Extracting sublists that contain similar elementsPrepend 0 to sublistsComparing elements of the $n^textth$ sublist in a ragged list with the $n^textth$ member of a sequenceSelect sublists with second element nearest to zeroAdd elements of one list to sublists of another listSplit list into overlapping sublistsEliminating elements from sublists under a global conditionSort columns in a TableFormList manipulation with random elements under limiting conditionsHow to split sublists into sub-sublists without merging the sublistsProblems in dealing with sublists
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Extracting sublists that contain similar elements
Prepend 0 to sublistsComparing elements of the $n^textth$ sublist in a ragged list with the $n^textth$ member of a sequenceSelect sublists with second element nearest to zeroAdd elements of one list to sublists of another listSplit list into overlapping sublistsEliminating elements from sublists under a global conditionSort columns in a TableFormList manipulation with random elements under limiting conditionsHow to split sublists into sub-sublists without merging the sublistsProblems in dealing with sublists
$begingroup$
Let's say I have a list containing different sized sublists of string pairs structured like this
"A", "B", "C", "D", "E", "D", "G","B", "H", "B", "I", "B"
where the first and second sublists each contain two string pairs and the third and fourth sublists each contain 1 string pair.
I want to get rid of every string pair where the second string of the string pair does not appear in each sublist. For example, there exists string pairs in each sublist where the second string of the string pair is "B". Thus, I only want to keep those string pairs. The final list should look like this
"A", "B", "G","B", "H", "B", "I", "B"
Although each of the first two sublists contain a pair of strings where there is a "D" in the second element, these need to be removed since there aren't string pairs containing "D"'s in the second element of the string pairs in the third and fourth sublists.
I would like for this function to work no matter the number of sublists and no matter the number of elements contained in those sublists. I know how to do this if the list was completely flattened, but it is not in this case. There is no need for this to be an efficient process, so anything that can produce these results would be helpful. Any ideas?
list-manipulation
asked May 11 at 20:59
EricEric
411
$endgroup$
add a comment |
$begingroup$
Let's say I have a list containing different sized sublists of string pairs structured like this
"A", "B", "C", "D", "E", "D", "G","B", "H", "B", "I", "B"
where the first and second sublists each contain two string pairs and the third and fourth sublists each contain 1 string pair.
I want to get rid of every string pair where the second string of the string pair does not appear in each sublist. For example, there exists string pairs in each sublist where the second string of the string pair is "B". Thus, I only want to keep those string pairs. The final list should look like this
"A", "B", "G","B", "H", "B", "I", "B"
Although each of the first two sublists contain a pair of strings where there is a "D" in the second element, these need to be removed since there aren't string pairs containing "D"'s in the second element of the string pairs in the third and fourth sublists.
I would like for this function to work no matter the number of sublists and no matter the number of elements contained in those sublists. I know how to do this if the list was completely flattened, but it is not in this case. There is no need for this to be an efficient process, so anything that can produce these results would be helpful. Any ideas?
list-manipulation
asked May 11 at 20:59
EricEric
411
$endgroup$
add a comment |
$begingroup$
Let's say I have a list containing different sized sublists of string pairs structured like this
"A", "B", "C", "D", "E", "D", "G","B", "H", "B", "I", "B"
where the first and second sublists each contain two string pairs and the third and fourth sublists each contain 1 string pair.
I want to get rid of every string pair where the second string of the string pair does not appear in each sublist. For example, there exists string pairs in each sublist where the second string of the string pair is "B". Thus, I only want to keep those string pairs. The final list should look like this
"A", "B", "G","B", "H", "B", "I", "B"
Although each of the first two sublists contain a pair of strings where there is a "D" in the second element, these need to be removed since there aren't string pairs containing "D"'s in the second element of the string pairs in the third and fourth sublists.
I would like for this function to work no matter the number of sublists and no matter the number of elements contained in those sublists. I know how to do this if the list was completely flattened, but it is not in this case. There is no need for this to be an efficient process, so anything that can produce these results would be helpful. Any ideas?
list-manipulation
asked May 11 at 20:59
EricEric
411
$endgroup$
Let's say I have a list containing different sized sublists of string pairs structured like this
"A", "B", "C", "D", "E", "D", "G","B", "H", "B", "I", "B"
where the first and second sublists each contain two string pairs and the third and fourth sublists each contain 1 string pair.
I want to get rid of every string pair where the second string of the string pair does not appear in each sublist. For example, there exists string pairs in each sublist where the second string of the string pair is "B". Thus, I only want to keep those string pairs. The final list should look like this
"A", "B", "G","B", "H", "B", "I", "B"
Although each of the first two sublists contain a pair of strings where there is a "D" in the second element, these need to be removed since there aren't string pairs containing "D"'s in the second element of the string pairs in the third and fourth sublists.
I would like for this function to work no matter the number of sublists and no matter the number of elements contained in those sublists. I know how to do this if the list was completely flattened, but it is not in this case. There is no need for this to be an efficient process, so anything that can produce these results would be helpful. Any ideas?
list-manipulation
list-manipulation
asked May 11 at 20:59
EricEric
411
asked May 11 at 20:59
EricEric
411
asked May 11 at 20:59
EricEric
411
asked May 11 at 20:59
EricEric
411
asked May 11 at 20:59
EricEric
411
411
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
I assume that it's always the second element that determines sublist eligibility:
L = "A", "B", "C", "D", "E", "D", "G", "B", "H", "B", "I", "B";
Cases[_, Alternatives @@ Intersection @@ L[[All, All, 2]], ___] /@ L
"A", "B", "G", "B", "H", "B", "I", "B"
If instead it's always the last element, then the pattern would be
Cases[___, Alternatives @@ Intersection @@ L[[All, All, -1]]] /@ L
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 11 at 21:14
RomanRoman
9,17011440
$endgroup$
$begingroup$
Sorry, I should have been more clear. Both work in my case. Thank you very much!
$endgroup$
– Eric
May 11 at 21:29
add a comment |
$begingroup$
Using GroupBy and KeyIntersection:
ClearAll[f1]
f1 = Apply[Join]@*Values@*KeyIntersection@*Map[GroupBy[Last]];
lst = "A", "B", "C", "D", "E", "D", "G", "B", "H", "B", "I", "B";
f1 @ lst
"A", "B", "G", "B", "H", "B", "I", "B"
DeleteCases seems to be faster:
ClearAll[f2]
f2 = DeleteCases[#, Except@___, Alternatives @@ Intersection @@ #[[All, All, -1]], 2]&;
f2 @ lst
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 12 at 4:22
kglrkglr
194k10214435
$endgroup$
1
$begingroup$
By applying my method atCases-level 2 and then re-Listing, you lose information when multiple list members qualify: tryL = "A", "B", "C", "D", "X", "B", "E", "D", "G", "B", "H", "B", "I", "B", where your method takes the first list apart into"A", "B", "X", "B"instead of keeping its structure as"A", "B", "X", "B". Not sure what's the right behavior in this question though.
$endgroup$
– Roman
May 12 at 6:34
1
$begingroup$
@Roman, very good point. I think the result for your example should be"A", "B", "X", "B". I removedCases[... ,2]and replaced it withDeleteCases.
$endgroup$
– kglr
May 12 at 7:40
add a comment |
$begingroup$
list="A","B","C","D","E","D","G","B","H","B","I","B";
List/@Select[Flatten[list,1],MemberQ[Last@@@Select[list,Length@#==1&],#[[2]]]&]
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 11 at 21:08
J42161217J42161217
5,335525
$endgroup$
add a comment |
$begingroup$
a = "A", "B", "C", "D", "E", "D", "G", "B", "H","B", "I", "B"
f = Select[ContainsAny[Intersection @@ Union @@@ #]@*List@*Last] /@ # &;
f[a]
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 11 at 21:08
Henrik SchumacherHenrik Schumacher
63.2k587176
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I assume that it's always the second element that determines sublist eligibility:
L = "A", "B", "C", "D", "E", "D", "G", "B", "H", "B", "I", "B";
Cases[_, Alternatives @@ Intersection @@ L[[All, All, 2]], ___] /@ L
"A", "B", "G", "B", "H", "B", "I", "B"
If instead it's always the last element, then the pattern would be
Cases[___, Alternatives @@ Intersection @@ L[[All, All, -1]]] /@ L
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 11 at 21:14
RomanRoman
9,17011440
$endgroup$
$begingroup$
Sorry, I should have been more clear. Both work in my case. Thank you very much!
$endgroup$
– Eric
May 11 at 21:29
add a comment |
$begingroup$
I assume that it's always the second element that determines sublist eligibility:
L = "A", "B", "C", "D", "E", "D", "G", "B", "H", "B", "I", "B";
Cases[_, Alternatives @@ Intersection @@ L[[All, All, 2]], ___] /@ L
"A", "B", "G", "B", "H", "B", "I", "B"
If instead it's always the last element, then the pattern would be
Cases[___, Alternatives @@ Intersection @@ L[[All, All, -1]]] /@ L
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 11 at 21:14
RomanRoman
9,17011440
$endgroup$
$begingroup$
Sorry, I should have been more clear. Both work in my case. Thank you very much!
$endgroup$
– Eric
May 11 at 21:29
add a comment |
$begingroup$
I assume that it's always the second element that determines sublist eligibility:
L = "A", "B", "C", "D", "E", "D", "G", "B", "H", "B", "I", "B";
Cases[_, Alternatives @@ Intersection @@ L[[All, All, 2]], ___] /@ L
"A", "B", "G", "B", "H", "B", "I", "B"
If instead it's always the last element, then the pattern would be
Cases[___, Alternatives @@ Intersection @@ L[[All, All, -1]]] /@ L
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 11 at 21:14
RomanRoman
9,17011440
$endgroup$
I assume that it's always the second element that determines sublist eligibility:
L = "A", "B", "C", "D", "E", "D", "G", "B", "H", "B", "I", "B";
Cases[_, Alternatives @@ Intersection @@ L[[All, All, 2]], ___] /@ L
"A", "B", "G", "B", "H", "B", "I", "B"
If instead it's always the last element, then the pattern would be
Cases[___, Alternatives @@ Intersection @@ L[[All, All, -1]]] /@ L
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 11 at 21:14
RomanRoman
9,17011440
edited May 11 at 21:21
answered May 11 at 21:14
RomanRoman
9,17011440
answered May 11 at 21:14
RomanRoman
9,17011440
answered May 11 at 21:14
RomanRoman
9,17011440
9,17011440
$begingroup$
Sorry, I should have been more clear. Both work in my case. Thank you very much!
$endgroup$
– Eric
May 11 at 21:29
add a comment |
$begingroup$
Sorry, I should have been more clear. Both work in my case. Thank you very much!
$endgroup$
– Eric
May 11 at 21:29
$begingroup$
Sorry, I should have been more clear. Both work in my case. Thank you very much!
$endgroup$
– Eric
May 11 at 21:29
$begingroup$
Sorry, I should have been more clear. Both work in my case. Thank you very much!
$endgroup$
– Eric
May 11 at 21:29
add a comment |
$begingroup$
Using GroupBy and KeyIntersection:
ClearAll[f1]
f1 = Apply[Join]@*Values@*KeyIntersection@*Map[GroupBy[Last]];
lst = "A", "B", "C", "D", "E", "D", "G", "B", "H", "B", "I", "B";
f1 @ lst
"A", "B", "G", "B", "H", "B", "I", "B"
DeleteCases seems to be faster:
ClearAll[f2]
f2 = DeleteCases[#, Except@___, Alternatives @@ Intersection @@ #[[All, All, -1]], 2]&;
f2 @ lst
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 12 at 4:22
kglrkglr
194k10214435
$endgroup$
1
$begingroup$
By applying my method atCases-level 2 and then re-Listing, you lose information when multiple list members qualify: tryL = "A", "B", "C", "D", "X", "B", "E", "D", "G", "B", "H", "B", "I", "B", where your method takes the first list apart into"A", "B", "X", "B"instead of keeping its structure as"A", "B", "X", "B". Not sure what's the right behavior in this question though.
$endgroup$
– Roman
May 12 at 6:34
1
$begingroup$
@Roman, very good point. I think the result for your example should be"A", "B", "X", "B". I removedCases[... ,2]and replaced it withDeleteCases.
$endgroup$
– kglr
May 12 at 7:40
add a comment |
$begingroup$
Using GroupBy and KeyIntersection:
ClearAll[f1]
f1 = Apply[Join]@*Values@*KeyIntersection@*Map[GroupBy[Last]];
lst = "A", "B", "C", "D", "E", "D", "G", "B", "H", "B", "I", "B";
f1 @ lst
"A", "B", "G", "B", "H", "B", "I", "B"
DeleteCases seems to be faster:
ClearAll[f2]
f2 = DeleteCases[#, Except@___, Alternatives @@ Intersection @@ #[[All, All, -1]], 2]&;
f2 @ lst
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 12 at 4:22
kglrkglr
194k10214435
$endgroup$
1
$begingroup$
By applying my method atCases-level 2 and then re-Listing, you lose information when multiple list members qualify: tryL = "A", "B", "C", "D", "X", "B", "E", "D", "G", "B", "H", "B", "I", "B", where your method takes the first list apart into"A", "B", "X", "B"instead of keeping its structure as"A", "B", "X", "B". Not sure what's the right behavior in this question though.
$endgroup$
– Roman
May 12 at 6:34
1
$begingroup$
@Roman, very good point. I think the result for your example should be"A", "B", "X", "B". I removedCases[... ,2]and replaced it withDeleteCases.
$endgroup$
– kglr
May 12 at 7:40
add a comment |
$begingroup$
Using GroupBy and KeyIntersection:
ClearAll[f1]
f1 = Apply[Join]@*Values@*KeyIntersection@*Map[GroupBy[Last]];
lst = "A", "B", "C", "D", "E", "D", "G", "B", "H", "B", "I", "B";
f1 @ lst
"A", "B", "G", "B", "H", "B", "I", "B"
DeleteCases seems to be faster:
ClearAll[f2]
f2 = DeleteCases[#, Except@___, Alternatives @@ Intersection @@ #[[All, All, -1]], 2]&;
f2 @ lst
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 12 at 4:22
kglrkglr
194k10214435
$endgroup$
Using GroupBy and KeyIntersection:
ClearAll[f1]
f1 = Apply[Join]@*Values@*KeyIntersection@*Map[GroupBy[Last]];
lst = "A", "B", "C", "D", "E", "D", "G", "B", "H", "B", "I", "B";
f1 @ lst
"A", "B", "G", "B", "H", "B", "I", "B"
DeleteCases seems to be faster:
ClearAll[f2]
f2 = DeleteCases[#, Except@___, Alternatives @@ Intersection @@ #[[All, All, -1]], 2]&;
f2 @ lst
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 12 at 4:22
kglrkglr
194k10214435
edited May 12 at 7:38
answered May 12 at 4:22
kglrkglr
194k10214435
answered May 12 at 4:22
kglrkglr
194k10214435
answered May 12 at 4:22
kglrkglr
194k10214435
194k10214435
1
$begingroup$
By applying my method atCases-level 2 and then re-Listing, you lose information when multiple list members qualify: tryL = "A", "B", "C", "D", "X", "B", "E", "D", "G", "B", "H", "B", "I", "B", where your method takes the first list apart into"A", "B", "X", "B"instead of keeping its structure as"A", "B", "X", "B". Not sure what's the right behavior in this question though.
$endgroup$
– Roman
May 12 at 6:34
1
$begingroup$
@Roman, very good point. I think the result for your example should be"A", "B", "X", "B". I removedCases[... ,2]and replaced it withDeleteCases.
$endgroup$
– kglr
May 12 at 7:40
add a comment |
1
$begingroup$
By applying my method atCases-level 2 and then re-Listing, you lose information when multiple list members qualify: tryL = "A", "B", "C", "D", "X", "B", "E", "D", "G", "B", "H", "B", "I", "B", where your method takes the first list apart into"A", "B", "X", "B"instead of keeping its structure as"A", "B", "X", "B". Not sure what's the right behavior in this question though.
$endgroup$
– Roman
May 12 at 6:34
1
$begingroup$
@Roman, very good point. I think the result for your example should be"A", "B", "X", "B". I removedCases[... ,2]and replaced it withDeleteCases.
$endgroup$
– kglr
May 12 at 7:40
1
1
$begingroup$
By applying my method at
Cases-level 2 and then re-Listing, you lose information when multiple list members qualify: try L = "A", "B", "C", "D", "X", "B", "E", "D", "G", "B", "H", "B", "I", "B", where your method takes the first list apart into "A", "B", "X", "B" instead of keeping its structure as "A", "B", "X", "B". Not sure what's the right behavior in this question though.$endgroup$
– Roman
May 12 at 6:34
$begingroup$
By applying my method at
Cases-level 2 and then re-Listing, you lose information when multiple list members qualify: try L = "A", "B", "C", "D", "X", "B", "E", "D", "G", "B", "H", "B", "I", "B", where your method takes the first list apart into "A", "B", "X", "B" instead of keeping its structure as "A", "B", "X", "B". Not sure what's the right behavior in this question though.$endgroup$
– Roman
May 12 at 6:34
1
1
$begingroup$
@Roman, very good point. I think the result for your example should be
"A", "B", "X", "B". I removed Cases[... ,2] and replaced it with DeleteCases.$endgroup$
– kglr
May 12 at 7:40
$begingroup$
@Roman, very good point. I think the result for your example should be
"A", "B", "X", "B". I removed Cases[... ,2] and replaced it with DeleteCases.$endgroup$
– kglr
May 12 at 7:40
add a comment |
$begingroup$
list="A","B","C","D","E","D","G","B","H","B","I","B";
List/@Select[Flatten[list,1],MemberQ[Last@@@Select[list,Length@#==1&],#[[2]]]&]
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 11 at 21:08
J42161217J42161217
5,335525
$endgroup$
add a comment |
$begingroup$
list="A","B","C","D","E","D","G","B","H","B","I","B";
List/@Select[Flatten[list,1],MemberQ[Last@@@Select[list,Length@#==1&],#[[2]]]&]
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 11 at 21:08
J42161217J42161217
5,335525
$endgroup$
add a comment |
$begingroup$
list="A","B","C","D","E","D","G","B","H","B","I","B";
List/@Select[Flatten[list,1],MemberQ[Last@@@Select[list,Length@#==1&],#[[2]]]&]
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 11 at 21:08
J42161217J42161217
5,335525
$endgroup$
list="A","B","C","D","E","D","G","B","H","B","I","B";
List/@Select[Flatten[list,1],MemberQ[Last@@@Select[list,Length@#==1&],#[[2]]]&]
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 11 at 21:08
J42161217J42161217
5,335525
answered May 11 at 21:08
J42161217J42161217
5,335525
answered May 11 at 21:08
J42161217J42161217
5,335525
answered May 11 at 21:08
J42161217J42161217
5,335525
5,335525
add a comment |
add a comment |
$begingroup$
a = "A", "B", "C", "D", "E", "D", "G", "B", "H","B", "I", "B"
f = Select[ContainsAny[Intersection @@ Union @@@ #]@*List@*Last] /@ # &;
f[a]
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 11 at 21:08
Henrik SchumacherHenrik Schumacher
63.2k587176
$endgroup$
add a comment |
$begingroup$
a = "A", "B", "C", "D", "E", "D", "G", "B", "H","B", "I", "B"
f = Select[ContainsAny[Intersection @@ Union @@@ #]@*List@*Last] /@ # &;
f[a]
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 11 at 21:08
Henrik SchumacherHenrik Schumacher
63.2k587176
$endgroup$
add a comment |
$begingroup$
a = "A", "B", "C", "D", "E", "D", "G", "B", "H","B", "I", "B"
f = Select[ContainsAny[Intersection @@ Union @@@ #]@*List@*Last] /@ # &;
f[a]
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 11 at 21:08
Henrik SchumacherHenrik Schumacher
63.2k587176
$endgroup$
a = "A", "B", "C", "D", "E", "D", "G", "B", "H","B", "I", "B"
f = Select[ContainsAny[Intersection @@ Union @@@ #]@*List@*Last] /@ # &;
f[a]
"A", "B", "G", "B", "H", "B", "I", "B"
answered May 11 at 21:08
Henrik SchumacherHenrik Schumacher
63.2k587176
edited May 11 at 21:14
answered May 11 at 21:08
Henrik SchumacherHenrik Schumacher
63.2k587176
answered May 11 at 21:08
Henrik SchumacherHenrik Schumacher
63.2k587176
answered May 11 at 21:08
Henrik SchumacherHenrik Schumacher
63.2k587176
63.2k587176
add a comment |
add a comment |
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