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Sort a list by elements of another list
Ordered indices of a multiple product or sumThe efficiency compare between Flatten[#, 1] & and Join @@ # &Lexicographic ordering of lists-of-lists?How-to select an entry from a list of pairs that meets a condition depending the 2nd element of each pairList of (sub-)lists - query sub-lists by names?Can we intelligently control evaluation in Thread?Sort list but keeping the number of swaps requiredFind positions in which list elements are equalHow can I check if elements between lists are equal?comparing lists of strings
$begingroup$
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 37 mins ago
MarcoB
37.9k556114
asked 2 hours ago
M.A.M.A.
645
$endgroup$
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
edited 37 mins ago
MarcoB
37.9k556114
asked 2 hours ago
M.A.M.A.
645
$endgroup$
$begingroup$
to be more specificlist2should be sorted according to the first column oflist1
$endgroup$
– M.A.
2 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
edited 37 mins ago
MarcoB
37.9k556114
asked 2 hours ago
M.A.M.A.
645
$endgroup$
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 37 mins ago
MarcoB
37.9k556114
asked 2 hours ago
M.A.M.A.
645
edited 37 mins ago
MarcoB
37.9k556114
asked 2 hours ago
M.A.M.A.
645
edited 37 mins ago
MarcoB
37.9k556114
edited 37 mins ago
MarcoB
37.9k556114
edited 37 mins ago
MarcoB
37.9k556114
37.9k556114
asked 2 hours ago
M.A.M.A.
645
asked 2 hours ago
M.A.M.A.
645
asked 2 hours ago
M.A.M.A.
645
645
$begingroup$
to be more specificlist2should be sorted according to the first column oflist1
$endgroup$
– M.A.
2 hours ago
add a comment |
$begingroup$
to be more specificlist2should be sorted according to the first column oflist1
$endgroup$
– M.A.
2 hours ago
$begingroup$
to be more specific
list2should be sorted according to the first column of list1$endgroup$
– M.A.
2 hours ago
$begingroup$
to be more specific
list2should be sorted according to the first column of list1$endgroup$
– M.A.
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 2 hours ago
Henrik SchumacherHenrik Schumacher
58.1k580160
$endgroup$
add a comment |
$begingroup$
Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]
A, 4, B, 5, C, 1
answered 52 mins ago
MikeYMikeY
3,503714
$endgroup$
$begingroup$
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
25 mins ago
1
$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
8 mins ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 2 hours ago
Henrik SchumacherHenrik Schumacher
58.1k580160
$endgroup$
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
answered 2 hours ago
Henrik SchumacherHenrik Schumacher
58.1k580160
$endgroup$
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
answered 2 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
A, 4, B, 5, C, 1, D, 11
answered 2 hours ago
Henrik SchumacherHenrik Schumacher
58.1k580160
edited 28 mins ago
answered 2 hours ago
Henrik SchumacherHenrik Schumacher
58.1k580160
answered 2 hours ago
Henrik SchumacherHenrik Schumacher
58.1k580160
answered 2 hours ago
Henrik SchumacherHenrik Schumacher
58.1k580160
58.1k580160
add a comment |
add a comment |
$begingroup$
Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]
A, 4, B, 5, C, 1
answered 52 mins ago
MikeYMikeY
3,503714
$endgroup$
$begingroup$
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
25 mins ago
1
$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
8 mins ago
add a comment |
$begingroup$
Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]
A, 4, B, 5, C, 1
answered 52 mins ago
MikeYMikeY
3,503714
$endgroup$
$begingroup$
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
25 mins ago
1
$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
8 mins ago
add a comment |
$begingroup$
Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]
A, 4, B, 5, C, 1
answered 52 mins ago
MikeYMikeY
3,503714
$endgroup$
Permute[list2, FindPermutation[ list2[[All,1]] , list1[[All,1]] ] ]
A, 4, B, 5, C, 1
answered 52 mins ago
MikeYMikeY
3,503714
edited 9 mins ago
answered 52 mins ago
MikeYMikeY
3,503714
answered 52 mins ago
MikeYMikeY
3,503714
answered 52 mins ago
MikeYMikeY
3,503714
3,503714
$begingroup$
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
25 mins ago
1
$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
8 mins ago
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
25 mins ago
1
$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
8 mins ago
$begingroup$
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
25 mins ago
$begingroup$
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
25 mins ago
1
1
$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
8 mins ago
$begingroup$
Doh...fixed it. Both ways give the same answer, which leads to sloppy debugging.
$endgroup$
– MikeY
8 mins 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.
2 hours ago