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}} *)
list-manipulation sorting
edited Mar 27 at 20:35
MarcoB
38.6k557115
asked Mar 27 at 18:56
M.A.M.A.
896
$endgroup$
add a comment |
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}} *)
list-manipulation sorting
edited Mar 27 at 20:35
MarcoB
38.6k557115
asked Mar 27 at 18:56
M.A.M.A.
896
$endgroup$
$begingroup$
to be more specificlist2should be sorted according to the first column oflist1
$endgroup$
– M.A.
Mar 27 at 18:58
1
$begingroup$
Is there a better example to show what you want? Doesn'tSort[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
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}} *)
list-manipulation sorting
edited Mar 27 at 20:35
MarcoB
38.6k557115
asked Mar 27 at 18:56
M.A.M.A.
896
$endgroup$
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
list-manipulation sorting
edited Mar 27 at 20:35
MarcoB
38.6k557115
asked Mar 27 at 18:56
M.A.M.A.
896
edited Mar 27 at 20:35
MarcoB
38.6k557115
asked Mar 27 at 18:56
M.A.M.A.
896
edited Mar 27 at 20:35
MarcoB
38.6k557115
edited Mar 27 at 20:35
MarcoB
38.6k557115
edited Mar 27 at 20:35
MarcoB
38.6k557115
38.6k557115
asked Mar 27 at 18:56
M.A.M.A.
896
asked Mar 27 at 18:56
M.A.M.A.
896
asked Mar 27 at 18:56
M.A.M.A.
896
896
$begingroup$
to be more specificlist2should be sorted according to the first column oflist1
$endgroup$
– M.A.
Mar 27 at 18:58
1
$begingroup$
Is there a better example to show what you want? Doesn'tSort[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 |
$begingroup$
to be more specificlist2should be sorted according to the first column oflist1
$endgroup$
– M.A.
Mar 27 at 18:58
1
$begingroup$
Is there a better example to show what you want? Doesn'tSort[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
add a comment |
3 Answers
3
active
oldest
votes
7
$begingroup$
Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]
{{A, 4}, {B, 5}, {C, 1}}
answered Mar 27 at 20:21
MikeYMikeY
3,768916
$endgroup$
$begingroup$
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 bePermute[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}}
answered Mar 27 at 19:09
Henrik SchumacherHenrik Schumacher
59.4k582165
$endgroup$
$begingroup$
works well with the example lists. However, something goes wrong when I use other lists with Strings in the first columns instead ofA,BandC....
$endgroup$
– 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 *)
answered Mar 27 at 21:18
RomanRoman
4,51511127
$endgroup$
$begingroup$
Thanks for the benchmark. I started down theOrderingroad, 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 |
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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}}
answered Mar 27 at 20:21
MikeYMikeY
3,768916
$endgroup$
$begingroup$
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 bePermute[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}}
answered Mar 27 at 20:21
MikeYMikeY
3,768916
$endgroup$
$begingroup$
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 bePermute[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
7
7
$begingroup$
Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]
{{A, 4}, {B, 5}, {C, 1}}
answered Mar 27 at 20:21
MikeYMikeY
3,768916
$endgroup$
Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]
{{A, 4}, {B, 5}, {C, 1}}
answered Mar 27 at 20:21
MikeYMikeY
3,768916
edited Mar 27 at 21:03
answered Mar 27 at 20:21
MikeYMikeY
3,768916
answered Mar 27 at 20:21
MikeYMikeY
3,768916
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 bePermute[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 |
$begingroup$
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 bePermute[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
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}}
answered Mar 27 at 19:09
Henrik SchumacherHenrik Schumacher
59.4k582165
$endgroup$
$begingroup$
works well with the example lists. However, something goes wrong when I use other lists with Strings in the first columns instead ofA,BandC....
$endgroup$
– 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}}
answered Mar 27 at 19:09
Henrik SchumacherHenrik Schumacher
59.4k582165
$endgroup$
$begingroup$
works well with the example lists. However, something goes wrong when I use other lists with Strings in the first columns instead ofA,BandC....
$endgroup$
– M.A.
Mar 27 at 21:29
add a comment |
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}}
answered Mar 27 at 19:09
Henrik SchumacherHenrik Schumacher
59.4k582165
$endgroup$
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}}
answered Mar 27 at 19:09
Henrik SchumacherHenrik Schumacher
59.4k582165
edited Mar 27 at 20:44
answered Mar 27 at 19:09
Henrik SchumacherHenrik Schumacher
59.4k582165
answered Mar 27 at 19:09
Henrik SchumacherHenrik Schumacher
59.4k582165
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 ofA,BandC....
$endgroup$
– M.A.
Mar 27 at 21:29
add a comment |
$begingroup$
works well with the example lists. However, something goes wrong when I use other lists with Strings in the first columns instead ofA,BandC....
$endgroup$
– 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
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 *)
answered Mar 27 at 21:18
RomanRoman
4,51511127
$endgroup$
$begingroup$
Thanks for the benchmark. I started down theOrderingroad, 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 *)
answered Mar 27 at 21:18
RomanRoman
4,51511127
$endgroup$
$begingroup$
Thanks for the benchmark. I started down theOrderingroad, 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
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 *)
answered Mar 27 at 21:18
RomanRoman
4,51511127
$endgroup$
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 *)
answered Mar 27 at 21:18
RomanRoman
4,51511127
edited Mar 28 at 12:24
answered Mar 27 at 21:18
RomanRoman
4,51511127
answered Mar 27 at 21:18
RomanRoman
4,51511127
answered Mar 27 at 21:18
RomanRoman
4,51511127
4,51511127
$begingroup$
Thanks for the benchmark. I started down theOrderingroad, 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 |
$begingroup$
Thanks for the benchmark. I started down theOrderingroad, 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
add a comment |
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$begingroup$
to be more specific
list2should be sorted according to the first column oflist1$endgroup$
– 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