Terse Method to Swap Lowest for Highest?












12












$begingroup$


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.






list-manipulation performance-tuning sorting permutation







share|improve this question











$endgroup$

















    12












    $begingroup$


    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.






    list-manipulation performance-tuning sorting permutation







    share|improve this question











    $endgroup$















      12












      12








      12





      $begingroup$


      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.






      list-manipulation performance-tuning sorting permutation







      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




      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♦

      98.8k10311467




      98.8k10311467










      asked Mar 22 at 20:57









      EdmundEdmund

      26.7k330103




      26.7k330103






















          2 Answers
          2






          active

          oldest

          votes


















          15












          $begingroup$

          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
            Mar 23 at 16:01












          • $begingroup$
            @Rabbit, what version number of Mathematica is giving that result?
            $endgroup$
            – J. M. is slightly pensive♦
            Mar 23 at 16:10










          • $begingroup$
            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
            Mar 23 at 16:16










          • $begingroup$
            @Rabbit, can you try with With[{ord = Ordering[test]}, test[[PermutationProduct[InversePermutation[ord], Reverse[ord]]]]]?
            $endgroup$
            – J. M. is slightly pensive♦
            Mar 23 at 17:29












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          2 Answers
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          2 Answers
          2






          active

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          active

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          15












          $begingroup$

          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
















          15












          $begingroup$

          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














          15












          15








          15





          $begingroup$

          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

          72.1k395186




          72.1k395186








          • 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
            Mar 23 at 16:01












          • $begingroup$
            @Rabbit, what version number of Mathematica is giving that result?
            $endgroup$
            – J. M. is slightly pensive♦
            Mar 23 at 16:10










          • $begingroup$
            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
            Mar 23 at 16:16










          • $begingroup$
            @Rabbit, can you try with With[{ord = Ordering[test]}, test[[PermutationProduct[InversePermutation[ord], Reverse[ord]]]]]?
            $endgroup$
            – J. M. is slightly pensive♦
            Mar 23 at 17:29
















          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
            Mar 23 at 16:01












          • $begingroup$
            @Rabbit, what version number of Mathematica is giving that result?
            $endgroup$
            – J. M. is slightly pensive♦
            Mar 23 at 16:10










          • $begingroup$
            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
            Mar 23 at 16:16










          • $begingroup$
            @Rabbit, can you try with With[{ord = Ordering[test]}, test[[PermutationProduct[InversePermutation[ord], Reverse[ord]]]]]?
            $endgroup$
            – J. M. is slightly pensive♦
            Mar 23 at 17:29














          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 pensive♦J. M. is slightly pensive

          98.8k10311467




          98.8k10311467












          • $begingroup$
            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
            Mar 23 at 16:01












          • $begingroup$
            @Rabbit, what version number of Mathematica is giving that result?
            $endgroup$
            – J. M. is slightly pensive♦
            Mar 23 at 16:10










          • $begingroup$
            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
            Mar 23 at 16:16










          • $begingroup$
            @Rabbit, can you try with With[{ord = Ordering[test]}, test[[PermutationProduct[InversePermutation[ord], Reverse[ord]]]]]?
            $endgroup$
            – J. M. is slightly pensive♦
            Mar 23 at 17:29


















          • $begingroup$
            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
            Mar 23 at 16:01












          • $begingroup$
            @Rabbit, what version number of Mathematica is giving that result?
            $endgroup$
            – J. M. is slightly pensive♦
            Mar 23 at 16:10










          • $begingroup$
            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
            Mar 23 at 16:16










          • $begingroup$
            @Rabbit, can you try with With[{ord = Ordering[test]}, test[[PermutationProduct[InversePermutation[ord], Reverse[ord]]]]]?
            $endgroup$
            – J. M. is slightly pensive♦
            Mar 23 at 17:29
















          $begingroup$
          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
          Mar 23 at 16:01






          $begingroup$
          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
          Mar 23 at 16:01














          $begingroup$
          @Rabbit, what version number of Mathematica is giving that result?
          $endgroup$
          – J. M. is slightly pensive♦
          Mar 23 at 16:10




          $begingroup$
          @Rabbit, what version number of Mathematica is giving that result?
          $endgroup$
          – J. M. is slightly pensive♦
          Mar 23 at 16:10












          $begingroup$
          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
          Mar 23 at 16:16




          $begingroup$
          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
          Mar 23 at 16:16












          $begingroup$
          @Rabbit, can you try with With[{ord = Ordering[test]}, test[[PermutationProduct[InversePermutation[ord], Reverse[ord]]]]]?
          $endgroup$
          – J. M. is slightly pensive♦
          Mar 23 at 17:29




          $begingroup$
          @Rabbit, can you try with With[{ord = Ordering[test]}, test[[PermutationProduct[InversePermutation[ord], Reverse[ord]]]]]?
          $endgroup$
          – J. M. is slightly pensive♦
          Mar 23 at 17:29


















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          Alcedinidae

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          Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout
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          RAC Tourist Trophy

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          Pour les articles homonymes, voir Tourist Trophy. Le RAC Tourist Trophy est une course automobile organisée par le Royal Automobile Club d'Angleterre. Créée en 1905, cette course a fait partie des épreuves comptant pour le championnat du monde des voitures de sport entre 1953 et 1955, en 1958 et 1959, et entre 1962 et 1965. Après une interruption de 17 ans (l'épreuve s'arrête après 1988), le RAC Tourist Trophy reprend en 2005 sous l'égide de la FIA GT. Palmarès | John Napier sur Arrol-Johnston, vainqueur du premier RAC Tourist Trophy, en 1905. L'honorable C.S. Rolls, vainqueur du RAC Tourist Trophy 1906. Ernest Curtis, sur Rover 20 hp, vainqueur du RAC Tourist Trophy 1907. William Watson, vainqueur du RAC Tourist Trophy 1908. Kenelm Lee Guinness, vainqueur du RAC Tourist Trophy 1914 sur Sunbeam. Départ du RAC Tourist Trophy 1928. Louis Gérard, vainqueur du RAC Tourist Trophy 1938 sur Delage. Année Circuit Vainqueur
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