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Sort a list by elements of another list
The Next CEO of Stack OverflowOrdered 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
$begingroup$
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 yesterday
MarcoB
38k556114
asked yesterday
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 yesterday
MarcoB
38k556114
asked yesterday
M.A.M.A.
896
$endgroup$
$begingroup$
to be more specificlist2should be sorted according to the first column oflist1
$endgroup$
– M.A.
yesterday
1
$begingroup$
Is there a better example to show what you want? Doesn'tSort[list2]give the desired output?
$endgroup$
– Jason B.
22 hours ago
4
$begingroup$
list2[[OrderingBy[list1, -#[[2]] &]]]in the next release...
$endgroup$
– Daniel Lichtblau
22 hours ago
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 yesterday
MarcoB
38k556114
asked yesterday
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 yesterday
MarcoB
38k556114
asked yesterday
M.A.M.A.
896
edited yesterday
MarcoB
38k556114
asked yesterday
M.A.M.A.
896
edited yesterday
MarcoB
38k556114
edited yesterday
MarcoB
38k556114
edited yesterday
MarcoB
38k556114
38k556114
asked yesterday
M.A.M.A.
896
asked yesterday
M.A.M.A.
896
asked yesterday
M.A.M.A.
896
896
$begingroup$
to be more specificlist2should be sorted according to the first column oflist1
$endgroup$
– M.A.
yesterday
1
$begingroup$
Is there a better example to show what you want? Doesn'tSort[list2]give the desired output?
$endgroup$
– Jason B.
22 hours ago
4
$begingroup$
list2[[OrderingBy[list1, -#[[2]] &]]]in the next release...
$endgroup$
– Daniel Lichtblau
22 hours ago
add a comment |
$begingroup$
to be more specificlist2should be sorted according to the first column oflist1
$endgroup$
– M.A.
yesterday
1
$begingroup$
Is there a better example to show what you want? Doesn'tSort[list2]give the desired output?
$endgroup$
– Jason B.
22 hours ago
4
$begingroup$
list2[[OrderingBy[list1, -#[[2]] &]]]in the next release...
$endgroup$
– Daniel Lichtblau
22 hours ago
$begingroup$
to be more specific
list2should be sorted according to the first column of list1$endgroup$
– M.A.
yesterday
$begingroup$
to be more specific
list2should be sorted according to the first column of list1$endgroup$
– M.A.
yesterday
1
1
$begingroup$
Is there a better example to show what you want? Doesn't
Sort[list2] give the desired output?$endgroup$
– Jason B.
22 hours ago
$begingroup$
Is there a better example to show what you want? Doesn't
Sort[list2] give the desired output?$endgroup$
– Jason B.
22 hours ago
4
4
$begingroup$
list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...$endgroup$
– Daniel Lichtblau
22 hours ago
$begingroup$
list2[[OrderingBy[list1, -#[[2]] &]]] in the next release...$endgroup$
– Daniel Lichtblau
22 hours ago
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 yesterday
MikeYMikeY
3,548714
$endgroup$
$begingroup$
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
yesterday
1
$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
yesterday
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 yesterday
Henrik SchumacherHenrik Schumacher
58.3k580160
$endgroup$
$begingroup$
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.
yesterday
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 yesterday
RomanRoman
3,8501020
$endgroup$
$begingroup$
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
22 hours ago
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 yesterday
MikeYMikeY
3,548714
$endgroup$
$begingroup$
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
yesterday
1
$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
yesterday
add a comment |
7
$begingroup$
Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]
A, 4, B, 5, C, 1
answered yesterday
MikeYMikeY
3,548714
$endgroup$
$begingroup$
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
yesterday
1
$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
yesterday
add a comment |
7
7
7
$begingroup$
Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]
A, 4, B, 5, C, 1
answered yesterday
MikeYMikeY
3,548714
$endgroup$
Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]
A, 4, B, 5, C, 1
answered yesterday
MikeYMikeY
3,548714
edited yesterday
answered yesterday
MikeYMikeY
3,548714
answered yesterday
MikeYMikeY
3,548714
answered yesterday
MikeYMikeY
3,548714
3,548714
$begingroup$
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
yesterday
1
$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
yesterday
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
yesterday
1
$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
1
1
$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
yesterday
$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
yesterday
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 yesterday
Henrik SchumacherHenrik Schumacher
58.3k580160
$endgroup$
$begingroup$
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.
yesterday
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 yesterday
Henrik SchumacherHenrik Schumacher
58.3k580160
$endgroup$
$begingroup$
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.
yesterday
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 yesterday
Henrik SchumacherHenrik Schumacher
58.3k580160
$endgroup$
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 yesterday
Henrik SchumacherHenrik Schumacher
58.3k580160
edited yesterday
answered yesterday
Henrik SchumacherHenrik Schumacher
58.3k580160
answered yesterday
Henrik SchumacherHenrik Schumacher
58.3k580160
answered yesterday
Henrik SchumacherHenrik Schumacher
58.3k580160
58.3k580160
$begingroup$
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.
yesterday
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.
yesterday
$begingroup$
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.
yesterday
$begingroup$
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.
yesterday
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 yesterday
RomanRoman
3,8501020
$endgroup$
$begingroup$
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
22 hours ago
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 yesterday
RomanRoman
3,8501020
$endgroup$
$begingroup$
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
22 hours ago
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 yesterday
RomanRoman
3,8501020
$endgroup$
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 yesterday
RomanRoman
3,8501020
edited 9 hours ago
answered yesterday
RomanRoman
3,8501020
answered yesterday
RomanRoman
3,8501020
answered yesterday
RomanRoman
3,8501020
3,8501020
$begingroup$
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
22 hours ago
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
22 hours ago
$begingroup$
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
22 hours ago
$begingroup$
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
22 hours ago
add a comment |
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$begingroup$
to be more specific
list2should be sorted according to the first column oflist1$endgroup$
– M.A.
yesterday
1
$begingroup$
Is there a better example to show what you want? Doesn't
Sort[list2]give the desired output?$endgroup$
– Jason B.
22 hours ago
4
$begingroup$
list2[[OrderingBy[list1, -#[[2]] &]]]in the next release...$endgroup$
– Daniel Lichtblau
22 hours ago