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Sort a list by elements of another list

Ordered 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

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

edited 7 hours ago

MarcoB

37.9k556114

asked 9 hours ago

M.A.M.A.

896

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

edited 7 hours ago

MarcoB

37.9k556114

asked 9 hours ago

M.A.M.A.

896

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

edited 7 hours ago

MarcoB

37.9k556114

asked 9 hours ago

M.A.M.A.

896

$endgroup$

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*)

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

share|improve this question

edited 7 hours ago

MarcoB

37.9k556114

asked 9 hours ago

M.A.M.A.

896

share|improve this question

edited 7 hours ago

MarcoB

37.9k556114

asked 9 hours ago

M.A.M.A.

896

edited 7 hours ago

MarcoB

37.9k556114

edited 7 hours ago

MarcoB

37.9k556114

asked 9 hours ago

M.A.M.A.

896

asked 9 hours ago

M.A.M.A.

896

3 Answers
3

active

oldest

votes

6

$begingroup$

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

A, 4, B, 5, C, 1

share|improve this answer

edited 7 hours ago

answered 7 hours ago

MikeYMikeY

3,538714

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

edited 7 hours ago

answered 9 hours ago

Henrik SchumacherHenrik Schumacher

58.1k580160

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  6 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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[PermutationReplace[Range[s], RandomPermutation[s]],
 RandomInteger[0, 10, s]];
list2 = Transpose[PermutationReplace[Range[s], 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

edited 4 hours ago

answered 6 hours ago

RomanRoman

3,7251020

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 hours ago
  

  
  
  
  

add a comment | 

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6

$begingroup$

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

A, 4, B, 5, C, 1

share|improve this answer

edited 7 hours ago

answered 7 hours ago

MikeYMikeY

3,538714

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

6

$begingroup$

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

A, 4, B, 5, C, 1

share|improve this answer

edited 7 hours ago

answered 7 hours ago

MikeYMikeY

3,538714

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

A, 4, B, 5, C, 1

share|improve this answer

edited 7 hours ago

answered 7 hours ago

MikeYMikeY

3,538714

$endgroup$

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

share|improve this answer

edited 7 hours ago

answered 7 hours ago

MikeYMikeY

3,538714

answered 7 hours ago

MikeYMikeY

3,538714

answered 7 hours ago

MikeYMikeY

3,538714

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

edited 7 hours ago

answered 9 hours ago

Henrik SchumacherHenrik Schumacher

58.1k580160

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  6 hours ago
  

  
  
  
  

add a comment | 

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

edited 7 hours ago

answered 9 hours ago

Henrik SchumacherHenrik Schumacher

58.1k580160

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 7 hours ago

answered 9 hours ago

Henrik SchumacherHenrik Schumacher

58.1k580160

$endgroup$

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

share|improve this answer

edited 7 hours ago

answered 9 hours ago

Henrik SchumacherHenrik Schumacher

58.1k580160

answered 9 hours ago

Henrik SchumacherHenrik Schumacher

58.1k580160

answered 9 hours ago

Henrik SchumacherHenrik Schumacher

58.1k580160

4

$begingroup$

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[PermutationReplace[Range[s], RandomPermutation[s]],
 RandomInteger[0, 10, s]];
list2 = Transpose[PermutationReplace[Range[s], 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

edited 4 hours ago

answered 6 hours ago

RomanRoman

3,7251020

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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[PermutationReplace[Range[s], RandomPermutation[s]],
 RandomInteger[0, 10, s]];
list2 = Transpose[PermutationReplace[Range[s], 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

edited 4 hours ago

answered 6 hours ago

RomanRoman

3,7251020

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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[PermutationReplace[Range[s], RandomPermutation[s]],
 RandomInteger[0, 10, s]];
list2 = Transpose[PermutationReplace[Range[s], 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

edited 4 hours ago

answered 6 hours ago

RomanRoman

3,7251020

$endgroup$

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

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

s = 10^7;
list1 = Transpose[PermutationReplace[Range[s], RandomPermutation[s]],
 RandomInteger[0, 10, s]];
list2 = Transpose[PermutationReplace[Range[s], 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

edited 4 hours ago

answered 6 hours ago

RomanRoman

3,7251020

answered 6 hours ago

RomanRoman

3,7251020

answered 6 hours ago

RomanRoman

3,7251020