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ls Ordering[Ordering[list]] optimal?

Sort+Union on a listEquivalence under common permutationsMax element of a list with a custom ordering functionReplacing members of a list with elements sampled without replacement from another listFinding pairs where the intersection of them is empty set from a nested listLength of Union Incorrect?Changing the order of elements in a listHow to efficiently implement Disjoint Set/Union Find data structure?Fast find sublist in a list procedure (for packed arrays of integers)Finding $n^text{th}$ largest/smallest element

4

1

$begingroup$

Given a list list with unique elements, the task is to replace each element by its position in Sort[list]. For example,

list = {"A", "B", "D", "C", "Z", "W"};

Position[Sort[list], #][[1, 1]] & /@ list

{1, 2, 4, 3, 6, 5}

Much more efficient is to call Ordering twice:

Ordering[Ordering[list]]

{1, 2, 4, 3, 6, 5}

When applied on a permutation of Range[length] this operation does nothing:

list = {2, 10, 1, 4, 8, 6, 3, 9, 5, 7};

Ordering[Ordering[list]]

{2, 10, 1, 4, 8, 6, 3, 9, 5, 7}

Question: is there a more efficient way of doing this operation, making a single function call instead of calling Ordering twice?

---

benchmarks

Solutions are given from fastest to slowest:

L = RandomReal[{0, 1}, 10^7];



(* Henrik Schumacher *)

R1[[Ordering[L]]] = R1 = Range[Length[L]]; //AbsoluteTiming//First

(* 2.45501 *)



(* orignal post *)

R2 = Ordering[Ordering[L]]; //AbsoluteTiming//First

(* 4.2129 *)



(* J.M. *)

R3 = PermutationList[InversePermutation[FindPermutation[L]]]; //AbsoluteTiming//First

(* 4.75192 *)



(* check *)

R1 == R2 == R3

(* True *)

list-manipulation performance-tuning sorting permutation

share|improve this question

edited 4 mins ago

Roman

asked 1 hour ago

RomanRoman

3,7951020

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried out PermutationList[FindPermutation[list]] or InversePermutation[Ordering[list]]?
  $endgroup$
  – J. M. is slightly pensive♦
  49 mins ago
  
  
  

  
  
  
  

add a comment | 

4

1

$begingroup$

Given a list list with unique elements, the task is to replace each element by its position in Sort[list]. For example,

list = {"A", "B", "D", "C", "Z", "W"};

Position[Sort[list], #][[1, 1]] & /@ list

{1, 2, 4, 3, 6, 5}

Much more efficient is to call Ordering twice:

Ordering[Ordering[list]]

{1, 2, 4, 3, 6, 5}

When applied on a permutation of Range[length] this operation does nothing:

list = {2, 10, 1, 4, 8, 6, 3, 9, 5, 7};

Ordering[Ordering[list]]

{2, 10, 1, 4, 8, 6, 3, 9, 5, 7}

Question: is there a more efficient way of doing this operation, making a single function call instead of calling Ordering twice?

---

benchmarks

Solutions are given from fastest to slowest:

L = RandomReal[{0, 1}, 10^7];



(* Henrik Schumacher *)

R1[[Ordering[L]]] = R1 = Range[Length[L]]; //AbsoluteTiming//First

(* 2.45501 *)



(* orignal post *)

R2 = Ordering[Ordering[L]]; //AbsoluteTiming//First

(* 4.2129 *)



(* J.M. *)

R3 = PermutationList[InversePermutation[FindPermutation[L]]]; //AbsoluteTiming//First

(* 4.75192 *)



(* check *)

R1 == R2 == R3

(* True *)

list-manipulation performance-tuning sorting permutation

share|improve this question

edited 4 mins ago

Roman

asked 1 hour ago

RomanRoman

3,7951020

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried out PermutationList[FindPermutation[list]] or InversePermutation[Ordering[list]]?
  $endgroup$
  – J. M. is slightly pensive♦
  49 mins ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Given a list list with unique elements, the task is to replace each element by its position in Sort[list]. For example,

list = {"A", "B", "D", "C", "Z", "W"};

Position[Sort[list], #][[1, 1]] & /@ list

{1, 2, 4, 3, 6, 5}

Much more efficient is to call Ordering twice:

Ordering[Ordering[list]]

{1, 2, 4, 3, 6, 5}

When applied on a permutation of Range[length] this operation does nothing:

list = {2, 10, 1, 4, 8, 6, 3, 9, 5, 7};

Ordering[Ordering[list]]

{2, 10, 1, 4, 8, 6, 3, 9, 5, 7}

Question: is there a more efficient way of doing this operation, making a single function call instead of calling Ordering twice?

---

benchmarks

Solutions are given from fastest to slowest:

L = RandomReal[{0, 1}, 10^7];



(* Henrik Schumacher *)

R1[[Ordering[L]]] = R1 = Range[Length[L]]; //AbsoluteTiming//First

(* 2.45501 *)



(* orignal post *)

R2 = Ordering[Ordering[L]]; //AbsoluteTiming//First

(* 4.2129 *)



(* J.M. *)

R3 = PermutationList[InversePermutation[FindPermutation[L]]]; //AbsoluteTiming//First

(* 4.75192 *)



(* check *)

R1 == R2 == R3

(* True *)

list-manipulation performance-tuning sorting permutation

share|improve this question

edited 4 mins ago

Roman

asked 1 hour ago

RomanRoman

3,7951020

$endgroup$

list = {"A", "B", "D", "C", "Z", "W"};

Position[Sort[list], #][[1, 1]] & /@ list

Ordering[Ordering[list]]

list = {2, 10, 1, 4, 8, 6, 3, 9, 5, 7};

Ordering[Ordering[list]]

L = RandomReal[{0, 1}, 10^7];



(* Henrik Schumacher *)

R1[[Ordering[L]]] = R1 = Range[Length[L]]; //AbsoluteTiming//First

(* 2.45501 *)



(* orignal post *)

R2 = Ordering[Ordering[L]]; //AbsoluteTiming//First

(* 4.2129 *)



(* J.M. *)

R3 = PermutationList[InversePermutation[FindPermutation[L]]]; //AbsoluteTiming//First

(* 4.75192 *)



(* check *)

R1 == R2 == R3

(* True *)

share|improve this question

edited 4 mins ago

Roman

asked 1 hour ago

RomanRoman

3,7951020

share|improve this question

edited 4 mins ago

Roman

asked 1 hour ago

RomanRoman

3,7951020

asked 1 hour ago

RomanRoman

3,7951020

asked 1 hour ago

RomanRoman

3,7951020

1 Answer
1

active

oldest

votes

3

$begingroup$

No, Ordering[Ordering[list]] not optimal. And yes, there is a faster method:

list = RandomReal[{-1, 1}, 1000000];



First@RepeatedTiming[

  a[[Ordering[list]]] = a = Range[Length[list]];

  ]



First@RepeatedTiming[

  b = Ordering[Ordering[list]]

  ]



a == b

0.13

0.236

True

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Henrik SchumacherHenrik Schumacher

58.1k580160

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This solution cuts the effort down from two function calls to one function call plus one indexed substitution (very fast). 90% of what I was looking for, thanks!
  $endgroup$
  – Roman
  2 mins ago
  

  
  
  
  

add a comment | 

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3

$begingroup$

No, Ordering[Ordering[list]] not optimal. And yes, there is a faster method:

list = RandomReal[{-1, 1}, 1000000];



First@RepeatedTiming[

  a[[Ordering[list]]] = a = Range[Length[list]];

  ]



First@RepeatedTiming[

  b = Ordering[Ordering[list]]

  ]



a == b

0.13

0.236

True

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Henrik SchumacherHenrik Schumacher

58.1k580160

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This solution cuts the effort down from two function calls to one function call plus one indexed substitution (very fast). 90% of what I was looking for, thanks!
  $endgroup$
  – Roman
  2 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

No, Ordering[Ordering[list]] not optimal. And yes, there is a faster method:

list = RandomReal[{-1, 1}, 1000000];



First@RepeatedTiming[

  a[[Ordering[list]]] = a = Range[Length[list]];

  ]



First@RepeatedTiming[

  b = Ordering[Ordering[list]]

  ]



a == b

0.13

0.236

True

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Henrik SchumacherHenrik Schumacher

58.1k580160

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This solution cuts the effort down from two function calls to one function call plus one indexed substitution (very fast). 90% of what I was looking for, thanks!
  $endgroup$
  – Roman
  2 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

No, Ordering[Ordering[list]] not optimal. And yes, there is a faster method:

list = RandomReal[{-1, 1}, 1000000];



First@RepeatedTiming[

  a[[Ordering[list]]] = a = Range[Length[list]];

  ]



First@RepeatedTiming[

  b = Ordering[Ordering[list]]

  ]



a == b

0.13

0.236

True

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Henrik SchumacherHenrik Schumacher

58.1k580160

$endgroup$

list = RandomReal[{-1, 1}, 1000000];



First@RepeatedTiming[

  a[[Ordering[list]]] = a = Range[Length[list]];

  ]



First@RepeatedTiming[

  b = Ordering[Ordering[list]]

  ]



a == b

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Henrik SchumacherHenrik Schumacher

58.1k580160

answered 1 hour ago

Henrik SchumacherHenrik Schumacher

58.1k580160

answered 1 hour ago

Henrik SchumacherHenrik Schumacher

58.1k580160