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Sort a list by elements of another list



The Next CEO of Stack OverflowOrdered indices of a multiple product or sumImporting, sorting and exporting listsThe efficiency compare between Flatten[#, 1] & and Join @@ # &Lexicographic ordering of lists-of-lists?Problem with Custom Sort/Split/GatherApplying multiple functions to a single column in a tableList of (sub-)lists - query sub-lists by names?Find positions in which list elements are equalHow can I check if elements between lists are equal?comparing lists of strings










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.
    2 days ago






  • 1




    $begingroup$
    Is there a better example to show what you want? Doesn't Sort[list2] give the desired output?
    $endgroup$
    – Jason B.
    2 days ago






  • 4




    $begingroup$
    list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...
    $endgroup$
    – Daniel Lichtblau
    2 days ago















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.
    2 days ago






  • 1




    $begingroup$
    Is there a better example to show what you want? Doesn't Sort[list2] give the desired output?
    $endgroup$
    – Jason B.
    2 days ago






  • 4




    $begingroup$
    list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...
    $endgroup$
    – Daniel Lichtblau
    2 days ago













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 2 days ago









MarcoB

38.1k556114




38.1k556114










asked 2 days ago









M.A.M.A.

896




896











  • $begingroup$
    to be more specific list2should be sorted according to the first column of list1
    $endgroup$
    – M.A.
    2 days ago






  • 1




    $begingroup$
    Is there a better example to show what you want? Doesn't Sort[list2] give the desired output?
    $endgroup$
    – Jason B.
    2 days ago






  • 4




    $begingroup$
    list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...
    $endgroup$
    – Daniel Lichtblau
    2 days ago
















  • $begingroup$
    to be more specific list2should be sorted according to the first column of list1
    $endgroup$
    – M.A.
    2 days ago






  • 1




    $begingroup$
    Is there a better example to show what you want? Doesn't Sort[list2] give the desired output?
    $endgroup$
    – Jason B.
    2 days ago






  • 4




    $begingroup$
    list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...
    $endgroup$
    – Daniel Lichtblau
    2 days ago















$begingroup$
to be more specific list2should be sorted according to the first column of list1
$endgroup$
– M.A.
2 days ago




$begingroup$
to be more specific list2should be sorted according to the first column of list1
$endgroup$
– M.A.
2 days ago




1




1




$begingroup$
Is there a better example to show what you want? Doesn't Sort[list2] give the desired output?
$endgroup$
– Jason B.
2 days ago




$begingroup$
Is there a better example to show what you want? Doesn't Sort[list2] give the desired output?
$endgroup$
– Jason B.
2 days ago




4




4




$begingroup$
list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...
$endgroup$
– Daniel Lichtblau
2 days ago




$begingroup$
list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...
$endgroup$
– Daniel Lichtblau
2 days ago










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
    2 days ago






  • 1




    $begingroup$
    Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
    $endgroup$
    – MikeY
    2 days ago


















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.
    2 days ago


















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
    2 days ago











Your Answer





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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
    2 days ago






  • 1




    $begingroup$
    Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
    $endgroup$
    – MikeY
    2 days ago















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
    2 days ago






  • 1




    $begingroup$
    Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
    $endgroup$
    – MikeY
    2 days ago













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 2 days ago

























answered 2 days ago









MikeYMikeY

3,553714




3,553714











  • $begingroup$
    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
    2 days ago






  • 1




    $begingroup$
    Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
    $endgroup$
    – MikeY
    2 days ago
















  • $begingroup$
    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
    2 days ago






  • 1




    $begingroup$
    Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
    $endgroup$
    – MikeY
    2 days ago















$begingroup$
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
2 days ago




$begingroup$
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
2 days ago




1




1




$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
2 days ago




$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
2 days ago











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.
    2 days ago















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.
    2 days ago













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 2 days ago

























answered 2 days ago









Henrik SchumacherHenrik Schumacher

58.4k581161




58.4k581161











  • $begingroup$
    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.
    2 days ago
















  • $begingroup$
    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.
    2 days ago















$begingroup$
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.
2 days ago




$begingroup$
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.
2 days ago











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
    2 days ago















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
    2 days ago













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 *)






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edited yesterday

























answered 2 days ago









RomanRoman

3,9661022




3,9661022











  • $begingroup$
    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
    2 days ago
















  • $begingroup$
    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
    2 days ago















$begingroup$
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
2 days ago




$begingroup$
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
2 days ago

















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