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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












$begingroup$


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











$endgroup$








  • 1




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


















4












$begingroup$


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











$endgroup$








  • 1




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
















4












4








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











$endgroup$




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 4 mins ago







Roman

















asked 1 hour ago









RomanRoman

3,7951020




3,7951020








  • 1




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
















  • 1




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










1




1




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






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












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











$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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











$endgroup$













  • $begingroup$
    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
















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











$endgroup$













  • $begingroup$
    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














3












3








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











$endgroup$



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 1 hour ago

























answered 1 hour ago









Henrik SchumacherHenrik Schumacher

58.1k580160




58.1k580160












  • $begingroup$
    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


















  • $begingroup$
    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
















$begingroup$
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




$begingroup$
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


















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