3.0.0 (5944644, 2018030701) Win 10. I did a trace, which might have had the needed info but I didn't catch it. Started w/ fresh kernel, & repeated, w/ same result. Baffled.
            $endgroup$
            – Rabbit
            2 days ago










          • $begingroup$
            @Rabbit, can you try with With[ord = Ordering[test], test[[PermutationProduct[InversePermutation[ord], Reverse[ord]]]]]?
            $endgroup$
            – J. M. is slightly pensive
            2 days ago










          Your Answer





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          2 Answers
          2






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          2 Answers
          2






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          13












          $begingroup$

          How about:



          Module[tmp = test,
          With[ord=Ordering[tmp],
          tmp[[ord]] = Reverse @ tmp[[ord]]];
          tmp
          ]



          56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1







          share|improve this answer









          $endgroup$








          • 1




            $begingroup$
            That is so obvious I want to cry. Thanks (+1).
            $endgroup$
            – Edmund
            Mar 22 at 21:15















          13












          $begingroup$

          How about:



          Module[tmp = test,
          With[ord=Ordering[tmp],
          tmp[[ord]] = Reverse @ tmp[[ord]]];
          tmp
          ]



          56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1







          share|improve this answer









          $endgroup$








          • 1




            $begingroup$
            That is so obvious I want to cry. Thanks (+1).
            $endgroup$
            – Edmund
            Mar 22 at 21:15













          13












          13








          13





          $begingroup$

          How about:



          Module[tmp = test,
          With[ord=Ordering[tmp],
          tmp[[ord]] = Reverse @ tmp[[ord]]];
          tmp
          ]



          56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1







          share|improve this answer









          $endgroup$



          How about:



          Module[tmp = test,
          With[ord=Ordering[tmp],
          tmp[[ord]] = Reverse @ tmp[[ord]]];
          tmp
          ]



          56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Mar 22 at 21:11









          Carl WollCarl Woll

          71.6k394186




          71.6k394186







          • 1




            $begingroup$
            That is so obvious I want to cry. Thanks (+1).
            $endgroup$
            – Edmund
            Mar 22 at 21:15












          • 1




            $begingroup$
            That is so obvious I want to cry. Thanks (+1).
            $endgroup$
            – Edmund
            Mar 22 at 21:15







          1




          1




          $begingroup$
          That is so obvious I want to cry. Thanks (+1).
          $endgroup$
          – Edmund
          Mar 22 at 21:15




          $begingroup$
          That is so obvious I want to cry. Thanks (+1).
          $endgroup$
          – Edmund
          Mar 22 at 21:15











          7












          $begingroup$

          This is equivalent to Carl's procedure, except that it uses one less scratch list:



          With[ord = Ordering[test],
          test[[PermutationProduct[Reverse[ord], InversePermutation[ord]]]]]
          56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1


          Recall that list[[perm]] = list is equivalent to list = list[[InversePermutation[perm]]], where perm is a permutation list. (The situation is equivalent to list.pmat being the same as Transpose[pmat].list if pmat is a permutation matrix.) You can then use PermutationProduct[] to compose successive permutations.



          (This was supposed to be a comment, but it got too long.)






          share|improve this answer











          $endgroup$












          • $begingroup$
            This solution doesn't copy the list so may be faster than Carl's. (+1).
            $endgroup$
            – Edmund
            Mar 23 at 3:38










          • $begingroup$
            FWIW, I consistently get 56, -2, 6, -4, 5, 1, 59, 3, -3, -5, 4, 9, 10, 58, 57, 60, 8, 7, 2, -1 from this.
            $endgroup$
            – Rabbit
            2 days ago











          • $begingroup$
            @Rabbit, what version number of Mathematica is giving that result?
            $endgroup$
            – J. M. is slightly pensive
            2 days ago










          • $begingroup$
            11.3.0.0 (5944644, 2018030701) Win 10. I did a trace, which might have had the needed info but I didn't catch it. Started w/ fresh kernel, & repeated, w/ same result. Baffled.
            $endgroup$
            – Rabbit
            2 days ago










          • $begingroup$
            @Rabbit, can you try with With[ord = Ordering[test], test[[PermutationProduct[InversePermutation[ord], Reverse[ord]]]]]?
            $endgroup$
            – J. M. is slightly pensive
            2 days ago















          7












          $begingroup$

          This is equivalent to Carl's procedure, except that it uses one less scratch list:



          With[ord = Ordering[test],
          test[[PermutationProduct[Reverse[ord], InversePermutation[ord]]]]]
          56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1


          Recall that list[[perm]] = list is equivalent to list = list[[InversePermutation[perm]]], where perm is a permutation list. (The situation is equivalent to list.pmat being the same as Transpose[pmat].list if pmat is a permutation matrix.) You can then use PermutationProduct[] to compose successive permutations.



          (This was supposed to be a comment, but it got too long.)






          share|improve this answer











          $endgroup$












          • $begingroup$
            This solution doesn't copy the list so may be faster than Carl's. (+1).
            $endgroup$
            – Edmund
            Mar 23 at 3:38










          • $begingroup$
            FWIW, I consistently get 56, -2, 6, -4, 5, 1, 59, 3, -3, -5, 4, 9, 10, 58, 57, 60, 8, 7, 2, -1 from this.
            $endgroup$
            – Rabbit
            2 days ago











          • $begingroup$
            @Rabbit, what version number of Mathematica is giving that result?
            $endgroup$
            – J. M. is slightly pensive
            2 days ago










          • $begingroup$
            11.3.0.0 (5944644, 2018030701) Win 10. I did a trace, which might have had the needed info but I didn't catch it. Started w/ fresh kernel, & repeated, w/ same result. Baffled.
            $endgroup$
            – Rabbit
            2 days ago










          • $begingroup$
            @Rabbit, can you try with With[ord = Ordering[test], test[[PermutationProduct[InversePermutation[ord], Reverse[ord]]]]]?
            $endgroup$
            – J. M. is slightly pensive
            2 days ago













          7












          7








          7





          $begingroup$

          This is equivalent to Carl's procedure, except that it uses one less scratch list:



          With[ord = Ordering[test],
          test[[PermutationProduct[Reverse[ord], InversePermutation[ord]]]]]
          56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1


          Recall that list[[perm]] = list is equivalent to list = list[[InversePermutation[perm]]], where perm is a permutation list. (The situation is equivalent to list.pmat being the same as Transpose[pmat].list if pmat is a permutation matrix.) You can then use PermutationProduct[] to compose successive permutations.



          (This was supposed to be a comment, but it got too long.)






          share|improve this answer











          $endgroup$



          This is equivalent to Carl's procedure, except that it uses one less scratch list:



          With[ord = Ordering[test],
          test[[PermutationProduct[Reverse[ord], InversePermutation[ord]]]]]
          56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1


          Recall that list[[perm]] = list is equivalent to list = list[[InversePermutation[perm]]], where perm is a permutation list. (The situation is equivalent to list.pmat being the same as Transpose[pmat].list if pmat is a permutation matrix.) You can then use PermutationProduct[] to compose successive permutations.



          (This was supposed to be a comment, but it got too long.)







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Mar 23 at 2:27

























          answered Mar 23 at 2:14









          J. M. is slightly pensiveJ. M. is slightly pensive

          98.5k10308466




          98.5k10308466











          • $begingroup$
            This solution doesn't copy the list so may be faster than Carl's. (+1).
            $endgroup$
            – Edmund
            Mar 23 at 3:38










          • $begingroup$
            FWIW, I consistently get 56, -2, 6, -4, 5, 1, 59, 3, -3, -5, 4, 9, 10, 58, 57, 60, 8, 7, 2, -1 from this.
            $endgroup$
            – Rabbit
            2 days ago











          • $begingroup$
            @Rabbit, what version number of Mathematica is giving that result?
            $endgroup$
            – J. M. is slightly pensive
            2 days ago










          • $begingroup$
            11.3.0.0 (5944644, 2018030701) Win 10. I did a trace, which might have had the needed info but I didn't catch it. Started w/ fresh kernel, & repeated, w/ same result. Baffled.
            $endgroup$
            – Rabbit
            2 days ago










          • $begingroup$
            @Rabbit, can you try with With[ord = Ordering[test], test[[PermutationProduct[InversePermutation[ord], Reverse[ord]]]]]?
            $endgroup$
            – J. M. is slightly pensive
            2 days ago
















          • $begingroup$
            This solution doesn't copy the list so may be faster than Carl's. (+1).
            $endgroup$
            – Edmund
            Mar 23 at 3:38










          • $begingroup$
            FWIW, I consistently get 56, -2, 6, -4, 5, 1, 59, 3, -3, -5, 4, 9, 10, 58, 57, 60, 8, 7, 2, -1 from this.
            $endgroup$
            – Rabbit
            2 days ago











          • $begingroup$
            @Rabbit, what version number of Mathematica is giving that result?
            $endgroup$
            – J. M. is slightly pensive
            2 days ago










          • $begingroup$
            11.3.0.0 (5944644, 2018030701) Win 10. I did a trace, which might have had the needed info but I didn't catch it. Started w/ fresh kernel, & repeated, w/ same result. Baffled.
            $endgroup$
            – Rabbit
            2 days ago










          • $begingroup$
            @Rabbit, can you try with With[ord = Ordering[test], test[[PermutationProduct[InversePermutation[ord], Reverse[ord]]]]]?
            $endgroup$
            – J. M. is slightly pensive
            2 days ago















          $begingroup$
          This solution doesn't copy the list so may be faster than Carl's. (+1).
          $endgroup$
          – Edmund
          Mar 23 at 3:38




          $begingroup$
          This solution doesn't copy the list so may be faster than Carl's. (+1).
          $endgroup$
          – Edmund
          Mar 23 at 3:38












          $begingroup$
          FWIW, I consistently get 56, -2, 6, -4, 5, 1, 59, 3, -3, -5, 4, 9, 10, 58, 57, 60, 8, 7, 2, -1 from this.
          $endgroup$
          – Rabbit
          2 days ago





          $begingroup$
          FWIW, I consistently get 56, -2, 6, -4, 5, 1, 59, 3, -3, -5, 4, 9, 10, 58, 57, 60, 8, 7, 2, -1 from this.
          $endgroup$
          – Rabbit
          2 days ago













          $begingroup$
          @Rabbit, what version number of Mathematica is giving that result?
          $endgroup$
          – J. M. is slightly pensive
          2 days ago




          $begingroup$
          @Rabbit, what version number of Mathematica is giving that result?
          $endgroup$
          – J. M. is slightly pensive
          2 days ago












          $begingroup$
          11.3.0.0 (5944644, 2018030701) Win 10. I did a trace, which might have had the needed info but I didn't catch it. Started w/ fresh kernel, & repeated, w/ same result. Baffled.
          $endgroup$
          – Rabbit
          2 days ago




          $begingroup$
          11.3.0.0 (5944644, 2018030701) Win 10. I did a trace, which might have had the needed info but I didn't catch it. Started w/ fresh kernel, & repeated, w/ same result. Baffled.
          $endgroup$
          – Rabbit
          2 days ago












          $begingroup$
          @Rabbit, can you try with With[ord = Ordering[test], test[[PermutationProduct[InversePermutation[ord], Reverse[ord]]]]]?
          $endgroup$
          – J. M. is slightly pensive
          2 days ago




          $begingroup$
          @Rabbit, can you try with With[ord = Ordering[test], test[[PermutationProduct[InversePermutation[ord], Reverse[ord]]]]]?
          $endgroup$
          – J. M. is slightly pensive
          2 days ago

















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