Terse Method to Swap Lowest for Highest?












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.






list-manipulation performance-tuning sorting permutation







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.






    list-manipulation performance-tuning sorting permutation







    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.






      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 2 days ago









      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
            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










          • $begingroup$
            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











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          2 Answers
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          2 Answers
          2






          active

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          active

          oldest

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          active

          oldest

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          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
            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










          • $begingroup$
            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
            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










          • $begingroup$
            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 2 days ago

























          answered 2 days ago









          J. M. is slightly pensive♦J. 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
            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










          • $begingroup$
            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
            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










          • $begingroup$
            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
          2 days ago




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






          $begingroup$
          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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          Bunad

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          Tradisjonelle folkedrakter frå Aust-Telemark, truleg frå 1880-åra. Drakter som dette er grunnlaget for moderne bunadar. Foto: Axel Lindahl Bunad frå Eggedal i Buskerud fotografert i 1906. Bunad (frå kledebunad , der norrønt «búnaðr» har same rot som (føre)bu) er ei samnemning brukt om klesdrakter i tradisjonell nordisk stil — ofte utvikla frå folkedrakter, men likevel i ein eigen tradisjon. Ein bunad er typisk ei forseggjort og kostbar drakt, med mykje broderi, og med handsmidde smykke i gull eller sølv, særleg dei særeigne søljene, som obligatorisk prynad i tillegg. Bunad og bunadbruk er eit debattemne med mange ulike oppfatningar, til dømes om kva som er ein bunad, og om ein kan endra på utforminga av draktene eller ikkje. Den norske bunadrørsla går tilbake til folkedansrørsla og føregangsfolk som Hulda Garborg og Klara Semb. I dag blir bunad brukt som nasjonaldrakt. Festbunad er likestilt med galla-antrekk, og ofte nytta ved særskilde høve, som konfirmasjonar eller d
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          Færeyskur hestur Heimild | Tengill | Tilvísanir | LeiðsagnarvalRossið - síða um færeyska hrossið á færeyskuGott ár hjá færeyska hestinum

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          FæreyjarHestar færeyskahjaltlandshestinumíslenski hesturinnBretlandskolanámumenskuWikipediafæreyskuWikipediaRossið - síða um færeyska hrossið á færeysku Færeyskur hestur Úr Wikipediu, frjálsa alfræðiritinu Jump to navigation Jump to search Rani og Grani. Grani í vetrarfeldi. Færeyski hesturinn (færeyska: Føroyska rossið ) er hestur sem hefur lifað í Færeyjum í hundruði ára. Hann er smágerður, 120-132 cm á hæð, og helst skyldur hjaltlandshestinum. Hlutverk hans var að draga vagna og plóg og bera klyfjar. Hann hefur fjórar gangtegundir eins og íslenski hesturinn (þar með talið tölt) og er með fjölda litaafbrigði. Nú til dags er hann notaður til frístunda. Í lok 19. aldar voru til um 800 hross. Mörg þeirra voru seld til Bretlands til að vinna í kolanámum og um 1960 voru aðeins nokkrir hestar eftir. Átak var gert í að varðveita hestinn og eru nú til 70-80 hross í Færeyjum. [1] Heimild |   Wikimedia Commons er með margmiðlunarefni sem te
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