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Terse Method to Swap Lowest for Highest?

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Terse Method to Swap Lowest for Highest?

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

4

$begingroup$

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

share|improve this question

edited 26 mins ago

m_goldberg

87.7k872198

asked 1 hour ago

EdmundEdmund

26.6k330102

$endgroup$

add a comment | 

4

$begingroup$

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

share|improve this question

edited 26 mins ago

m_goldberg

87.7k872198

asked 1 hour ago

EdmundEdmund

26.6k330102

$endgroup$

add a comment | 

$begingroup$

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

share|improve this question

edited 26 mins ago

m_goldberg

87.7k872198

asked 1 hour ago

EdmundEdmund

26.6k330102

$endgroup$

SeedRandom[987]

test = RandomSample@*Join @@ Range @@@ {{6, 10}, {56, 60}, {1, 5}, {-5, -1}}

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]]

share|improve this question

edited 26 mins ago

m_goldberg

87.7k872198

asked 1 hour ago

EdmundEdmund

26.6k330102

share|improve this question

edited 26 mins ago

m_goldberg

87.7k872198

asked 1 hour ago

EdmundEdmund

26.6k330102

edited 26 mins ago

m_goldberg

87.7k872198

edited 26 mins ago

m_goldberg

87.7k872198

asked 1 hour ago

EdmundEdmund

26.6k330102

asked 1 hour ago

EdmundEdmund

26.6k330102

1 Answer
1

active

oldest

votes

7

$begingroup$

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

answered 49 mins ago

Carl WollCarl Woll

71.2k394185

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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7

$begingroup$

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

answered 49 mins ago

Carl WollCarl Woll

71.2k394185

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

7

$begingroup$

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

answered 49 mins ago

Carl WollCarl Woll

71.2k394185

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

answered 49 mins ago

Carl WollCarl Woll

71.2k394185

$endgroup$

Module[{tmp = test},

    With[{ord=Ordering[tmp]},

        tmp[[ord]] = Reverse @ tmp[[ord]]];

    tmp

]

share|improve this answer

answered 49 mins ago

Carl WollCarl Woll

71.2k394185

answered 49 mins ago

Carl WollCarl Woll

71.2k394185

answered 49 mins ago

Carl WollCarl Woll

71.2k394185