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Terse Method to Swap Lowest for Highest?

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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

edited Mar 23 at 2:15

J. M. is away♦

99k10311470

asked Mar 22 at 20:57

EdmundEdmund

26.9k330103

$endgroup$

add a comment | 

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

edited Mar 23 at 2:15

J. M. is away♦

99k10311470

asked Mar 22 at 20:57

EdmundEdmund

26.9k330103

$endgroup$

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 permutation

share|improve this question

edited Mar 23 at 2:15

J. M. is away♦

99k10311470

asked Mar 22 at 20:57

EdmundEdmund

26.9k330103

$endgroup$

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 Mar 23 at 2:15

J. M. is away♦

99k10311470

asked Mar 22 at 20:57

EdmundEdmund

26.9k330103

share|improve this question

edited Mar 23 at 2:15

J. M. is away♦

99k10311470

asked Mar 22 at 20:57

EdmundEdmund

26.9k330103

edited Mar 23 at 2:15

J. M. is away♦

99k10311470

edited Mar 23 at 2:15

J. M. is away♦

99k10311470

asked Mar 22 at 20:57

EdmundEdmund

26.9k330103

asked Mar 22 at 20:57

EdmundEdmund

26.9k330103

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

answered Mar 22 at 21:11

Carl WollCarl Woll

76.8k3101201

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That is so obvious I want to cry. Thanks (+1).
  $endgroup$
  – Edmund
  Mar 22 at 21:15
  

  
  
  
  

add a comment | 

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

edited Mar 23 at 2:27

answered Mar 23 at 2:14

J. M. is away♦J. M. is away

99k10311470

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Christopher Lamb
  Mar 23 at 16:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  At least on my machine, the questioner's original (very verbose) proposed solution is faster than any other proposal. In order of presentation: 0.0000114366 v. 0.0000357762 v. 0.0000164219 (AbsoluteTiming in seconds).
  $endgroup$
  – CElliott
  Apr 3 at 14:10
  
  

  
  
  
  

 | 
show 3 more comments

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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

answered Mar 22 at 21:11

Carl WollCarl Woll

76.8k3101201

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That is so obvious I want to cry. Thanks (+1).
  $endgroup$
  – Edmund
  Mar 22 at 21:15
  

  
  
  
  

add a comment | 

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

answered Mar 22 at 21:11

Carl WollCarl Woll

76.8k3101201

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That is so obvious I want to cry. Thanks (+1).
  $endgroup$
  – Edmund
  Mar 22 at 21:15
  

  
  
  
  

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 Mar 22 at 21:11

Carl WollCarl Woll

76.8k3101201

$endgroup$

Module[tmp = test,
 With[ord=Ordering[tmp],
 tmp[[ord]] = Reverse @ tmp[[ord]]];
 tmp
]

share|improve this answer

answered Mar 22 at 21:11

Carl WollCarl Woll

76.8k3101201

answered Mar 22 at 21:11

Carl WollCarl Woll

76.8k3101201

answered Mar 22 at 21:11

Carl WollCarl Woll

76.8k3101201

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

edited Mar 23 at 2:27

answered Mar 23 at 2:14

J. M. is away♦J. M. is away

99k10311470

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Christopher Lamb
  Mar 23 at 16:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  At least on my machine, the questioner's original (very verbose) proposed solution is faster than any other proposal. In order of presentation: 0.0000114366 v. 0.0000357762 v. 0.0000164219 (AbsoluteTiming in seconds).
  $endgroup$
  – CElliott
  Apr 3 at 14:10
  
  

  
  
  
  

 | 
show 3 more comments

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

edited Mar 23 at 2:27

answered Mar 23 at 2:14

J. M. is away♦J. M. is away

99k10311470

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Christopher Lamb
  Mar 23 at 16:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  At least on my machine, the questioner's original (very verbose) proposed solution is faster than any other proposal. In order of presentation: 0.0000114366 v. 0.0000357762 v. 0.0000164219 (AbsoluteTiming in seconds).
  $endgroup$
  – CElliott
  Apr 3 at 14:10
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

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

edited Mar 23 at 2:27

answered Mar 23 at 2:14

J. M. is away♦J. M. is away

99k10311470

$endgroup$

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

share|improve this answer

edited Mar 23 at 2:27

answered Mar 23 at 2:14

J. M. is away♦J. M. is away

99k10311470

answered Mar 23 at 2:14

J. M. is away♦J. M. is away

99k10311470

answered Mar 23 at 2:14

J. M. is away♦J. M. is away

99k10311470