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ls Ordering[Ordering[list]] optimal?



The Next CEO of Stack OverflowOrdering problemHelp with PermutationsSort strings by natural orderingOrdering function with recognition of duplicatesNon-canonical re-ordering of nested listsAn efficient way to merge a pair of 4d arrays of non regular length listsMax element of a list with a custom ordering functionFinding pairs where the intersection of them is empty set from a nested listLexicographic ordering of lists-of-lists?Ordering elements of a list










9












$begingroup$


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











$endgroup$







  • 2




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
















9












$begingroup$


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











$endgroup$







  • 2




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














9












9








9


2



$begingroup$


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











$endgroup$




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




list-manipulation performance-tuning sorting permutation






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 10 hours ago







Roman

















asked 12 hours ago









RomanRoman

3,8501020




3,8501020







  • 2




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













  • 2




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








2




2




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





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











1 Answer
1






active

oldest

votes


















9












$begingroup$

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











$endgroup$












  • $begingroup$
    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
    10 hours ago










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    10 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









9












$begingroup$

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











$endgroup$












  • $begingroup$
    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
    10 hours ago










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    10 hours ago















9












$begingroup$

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











$endgroup$












  • $begingroup$
    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
    10 hours ago










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    10 hours ago













9












9








9





$begingroup$

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











$endgroup$



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














share|improve this answer



share|improve this answer








edited 11 hours ago

























answered 12 hours ago









Henrik SchumacherHenrik Schumacher

58.3k580160




58.3k580160











  • $begingroup$
    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
    10 hours ago










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    10 hours ago
















  • $begingroup$
    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
    10 hours ago










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    10 hours ago















$begingroup$
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
10 hours ago




$begingroup$
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
10 hours ago












$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
10 hours ago




$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
10 hours ago

















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