Sort a list by elements of another list












8












$begingroup$


I know there are a plenty of other questions here which appear to be similar, however I did not found anything which could give me a hint.



I have two lists:



list1 = {{A, 12}, {B, 10}, {C, 4}}; (*ordered according to the second column*)

list2 = {{B, 5}, {A, 4}, {C, 1}};   (*ordered according to the second column*)


Now I want to sort list2according to the list1-order so the output should be:



(* {{A, 4}, {B, 5}, {C, 1}} *)





list-manipulation sorting







share|improve this question











$endgroup$












  • $begingroup$
    to be more specific list2should be sorted according to the first column of list1
    $endgroup$
    – M.A.
    Mar 27 at 18:58






  • 1




    $begingroup$
    Is there a better example to show what you want? Doesn't Sort[list2] give the desired output?
    $endgroup$
    – Jason B.
    Mar 27 at 23:24






  • 4




    $begingroup$
    list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...
    $endgroup$
    – Daniel Lichtblau
    Mar 27 at 23:37
















8












$begingroup$


I know there are a plenty of other questions here which appear to be similar, however I did not found anything which could give me a hint.



I have two lists:



list1 = {{A, 12}, {B, 10}, {C, 4}}; (*ordered according to the second column*)

list2 = {{B, 5}, {A, 4}, {C, 1}};   (*ordered according to the second column*)


Now I want to sort list2according to the list1-order so the output should be:



(* {{A, 4}, {B, 5}, {C, 1}} *)





list-manipulation sorting







share|improve this question











$endgroup$












  • $begingroup$
    to be more specific list2should be sorted according to the first column of list1
    $endgroup$
    – M.A.
    Mar 27 at 18:58






  • 1




    $begingroup$
    Is there a better example to show what you want? Doesn't Sort[list2] give the desired output?
    $endgroup$
    – Jason B.
    Mar 27 at 23:24






  • 4




    $begingroup$
    list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...
    $endgroup$
    – Daniel Lichtblau
    Mar 27 at 23:37














8












8








8


1



$begingroup$


I know there are a plenty of other questions here which appear to be similar, however I did not found anything which could give me a hint.



I have two lists:



list1 = {{A, 12}, {B, 10}, {C, 4}}; (*ordered according to the second column*)

list2 = {{B, 5}, {A, 4}, {C, 1}};   (*ordered according to the second column*)


Now I want to sort list2according to the list1-order so the output should be:



(* {{A, 4}, {B, 5}, {C, 1}} *)





list-manipulation sorting







share|improve this question











$endgroup$




I know there are a plenty of other questions here which appear to be similar, however I did not found anything which could give me a hint.



I have two lists:



list1 = {{A, 12}, {B, 10}, {C, 4}}; (*ordered according to the second column*)

list2 = {{B, 5}, {A, 4}, {C, 1}};   (*ordered according to the second column*)


Now I want to sort list2according to the list1-order so the output should be:



(* {{A, 4}, {B, 5}, {C, 1}} *)





list-manipulation sorting




list-manipulation sorting






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 27 at 20:35









MarcoB

38.6k557115




38.6k557115










asked Mar 27 at 18:56









M.A.M.A.

896




896












  • $begingroup$
    to be more specific list2should be sorted according to the first column of list1
    $endgroup$
    – M.A.
    Mar 27 at 18:58






  • 1




    $begingroup$
    Is there a better example to show what you want? Doesn't Sort[list2] give the desired output?
    $endgroup$
    – Jason B.
    Mar 27 at 23:24






  • 4




    $begingroup$
    list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...
    $endgroup$
    – Daniel Lichtblau
    Mar 27 at 23:37


















  • $begingroup$
    to be more specific list2should be sorted according to the first column of list1
    $endgroup$
    – M.A.
    Mar 27 at 18:58






  • 1




    $begingroup$
    Is there a better example to show what you want? Doesn't Sort[list2] give the desired output?
    $endgroup$
    – Jason B.
    Mar 27 at 23:24






  • 4




    $begingroup$
    list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...
    $endgroup$
    – Daniel Lichtblau
    Mar 27 at 23:37
















$begingroup$
to be more specific list2should be sorted according to the first column of list1
$endgroup$
– M.A.
Mar 27 at 18:58




$begingroup$
to be more specific list2should be sorted according to the first column of list1
$endgroup$
– M.A.
Mar 27 at 18:58




1




1




$begingroup$
Is there a better example to show what you want? Doesn't Sort[list2] give the desired output?
$endgroup$
– Jason B.
Mar 27 at 23:24




$begingroup$
Is there a better example to show what you want? Doesn't Sort[list2] give the desired output?
$endgroup$
– Jason B.
Mar 27 at 23:24




4




4




$begingroup$
list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...
$endgroup$
– Daniel Lichtblau
Mar 27 at 23:37




$begingroup$
list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...
$endgroup$
– Daniel Lichtblau
Mar 27 at 23:37










3 Answers
3






active

oldest

votes


















7












$begingroup$

Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]



{{A, 4}, {B, 5}, {C, 1}}







share|improve this answer











$endgroup$













  • $begingroup$
    Actually, I like your solution much better than mine. By the way, when I found out that my former solution was incorrect, I also realized that your solution should better be Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]].
    $endgroup$
    – Henrik Schumacher
    Mar 27 at 20:47






  • 1




    $begingroup$
    Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
    $endgroup$
    – MikeY
    Mar 27 at 21:04



















6












$begingroup$

list1 = {{A, 12}, {B, 10}, {C, 4}, {D, 2}};

list2 = {{A, 4}, {D, 11}, {B, 5}, {C, 1}};



idx = Lookup[

   AssociationThread[list1[[All, 1]] -> Range[Length[list1]]],

   list2[[All, 1]]

   ];

result = list2;

result[[idx]] = list2;

result



{{A, 4}, {B, 5}, {C, 1}, {D, 11}}







share|improve this answer











$endgroup$













  • $begingroup$
    works well with the example lists. However, something goes wrong when I use other lists with Strings in the first columns instead of A, Band C....
    $endgroup$
    – M.A.
    Mar 27 at 21:29



















5












$begingroup$

ugly but fast:



list2[[Ordering[list2[[All, 1]]][[Ordering[Ordering[list1[[All, 1]]]]]]]]



{{A, 4}, {B, 5}, {C, 1}}




even faster:



result = list2;

result[[Ordering[list1[[All, 1]]]]] = SortBy[list2, First];

result



{{A, 4}, {B, 5}, {C, 1}}




benchmarks



s = 10^7;

list1 = Transpose[{PermutationList@RandomPermutation[s],

                   RandomInteger[{0, 10}, s]}];

list2 = Transpose[{PermutationList@RandomPermutation[s],

                   RandomInteger[{0, 10}, s]}];



(* my first solution *)

result1 = list2[[Ordering[list2[[All, 1]]][[Ordering[Ordering[list1[[All, 1]]]]]]]]; //AbsoluteTiming//First

(* 8.6416 *)



(* my second solution *)

result2 = Module[{L},

  L = list2;

  L[[Ordering[list1[[All, 1]]]]] = SortBy[list2, First];

  L]; //AbsoluteTiming//First

(* 6.89593 *)



(* MikeY's solution *)

result3 = Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]]; //AbsoluteTiming//First

(* 15.808 *)



(* Henrik Schumacher's solution *)

result4 = Module[{idx, L},

  idx = Lookup[AssociationThread[list1[[All, 1]] -> Range[Length[list1]]], list2[[All, 1]]];

  L = list2;

  L[[idx]] = list2;

  L]; //AbsoluteTiming//First

(* 31.7412 *)



(* make sure all methods agree *)

result1 == result2 == result3 == result4

(* True *)





share|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the benchmark. I started down the Ordering road, but went for parsimony of expression. Mild bummer that it is at least twice as slow as the best method.
    $endgroup$
    – MikeY
    Mar 27 at 23:34












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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]



{{A, 4}, {B, 5}, {C, 1}}







share|improve this answer











$endgroup$













  • $begingroup$
    Actually, I like your solution much better than mine. By the way, when I found out that my former solution was incorrect, I also realized that your solution should better be Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]].
    $endgroup$
    – Henrik Schumacher
    Mar 27 at 20:47






  • 1




    $begingroup$
    Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
    $endgroup$
    – MikeY
    Mar 27 at 21:04
















7












$begingroup$

Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]



{{A, 4}, {B, 5}, {C, 1}}







share|improve this answer











$endgroup$













  • $begingroup$
    Actually, I like your solution much better than mine. By the way, when I found out that my former solution was incorrect, I also realized that your solution should better be Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]].
    $endgroup$
    – Henrik Schumacher
    Mar 27 at 20:47






  • 1




    $begingroup$
    Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
    $endgroup$
    – MikeY
    Mar 27 at 21:04














7












7








7





$begingroup$

Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]



{{A, 4}, {B, 5}, {C, 1}}







share|improve this answer











$endgroup$



Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]



{{A, 4}, {B, 5}, {C, 1}}








share|improve this answer














share|improve this answer



share|improve this answer








edited Mar 27 at 21:03

























answered Mar 27 at 20:21









MikeYMikeY

3,768916




3,768916












  • $begingroup$
    Actually, I like your solution much better than mine. By the way, when I found out that my former solution was incorrect, I also realized that your solution should better be Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]].
    $endgroup$
    – Henrik Schumacher
    Mar 27 at 20:47






  • 1




    $begingroup$
    Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
    $endgroup$
    – MikeY
    Mar 27 at 21:04


















  • $begingroup$
    Actually, I like your solution much better than mine. By the way, when I found out that my former solution was incorrect, I also realized that your solution should better be Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]].
    $endgroup$
    – Henrik Schumacher
    Mar 27 at 20:47






  • 1




    $begingroup$
    Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
    $endgroup$
    – MikeY
    Mar 27 at 21:04
















$begingroup$
Actually, I like your solution much better than mine. By the way, when I found out that my former solution was incorrect, I also realized that your solution should better be Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]].
$endgroup$
– Henrik Schumacher
Mar 27 at 20:47




$begingroup$
Actually, I like your solution much better than mine. By the way, when I found out that my former solution was incorrect, I also realized that your solution should better be Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]].
$endgroup$
– Henrik Schumacher
Mar 27 at 20:47




1




1




$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
Mar 27 at 21:04




$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
Mar 27 at 21:04











6












$begingroup$

list1 = {{A, 12}, {B, 10}, {C, 4}, {D, 2}};

list2 = {{A, 4}, {D, 11}, {B, 5}, {C, 1}};



idx = Lookup[

   AssociationThread[list1[[All, 1]] -> Range[Length[list1]]],

   list2[[All, 1]]

   ];

result = list2;

result[[idx]] = list2;

result



{{A, 4}, {B, 5}, {C, 1}, {D, 11}}







share|improve this answer











$endgroup$













  • $begingroup$
    works well with the example lists. However, something goes wrong when I use other lists with Strings in the first columns instead of A, Band C....
    $endgroup$
    – M.A.
    Mar 27 at 21:29
















6












$begingroup$

list1 = {{A, 12}, {B, 10}, {C, 4}, {D, 2}};

list2 = {{A, 4}, {D, 11}, {B, 5}, {C, 1}};



idx = Lookup[

   AssociationThread[list1[[All, 1]] -> Range[Length[list1]]],

   list2[[All, 1]]

   ];

result = list2;

result[[idx]] = list2;

result



{{A, 4}, {B, 5}, {C, 1}, {D, 11}}







share|improve this answer











$endgroup$













  • $begingroup$
    works well with the example lists. However, something goes wrong when I use other lists with Strings in the first columns instead of A, Band C....
    $endgroup$
    – M.A.
    Mar 27 at 21:29














6












6








6





$begingroup$

list1 = {{A, 12}, {B, 10}, {C, 4}, {D, 2}};

list2 = {{A, 4}, {D, 11}, {B, 5}, {C, 1}};



idx = Lookup[

   AssociationThread[list1[[All, 1]] -> Range[Length[list1]]],

   list2[[All, 1]]

   ];

result = list2;

result[[idx]] = list2;

result



{{A, 4}, {B, 5}, {C, 1}, {D, 11}}







share|improve this answer











$endgroup$



list1 = {{A, 12}, {B, 10}, {C, 4}, {D, 2}};

list2 = {{A, 4}, {D, 11}, {B, 5}, {C, 1}};



idx = Lookup[

   AssociationThread[list1[[All, 1]] -> Range[Length[list1]]],

   list2[[All, 1]]

   ];

result = list2;

result[[idx]] = list2;

result



{{A, 4}, {B, 5}, {C, 1}, {D, 11}}








share|improve this answer














share|improve this answer



share|improve this answer








edited Mar 27 at 20:44

























answered Mar 27 at 19:09









Henrik SchumacherHenrik Schumacher

59.4k582165




59.4k582165












  • $begingroup$
    works well with the example lists. However, something goes wrong when I use other lists with Strings in the first columns instead of A, Band C....
    $endgroup$
    – M.A.
    Mar 27 at 21:29


















  • $begingroup$
    works well with the example lists. However, something goes wrong when I use other lists with Strings in the first columns instead of A, Band C....
    $endgroup$
    – M.A.
    Mar 27 at 21:29
















$begingroup$
works well with the example lists. However, something goes wrong when I use other lists with Strings in the first columns instead of A, Band C....
$endgroup$
– M.A.
Mar 27 at 21:29




$begingroup$
works well with the example lists. However, something goes wrong when I use other lists with Strings in the first columns instead of A, Band C....
$endgroup$
– M.A.
Mar 27 at 21:29











5












$begingroup$

ugly but fast:



list2[[Ordering[list2[[All, 1]]][[Ordering[Ordering[list1[[All, 1]]]]]]]]



{{A, 4}, {B, 5}, {C, 1}}




even faster:



result = list2;

result[[Ordering[list1[[All, 1]]]]] = SortBy[list2, First];

result



{{A, 4}, {B, 5}, {C, 1}}




benchmarks



s = 10^7;

list1 = Transpose[{PermutationList@RandomPermutation[s],

                   RandomInteger[{0, 10}, s]}];

list2 = Transpose[{PermutationList@RandomPermutation[s],

                   RandomInteger[{0, 10}, s]}];



(* my first solution *)

result1 = list2[[Ordering[list2[[All, 1]]][[Ordering[Ordering[list1[[All, 1]]]]]]]]; //AbsoluteTiming//First

(* 8.6416 *)



(* my second solution *)

result2 = Module[{L},

  L = list2;

  L[[Ordering[list1[[All, 1]]]]] = SortBy[list2, First];

  L]; //AbsoluteTiming//First

(* 6.89593 *)



(* MikeY's solution *)

result3 = Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]]; //AbsoluteTiming//First

(* 15.808 *)



(* Henrik Schumacher's solution *)

result4 = Module[{idx, L},

  idx = Lookup[AssociationThread[list1[[All, 1]] -> Range[Length[list1]]], list2[[All, 1]]];

  L = list2;

  L[[idx]] = list2;

  L]; //AbsoluteTiming//First

(* 31.7412 *)



(* make sure all methods agree *)

result1 == result2 == result3 == result4

(* True *)





share|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the benchmark. I started down the Ordering road, but went for parsimony of expression. Mild bummer that it is at least twice as slow as the best method.
    $endgroup$
    – MikeY
    Mar 27 at 23:34
















5












$begingroup$

ugly but fast:



list2[[Ordering[list2[[All, 1]]][[Ordering[Ordering[list1[[All, 1]]]]]]]]



{{A, 4}, {B, 5}, {C, 1}}




even faster:



result = list2;

result[[Ordering[list1[[All, 1]]]]] = SortBy[list2, First];

result



{{A, 4}, {B, 5}, {C, 1}}




benchmarks



s = 10^7;

list1 = Transpose[{PermutationList@RandomPermutation[s],

                   RandomInteger[{0, 10}, s]}];

list2 = Transpose[{PermutationList@RandomPermutation[s],

                   RandomInteger[{0, 10}, s]}];



(* my first solution *)

result1 = list2[[Ordering[list2[[All, 1]]][[Ordering[Ordering[list1[[All, 1]]]]]]]]; //AbsoluteTiming//First

(* 8.6416 *)



(* my second solution *)

result2 = Module[{L},

  L = list2;

  L[[Ordering[list1[[All, 1]]]]] = SortBy[list2, First];

  L]; //AbsoluteTiming//First

(* 6.89593 *)



(* MikeY's solution *)

result3 = Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]]; //AbsoluteTiming//First

(* 15.808 *)



(* Henrik Schumacher's solution *)

result4 = Module[{idx, L},

  idx = Lookup[AssociationThread[list1[[All, 1]] -> Range[Length[list1]]], list2[[All, 1]]];

  L = list2;

  L[[idx]] = list2;

  L]; //AbsoluteTiming//First

(* 31.7412 *)



(* make sure all methods agree *)

result1 == result2 == result3 == result4

(* True *)





share|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the benchmark. I started down the Ordering road, but went for parsimony of expression. Mild bummer that it is at least twice as slow as the best method.
    $endgroup$
    – MikeY
    Mar 27 at 23:34














5












5








5





$begingroup$

ugly but fast:



list2[[Ordering[list2[[All, 1]]][[Ordering[Ordering[list1[[All, 1]]]]]]]]



{{A, 4}, {B, 5}, {C, 1}}




even faster:



result = list2;

result[[Ordering[list1[[All, 1]]]]] = SortBy[list2, First];

result



{{A, 4}, {B, 5}, {C, 1}}




benchmarks



s = 10^7;

list1 = Transpose[{PermutationList@RandomPermutation[s],

                   RandomInteger[{0, 10}, s]}];

list2 = Transpose[{PermutationList@RandomPermutation[s],

                   RandomInteger[{0, 10}, s]}];



(* my first solution *)

result1 = list2[[Ordering[list2[[All, 1]]][[Ordering[Ordering[list1[[All, 1]]]]]]]]; //AbsoluteTiming//First

(* 8.6416 *)



(* my second solution *)

result2 = Module[{L},

  L = list2;

  L[[Ordering[list1[[All, 1]]]]] = SortBy[list2, First];

  L]; //AbsoluteTiming//First

(* 6.89593 *)



(* MikeY's solution *)

result3 = Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]]; //AbsoluteTiming//First

(* 15.808 *)



(* Henrik Schumacher's solution *)

result4 = Module[{idx, L},

  idx = Lookup[AssociationThread[list1[[All, 1]] -> Range[Length[list1]]], list2[[All, 1]]];

  L = list2;

  L[[idx]] = list2;

  L]; //AbsoluteTiming//First

(* 31.7412 *)



(* make sure all methods agree *)

result1 == result2 == result3 == result4

(* True *)





share|improve this answer











$endgroup$



ugly but fast:



list2[[Ordering[list2[[All, 1]]][[Ordering[Ordering[list1[[All, 1]]]]]]]]



{{A, 4}, {B, 5}, {C, 1}}




even faster:



result = list2;

result[[Ordering[list1[[All, 1]]]]] = SortBy[list2, First];

result



{{A, 4}, {B, 5}, {C, 1}}




benchmarks



s = 10^7;

list1 = Transpose[{PermutationList@RandomPermutation[s],

                   RandomInteger[{0, 10}, s]}];

list2 = Transpose[{PermutationList@RandomPermutation[s],

                   RandomInteger[{0, 10}, s]}];



(* my first solution *)

result1 = list2[[Ordering[list2[[All, 1]]][[Ordering[Ordering[list1[[All, 1]]]]]]]]; //AbsoluteTiming//First

(* 8.6416 *)



(* my second solution *)

result2 = Module[{L},

  L = list2;

  L[[Ordering[list1[[All, 1]]]]] = SortBy[list2, First];

  L]; //AbsoluteTiming//First

(* 6.89593 *)



(* MikeY's solution *)

result3 = Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]]; //AbsoluteTiming//First

(* 15.808 *)



(* Henrik Schumacher's solution *)

result4 = Module[{idx, L},

  idx = Lookup[AssociationThread[list1[[All, 1]] -> Range[Length[list1]]], list2[[All, 1]]];

  L = list2;

  L[[idx]] = list2;

  L]; //AbsoluteTiming//First

(* 31.7412 *)



(* make sure all methods agree *)

result1 == result2 == result3 == result4

(* True *)






share|improve this answer














share|improve this answer



share|improve this answer








edited Mar 28 at 12:24

























answered Mar 27 at 21:18









RomanRoman

4,51511127




4,51511127












  • $begingroup$
    Thanks for the benchmark. I started down the Ordering road, but went for parsimony of expression. Mild bummer that it is at least twice as slow as the best method.
    $endgroup$
    – MikeY
    Mar 27 at 23:34


















  • $begingroup$
    Thanks for the benchmark. I started down the Ordering road, but went for parsimony of expression. Mild bummer that it is at least twice as slow as the best method.
    $endgroup$
    – MikeY
    Mar 27 at 23:34
















$begingroup$
Thanks for the benchmark. I started down the Ordering road, but went for parsimony of expression. Mild bummer that it is at least twice as slow as the best method.
$endgroup$
– MikeY
Mar 27 at 23:34




$begingroup$
Thanks for the benchmark. I started down the Ordering road, but went for parsimony of expression. Mild bummer that it is at least twice as slow as the best method.
$endgroup$
– MikeY
Mar 27 at 23:34


















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