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

          oldest

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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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          Can an atomic nucleus contain both particles and antiparticles? [duplicate]

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          9 $begingroup$ This question already has an answer here: What happens if we put together a proton and an antineutron? 2 answers Is it theoretically possible to make a "deuterium" atom containing a proton and an antineutron in its nucleus? Would the strong nuclear force cause attraction between a proton and an antineutron? Would such a nucleus be stable, or would the proton somehow annihilate the antineutron when close enough? nuclear-physics antimatter share | cite | improve this question asked yesterday
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