Ggtkuk

Why isn't nylon as strong as kevlar?

Hyperlink on red background

Getting a W on your transcript for grad school applications

What is the most remote airport from the center of the city it supposedly serves?

Set collection doesn't always enforce uniqueness with the Date datatype? Does the following example seem correct?

How do I tell my manager that his code review comment is wrong?

Can hackers enable the camera after the user disabled it?

Using column size much larger than necessary

How can I close a gap between my fence and my neighbor's that's on his side of the property line?

How can I support myself financially as a 17 year old with a loan?

Missing Piece of Pie - Can you find it?

Why is "Vayechulu" said 3 times on Leil Shabbat?

How wide is a neg symbol, how to get the width for alignment?

How does this change to the opportunity attack rule impact combat?

How did Shepard's and Grissom's speeds compare with orbital velocity?

How can modem speed be 10 times slower than router?

How do I overfit?

What are the advantages of luxury car brands like Acura/Lexus over their sibling non-luxury brands Honda/Toyota?

Does a card have a keyword if it has the same effect as said keyword?

As matter approaches a black hole, does it speed up?

Why Isn’t SQL More Refactorable?

Why are prions in animal diets not destroyed by the digestive system?

In Avengers 1, why does Thanos need Loki?

What matters more when it comes to book covers? Is it ‘professional quality’ or relevancy?

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?

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

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193790%2fterse-method-to-swap-lowest-for-highest%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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