Styjun

Do sharpies or markers damage soft rock climbing gear?

Why is explainability not one of the criteria for publication?

Why is there not a willingness from the world to step in between Pakistan and India?

Why was this commercial plane highly delayed mid-flight?

Half filled water bottle

What is the meaning/usefulness of this exchange in Seinfeld's The Limo?

Will removing shelving screws from studs damage the studs?

Is the internet in Madagascar faster than in UK?

How to prevent a hosting company from accessing a VM's encryption keys?

Can I use coax outlets for cable modem?

Time difference between banns and marriage

Alternatives to Network Backup

Can I get a PhD for developing an educational software?

Can someone identify this unusual plane at airport?

Can I create something like a macro in Numbers?

What was the point of "Substance"?

Book featuring a child learning from a crowdsourced AI book

Can I take a boxed bicycle on a German train?

Is there any problem with a full installation on a USB drive?

How many petaflops does it take to land on the moon? What does Artemis need with an Aitken?

Why does the `ls` command sort files like this?

Group riding etiquette

Did the Apollo Guidance Computer really use 60% of the world's ICs in 1963?

Federal Pacific 200a main panel problem with oversized 100a 2pole breaker

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

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

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 Mar 27 at 20:35

MarcoB

39.1k55 gold badges5959 silver badges121121 bronze badges

asked Mar 27 at 18:56

M.A.M.A.

10377 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

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 Mar 27 at 20:35

MarcoB

39.1k55 gold badges5959 silver badges121121 bronze badges

asked Mar 27 at 18:56

M.A.M.A.

10377 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

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 Mar 27 at 20:35

MarcoB

39.1k55 gold badges5959 silver badges121121 bronze badges

asked Mar 27 at 18:56

M.A.M.A.

10377 bronze badges

$endgroup$

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 Mar 27 at 20:35

MarcoB

39.1k55 gold badges5959 silver badges121121 bronze badges

asked Mar 27 at 18:56

M.A.M.A.

10377 bronze badges

share|improve this question

edited Mar 27 at 20:35

MarcoB

39.1k55 gold badges5959 silver badges121121 bronze badges

asked Mar 27 at 18:56

M.A.M.A.

10377 bronze badges

edited Mar 27 at 20:35

MarcoB

39.1k55 gold badges5959 silver badges121121 bronze badges

edited Mar 27 at 20:35

MarcoB

39.1k55 gold badges5959 silver badges121121 bronze badges

asked Mar 27 at 18:56

M.A.M.A.

10377 bronze badges

asked Mar 27 at 18:56

M.A.M.A.

10377 bronze badges

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

edited Mar 27 at 21:03

answered Mar 27 at 20:21

MikeYMikeY

4,17899 silver badges1616 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

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 Mar 27 at 20:44

answered Mar 27 at 19:09

Henrik SchumacherHenrik Schumacher

67.8k55 gold badges9696 silver badges187187 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

edited Mar 28 at 12:24

answered Mar 27 at 21:18

RomanRoman

15.8k11 gold badge2121 silver badges5353 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194061%2fsort-a-list-by-elements-of-another-list%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

7

$begingroup$

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

A, 4, B, 5, C, 1

share|improve this answer

edited Mar 27 at 21:03

answered Mar 27 at 20:21

MikeYMikeY

4,17899 silver badges1616 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

7

$begingroup$

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

A, 4, B, 5, C, 1

share|improve this answer

edited Mar 27 at 21:03

answered Mar 27 at 20:21

MikeYMikeY

4,17899 silver badges1616 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

A, 4, B, 5, C, 1

share|improve this answer

edited Mar 27 at 21:03

answered Mar 27 at 20:21

MikeYMikeY

4,17899 silver badges1616 bronze badges

$endgroup$

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

share|improve this answer

edited Mar 27 at 21:03

answered Mar 27 at 20:21

MikeYMikeY

4,17899 silver badges1616 bronze badges

answered Mar 27 at 20:21

MikeYMikeY

4,17899 silver badges1616 bronze badges

answered Mar 27 at 20:21

MikeYMikeY

4,17899 silver badges1616 bronze badges

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 Mar 27 at 20:44

answered Mar 27 at 19:09

Henrik SchumacherHenrik Schumacher

67.8k55 gold badges9696 silver badges187187 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

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 Mar 27 at 20:44

answered Mar 27 at 19:09

Henrik SchumacherHenrik Schumacher

67.8k55 gold badges9696 silver badges187187 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

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 Mar 27 at 20:44

answered Mar 27 at 19:09

Henrik SchumacherHenrik Schumacher

67.8k55 gold badges9696 silver badges187187 bronze badges

$endgroup$

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 Mar 27 at 20:44

answered Mar 27 at 19:09

Henrik SchumacherHenrik Schumacher

67.8k55 gold badges9696 silver badges187187 bronze badges

answered Mar 27 at 19:09

Henrik SchumacherHenrik Schumacher

67.8k55 gold badges9696 silver badges187187 bronze badges

answered Mar 27 at 19:09

Henrik SchumacherHenrik Schumacher

67.8k55 gold badges9696 silver badges187187 bronze badges

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

edited Mar 28 at 12:24

answered Mar 27 at 21:18

RomanRoman

15.8k11 gold badge2121 silver badges5353 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

edited Mar 28 at 12:24

answered Mar 27 at 21:18

RomanRoman

15.8k11 gold badge2121 silver badges5353 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

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[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

edited Mar 28 at 12:24

answered Mar 27 at 21:18

RomanRoman

15.8k11 gold badge2121 silver badges5353 bronze badges

$endgroup$

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[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

edited Mar 28 at 12:24

answered Mar 27 at 21:18

RomanRoman

15.8k11 gold badge2121 silver badges5353 bronze badges

answered Mar 27 at 21:18

RomanRoman

15.8k11 gold badge2121 silver badges5353 bronze badges

answered Mar 27 at 21:18

RomanRoman

15.8k11 gold badge2121 silver badges5353 bronze badges