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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 to 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
asked 1 hour ago
M.A.M.A.
545
$endgroup$
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 to 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
asked 1 hour ago
M.A.M.A.
545
$endgroup$
$begingroup$
to be more specificlist2should be sorted according to the first column oflist1
$endgroup$
– M.A.
1 hour 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 to 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
asked 1 hour ago
M.A.M.A.
545
$endgroup$
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 to 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
list-manipulation
asked 1 hour ago
M.A.M.A.
545
asked 1 hour ago
M.A.M.A.
545
asked 1 hour ago
M.A.M.A.
545
asked 1 hour ago
M.A.M.A.
545
asked 1 hour ago
M.A.M.A.
545
545
$begingroup$
to be more specificlist2should be sorted according to the first column oflist1
$endgroup$
– M.A.
1 hour ago
add a comment |
$begingroup$
to be more specificlist2should be sorted according to the first column oflist1
$endgroup$
– M.A.
1 hour ago
$begingroup$
to be more specific
list2should be sorted according to the first column of list1$endgroup$
– M.A.
1 hour ago
$begingroup$
to be more specific
list2should be sorted according to the first column of list1$endgroup$
– M.A.
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
idx = Lookup[
AssociationThread[list1[[All, 1]] -> Range[Length[list1]]],
list2[[All, 1]]
];
list2[[idx]]
A, 4, B, 5, C, 1
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
58.1k580160
$endgroup$
add a comment |
$begingroup$
Permute[list2, FindPermutation[ list1[[All,1]] , list2[[All,1]] ] ]
A, 4, B, 5, C, 1
answered 11 mins ago
MikeYMikeY
3,473714
$endgroup$
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
$begingroup$
idx = Lookup[
AssociationThread[list1[[All, 1]] -> Range[Length[list1]]],
list2[[All, 1]]
];
list2[[idx]]
A, 4, B, 5, C, 1
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
58.1k580160
$endgroup$
add a comment |
$begingroup$
idx = Lookup[
AssociationThread[list1[[All, 1]] -> Range[Length[list1]]],
list2[[All, 1]]
];
list2[[idx]]
A, 4, B, 5, C, 1
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
58.1k580160
$endgroup$
add a comment |
$begingroup$
idx = Lookup[
AssociationThread[list1[[All, 1]] -> Range[Length[list1]]],
list2[[All, 1]]
];
list2[[idx]]
A, 4, B, 5, C, 1
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
58.1k580160
$endgroup$
idx = Lookup[
AssociationThread[list1[[All, 1]] -> Range[Length[list1]]],
list2[[All, 1]]
];
list2[[idx]]
A, 4, B, 5, C, 1
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
58.1k580160
edited 1 hour ago
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
58.1k580160
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
58.1k580160
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
58.1k580160
58.1k580160
add a comment |
add a comment |
$begingroup$
Permute[list2, FindPermutation[ list1[[All,1]] , list2[[All,1]] ] ]
A, 4, B, 5, C, 1
answered 11 mins ago
MikeYMikeY
3,473714
$endgroup$
add a comment |
$begingroup$
Permute[list2, FindPermutation[ list1[[All,1]] , list2[[All,1]] ] ]
A, 4, B, 5, C, 1
answered 11 mins ago
MikeYMikeY
3,473714
$endgroup$
add a comment |
$begingroup$
Permute[list2, FindPermutation[ list1[[All,1]] , list2[[All,1]] ] ]
A, 4, B, 5, C, 1
answered 11 mins ago
MikeYMikeY
3,473714
$endgroup$
Permute[list2, FindPermutation[ list1[[All,1]] , list2[[All,1]] ] ]
A, 4, B, 5, C, 1
answered 11 mins ago
MikeYMikeY
3,473714
answered 11 mins ago
MikeYMikeY
3,473714
answered 11 mins ago
MikeYMikeY
3,473714
answered 11 mins ago
MikeYMikeY
3,473714
3,473714
add a comment |
add a comment |
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$begingroup$
to be more specific
list2should be sorted according to the first column oflist1$endgroup$
– M.A.
1 hour ago