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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}} *)
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  • $\begingroup$ to be more specific list2should be sorted according to the first column of list1 $\endgroup$
    – M.A.
    Commented Mar 27, 2019 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.
    Commented Mar 27, 2019 at 23:24
  • 6
    $\begingroup$ list2[[OrderingBy[list1, -#[[2]] &]]] in the next release... $\endgroup$
    – Daniel Lichtblau
    Commented Mar 27, 2019 at 23:37

5 Answers 5

Reset to default
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Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]

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

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2
  • $\begingroup$ 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
    Commented Mar 27, 2019 at 20:47
  • 1
    $\begingroup$ Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging. $\endgroup$
    – MikeY
    Commented Mar 27, 2019 at 21:04
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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}}

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1
  • $\begingroup$ 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.
    Commented Mar 27, 2019 at 21:29
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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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1
  • $\begingroup$ 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
    Commented Mar 27, 2019 at 23:34
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p = {{a, 12}, {b, 10}, {c, 4}};

q = {{b, 5}, {a, 4}, {c, 1}};

n = Length[p];

Using SubsetMap (new in 12.0)

SubsetMap[Last /@ Sort[q] &, p, Table[{i, 2}, {i, n}]]

{{a, 4}, {b, 5}, {c, 1}}

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2
$\begingroup$ 
l1 = {{a, 12}, {b, 10}, {c, 4}};

l2 = {{b, 5}, {a, 4}, {c, 1}};

Using OrderingBy and RotateLeft:

l2[[RotateLeft@OrderingBy[l1, Last]]]

(*{{a, 4}, {b, 5}, {c, 1}}*)
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