Sort a list by elements of another list












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}} *)









share|improve this question











$endgroup$












  • $begingroup$
    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
















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}} *)









share|improve this question











$endgroup$












  • $begingroup$
    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














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}} *)









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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 27 at 20:35









MarcoB

38.6k557115




38.6k557115










asked Mar 27 at 18:56









M.A.M.A.

896




896












  • $begingroup$
    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


















  • $begingroup$
    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
















$begingroup$
to be more specific list2should be sorted according to the first column of list1
$endgroup$
– M.A.
Mar 27 at 18:58




$begingroup$
to be more specific list2should be sorted according to the first column of list1
$endgroup$
– M.A.
Mar 27 at 18:58




1




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




$begingroup$
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




4




$begingroup$
list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...
$endgroup$
– Daniel Lichtblau
Mar 27 at 23:37




$begingroup$
list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...
$endgroup$
– Daniel Lichtblau
Mar 27 at 23:37










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
    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



















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.
    Mar 27 at 21:29



















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
    Mar 27 at 23:34












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


















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















Required, but never shown

























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
    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
















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
    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














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 Mar 27 at 21:03

























answered Mar 27 at 20:21









MikeYMikeY

3,768916




3,768916












  • $begingroup$
    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


















  • $begingroup$
    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
















$begingroup$
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




$begingroup$
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




1




$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
Mar 27 at 21:04




$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
Mar 27 at 21:04











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.
    Mar 27 at 21:29
















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.
    Mar 27 at 21:29














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 Mar 27 at 20:44

























answered Mar 27 at 19:09









Henrik SchumacherHenrik Schumacher

59.4k582165




59.4k582165












  • $begingroup$
    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


















  • $begingroup$
    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
















$begingroup$
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




$begingroup$
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











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
    Mar 27 at 23:34
















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
    Mar 27 at 23:34














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 *)






share|improve this answer














share|improve this answer



share|improve this answer








edited Mar 28 at 12:24

























answered Mar 27 at 21:18









RomanRoman

4,51511127




4,51511127












  • $begingroup$
    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


















  • $begingroup$
    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
















$begingroup$
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




$begingroup$
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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematica Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














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















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Can I use Tabulator js library in my java Spring + Thymeleaf project?