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 yesterday









      J. M. is slightly pensive♦

      98.4k10307466




      98.4k10307466










      asked yesterday









      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
            yesterday



















          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
            yesterday










          • $begingroup$
            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
            yesterday












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










          • $begingroup$
            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
            yesterday










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











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          2 Answers
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          active

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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
            yesterday
















          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
            yesterday














          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 yesterday









          Carl WollCarl Woll

          71.4k394186




          71.4k394186








          • 1




            $begingroup$
            That is so obvious I want to cry. Thanks (+1).
            $endgroup$
            – Edmund
            yesterday














          • 1




            $begingroup$
            That is so obvious I want to cry. Thanks (+1).
            $endgroup$
            – Edmund
            yesterday








          1




          1




          $begingroup$
          That is so obvious I want to cry. Thanks (+1).
          $endgroup$
          – Edmund
          yesterday




          $begingroup$
          That is so obvious I want to cry. Thanks (+1).
          $endgroup$
          – Edmund
          yesterday











          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
            yesterday










          • $begingroup$
            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
            yesterday












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










          • $begingroup$
            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
            yesterday










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
















          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
            yesterday










          • $begingroup$
            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
            yesterday












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










          • $begingroup$
            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
            yesterday










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














          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 yesterday

























          answered yesterday









          J. M. is slightly pensive♦J. M. is slightly pensive

          98.4k10307466




          98.4k10307466












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










          • $begingroup$
            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
            yesterday












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










          • $begingroup$
            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
            yesterday










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


















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










          • $begingroup$
            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
            yesterday












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










          • $begingroup$
            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
            yesterday










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
















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




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












          $begingroup$
          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
          yesterday






          $begingroup$
          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
          yesterday














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




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












          $begingroup$
          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
          yesterday




          $begingroup$
          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
          yesterday












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




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


















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            本文介紹的是國際性音樂電視頻道Channel [V]的主條目。關於在臺灣地區前稱Channel [V]的子電視頻道，請見「 FOX (台灣頻道) 」。 body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} Channel [V] Channel [V] 所有者 星空傳媒 福斯傳媒集團 星空華文傳媒（中国大陆频道） 开播日期 1991年9月15日 口号 全球最強華語音樂頻道；亞洲第一音樂頻道 播出地区 亞洲各地 旧称 衛視音樂台 取代者 [V] by FOX (東南亞和中東國家版本) 姊妹频道 星空衛視 收看方法 有线电视   澳門澳門有線電視 [V]國際頻道：73頻道 [V]中國頻道：74頻道   中国深圳天威视讯 [V]中國頻道：尚未定頻   中国南京数字电视 [V]中國頻道：91頻道   中国中国大陸数字电视 [V]中國頻道：随地区而变 宽带电视   香港now寬頻電視 [V]國際頻道：534頻道 臺灣中華電信MOD [V]國際娛樂頻道HD：151頻道   新加坡Mio TV [V]國際娛樂頻道HD：352頻道 [V]中國頻道：547頻道 Channel [V] 是一個國際性的音樂電視頻道，隸屬於星空傳媒，在東亞、東南亞、南亞、中東地區等地多個國家播放。該頻道主要播放英文主流音樂和一些區域性語言的歌曲，其中在台灣與中國大陸主要播放華語流行音樂。2010年8月星空傳媒把中国大陆頻道控股權易手予上海市人民政府牽頭成立的華人文化產業投資基金 [1] 。 目录 1 V的含義 2 Channel [V]國際頻道 3 Channel [V]臺灣頻道 4 Channel [V]中國頻道 5 其它的Channel [V]頻道 6 参见 7 參考資料 8 外部链接 V的含義 官方並無提供字母 V 的確切含義，以下是流傳的說法： 最廣泛的說法是V代表胜利（victory）...
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