Mdtryhht

You cannot touch me, but I can touch you, who am I?

How did Doctor Strange see the winning outcome in Avengers: Infinity War?

Are student evaluations of teaching assistants read by others in the faculty?

Is there a good way to store credentials outside of a password manager?

How do scammers retract money, while you can’t?

What is the intuitive meaning of having a linear relationship between the logs of two variables?

Why not increase contact surface when reentering the atmosphere?

Unexpected indention in bibliography items (beamer)

Applicability of Single Responsibility Principle

Did Dumbledore lie to Harry about how long he had James Potter's invisibility cloak when he was examining it? If so, why?

How to draw lines on a tikz-cd diagram

What is the opposite of 'gravitas'?

Closest Prime Number

Italian words for tools

Is expanding the research of a group into machine learning as a PhD student risky?

Why, precisely, is argon used in neutrino experiments?

Failed to fetch jessie backports repository

How to write papers efficiently when English isn't my first language?

How can I kill an app using Terminal?

Why escape if the_content isnt?

Why Were Madagascar and New Zealand Discovered So Late?

How do I go from 300 unfinished/half written blog posts, to published posts?

CREATE opcode: what does it really do?

Why are there no referendums in the US?

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 1 hour ago

MarcoB

37.9k556114

asked 3 hours ago

M.A.M.A.

815

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  to be more specific list2should be sorted according to the first column of list1
  $endgroup$
  – M.A.
  3 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 1 hour ago

MarcoB

37.9k556114

asked 3 hours ago

M.A.M.A.

815

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  to be more specific list2should be sorted according to the first column of list1
  $endgroup$
  – M.A.
  3 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 1 hour ago

MarcoB

37.9k556114

asked 3 hours ago

M.A.M.A.

815

$endgroup$

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 1 hour ago

MarcoB

37.9k556114

asked 3 hours ago

M.A.M.A.

815

share|improve this question

edited 1 hour ago

MarcoB

37.9k556114

asked 3 hours ago

M.A.M.A.

815

edited 1 hour ago

MarcoB

37.9k556114

edited 1 hour ago

MarcoB

37.9k556114

asked 3 hours ago

M.A.M.A.

815

asked 3 hours ago

M.A.M.A.

815

3 Answers
3

active

oldest

votes

5

$begingroup$

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

A, 4, B, 5, C, 1

share|improve this answer

edited 1 hour ago

answered 1 hour ago

MikeYMikeY

3,528714

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
  $endgroup$
  – MikeY
  1 hour 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 1 hour ago

answered 3 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.
  51 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

ugly but fast:

list2[[Ordering[list2][[Ordering[Ordering[list1]]]]]]

A, 4, B, 5, C, 1

even faster:

result = list2;
result[[Ordering[list1]]] = Sort[list2];
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][[Ordering[Ordering[list1]]]]]]; //AbsoluteTiming//First
(* 10.5831 *)

(* my second solution *)
result2 = Module[L,
 L = list2;
 L[[Ordering[list1]]] = Sort[list2];
 L]; //AbsoluteTiming//First
(* 8.45556 *)

(* MikeY's solution *)
result3 = Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]]; //AbsoluteTiming//First
(* 16.273 *)

(* 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
(* 32.0212 *)

(* make sure all methods agree *)
result1 == result2 == result3 == result4
(* True *)

share|improve this answer

edited 21 mins ago

answered 1 hour ago

RomanRoman

3,6951020

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

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

5

$begingroup$

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

A, 4, B, 5, C, 1

share|improve this answer

edited 1 hour ago

answered 1 hour ago

MikeYMikeY

3,528714

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

5

$begingroup$

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

A, 4, B, 5, C, 1

share|improve this answer

edited 1 hour ago

answered 1 hour ago

MikeYMikeY

3,528714

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
  $endgroup$
  – MikeY
  1 hour 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 1 hour ago

answered 1 hour ago

MikeYMikeY

3,528714

$endgroup$

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

share|improve this answer

edited 1 hour ago

answered 1 hour ago

MikeYMikeY

3,528714

answered 1 hour ago

MikeYMikeY

3,528714

answered 1 hour ago

MikeYMikeY

3,528714

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 1 hour ago

answered 3 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.
  51 mins 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 1 hour ago

answered 3 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.
  51 mins 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 1 hour ago

answered 3 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 1 hour ago

answered 3 hours ago

Henrik SchumacherHenrik Schumacher

58.1k580160

answered 3 hours ago

Henrik SchumacherHenrik Schumacher

58.1k580160

answered 3 hours ago

Henrik SchumacherHenrik Schumacher

58.1k580160

1

$begingroup$

ugly but fast:

list2[[Ordering[list2][[Ordering[Ordering[list1]]]]]]

A, 4, B, 5, C, 1

even faster:

result = list2;
result[[Ordering[list1]]] = Sort[list2];
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][[Ordering[Ordering[list1]]]]]]; //AbsoluteTiming//First
(* 10.5831 *)

(* my second solution *)
result2 = Module[L,
 L = list2;
 L[[Ordering[list1]]] = Sort[list2];
 L]; //AbsoluteTiming//First
(* 8.45556 *)

(* MikeY's solution *)
result3 = Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]]; //AbsoluteTiming//First
(* 16.273 *)

(* 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
(* 32.0212 *)

(* make sure all methods agree *)
result1 == result2 == result3 == result4
(* True *)

share|improve this answer

edited 21 mins ago

answered 1 hour ago

RomanRoman

3,6951020

$endgroup$

add a comment | 

1

$begingroup$

ugly but fast:

list2[[Ordering[list2][[Ordering[Ordering[list1]]]]]]

A, 4, B, 5, C, 1

even faster:

result = list2;
result[[Ordering[list1]]] = Sort[list2];
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][[Ordering[Ordering[list1]]]]]]; //AbsoluteTiming//First
(* 10.5831 *)

(* my second solution *)
result2 = Module[L,
 L = list2;
 L[[Ordering[list1]]] = Sort[list2];
 L]; //AbsoluteTiming//First
(* 8.45556 *)

(* MikeY's solution *)
result3 = Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]]; //AbsoluteTiming//First
(* 16.273 *)

(* 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
(* 32.0212 *)

(* make sure all methods agree *)
result1 == result2 == result3 == result4
(* True *)

share|improve this answer

edited 21 mins ago

answered 1 hour ago

RomanRoman

3,6951020

$endgroup$

add a comment | 

$begingroup$

ugly but fast:

list2[[Ordering[list2][[Ordering[Ordering[list1]]]]]]

A, 4, B, 5, C, 1

even faster:

result = list2;
result[[Ordering[list1]]] = Sort[list2];
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][[Ordering[Ordering[list1]]]]]]; //AbsoluteTiming//First
(* 10.5831 *)

(* my second solution *)
result2 = Module[L,
 L = list2;
 L[[Ordering[list1]]] = Sort[list2];
 L]; //AbsoluteTiming//First
(* 8.45556 *)

(* MikeY's solution *)
result3 = Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]]; //AbsoluteTiming//First
(* 16.273 *)

(* 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
(* 32.0212 *)

(* make sure all methods agree *)
result1 == result2 == result3 == result4
(* True *)

share|improve this answer

edited 21 mins ago

answered 1 hour ago

RomanRoman

3,6951020

$endgroup$

list2[[Ordering[list2][[Ordering[Ordering[list1]]]]]]

result = list2;
result[[Ordering[list1]]] = Sort[list2];
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][[Ordering[Ordering[list1]]]]]]; //AbsoluteTiming//First
(* 10.5831 *)

(* my second solution *)
result2 = Module[L,
 L = list2;
 L[[Ordering[list1]]] = Sort[list2];
 L]; //AbsoluteTiming//First
(* 8.45556 *)

(* MikeY's solution *)
result3 = Permute[list2, FindPermutation[list2[[All, 1]], list1[[All, 1]]]]; //AbsoluteTiming//First
(* 16.273 *)

(* 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
(* 32.0212 *)

(* make sure all methods agree *)
result1 == result2 == result3 == result4
(* True *)

share|improve this answer

edited 21 mins ago

answered 1 hour ago

RomanRoman

3,6951020

answered 1 hour ago

RomanRoman

3,6951020

answered 1 hour ago

RomanRoman

3,6951020