selective deletions from list
I have a list:
lis = {{"b","x","d"},{"a","z","b"},{"a","x","b"},{"a","x","c"},{"b","z","d"}}
Certain consecutive elements of this list will have identical first and last members (in this example, "a" and "b" in lis[[2]] and lis[[3]]) and I would like to delete the element that has "x" as its middle element, to give:
res = {{"b","x","d"},{"a","z","b"},{"a","x","c"},{"b","z","d"}}
It seems like a job for SequenceCases, but am having no luck.
list-manipulation
asked Dec 11 at 23:30
Suite401
981312
add a comment |
I have a list:
lis = {{"b","x","d"},{"a","z","b"},{"a","x","b"},{"a","x","c"},{"b","z","d"}}
Certain consecutive elements of this list will have identical first and last members (in this example, "a" and "b" in lis[[2]] and lis[[3]]) and I would like to delete the element that has "x" as its middle element, to give:
res = {{"b","x","d"},{"a","z","b"},{"a","x","c"},{"b","z","d"}}
It seems like a job for SequenceCases, but am having no luck.
list-manipulation
asked Dec 11 at 23:30
Suite401
981312
2
elements 1 and 5 also have identical first and last members...Shouldn't you erase the first element, too?
– J42161217
Dec 11 at 23:35
Take a look atDeleteDuplicatesBy.
– Kuba♦
Dec 11 at 23:47
Another approach could be to start withSplit[lis, #1[[1]] == #2[[1]] && #1[[3]] == #2[[3]] &]to group the consecutive elements.
– That Gravity Guy
Dec 11 at 23:53
add a comment |
I have a list:
lis = {{"b","x","d"},{"a","z","b"},{"a","x","b"},{"a","x","c"},{"b","z","d"}}
Certain consecutive elements of this list will have identical first and last members (in this example, "a" and "b" in lis[[2]] and lis[[3]]) and I would like to delete the element that has "x" as its middle element, to give:
res = {{"b","x","d"},{"a","z","b"},{"a","x","c"},{"b","z","d"}}
It seems like a job for SequenceCases, but am having no luck.
list-manipulation
asked Dec 11 at 23:30
Suite401
981312
I have a list:
lis = {{"b","x","d"},{"a","z","b"},{"a","x","b"},{"a","x","c"},{"b","z","d"}}
Certain consecutive elements of this list will have identical first and last members (in this example, "a" and "b" in lis[[2]] and lis[[3]]) and I would like to delete the element that has "x" as its middle element, to give:
res = {{"b","x","d"},{"a","z","b"},{"a","x","c"},{"b","z","d"}}
It seems like a job for SequenceCases, but am having no luck.
list-manipulation
list-manipulation
asked Dec 11 at 23:30
Suite401
981312
asked Dec 11 at 23:30
Suite401
981312
asked Dec 11 at 23:30
Suite401
981312
asked Dec 11 at 23:30
Suite401
981312
asked Dec 11 at 23:30
Suite401
981312
981312
2
elements 1 and 5 also have identical first and last members...Shouldn't you erase the first element, too?
– J42161217
Dec 11 at 23:35
Take a look atDeleteDuplicatesBy.
– Kuba♦
Dec 11 at 23:47
Another approach could be to start withSplit[lis, #1[[1]] == #2[[1]] && #1[[3]] == #2[[3]] &]to group the consecutive elements.
– That Gravity Guy
Dec 11 at 23:53
add a comment |
2
elements 1 and 5 also have identical first and last members...Shouldn't you erase the first element, too?
– J42161217
Dec 11 at 23:35
Take a look atDeleteDuplicatesBy.
– Kuba♦
Dec 11 at 23:47
Another approach could be to start withSplit[lis, #1[[1]] == #2[[1]] && #1[[3]] == #2[[3]] &]to group the consecutive elements.
– That Gravity Guy
Dec 11 at 23:53
2
2
elements 1 and 5 also have identical first and last members...Shouldn't you erase the first element, too?
– J42161217
Dec 11 at 23:35
elements 1 and 5 also have identical first and last members...Shouldn't you erase the first element, too?
– J42161217
Dec 11 at 23:35
Take a look at
DeleteDuplicatesBy.– Kuba♦
Dec 11 at 23:47
Take a look at
DeleteDuplicatesBy.– Kuba♦
Dec 11 at 23:47
Another approach could be to start with
Split[lis, #1[[1]] == #2[[1]] && #1[[3]] == #2[[3]] &] to group the consecutive elements.– That Gravity Guy
Dec 11 at 23:53
Another approach could be to start with
Split[lis, #1[[1]] == #2[[1]] && #1[[3]] == #2[[3]] &] to group the consecutive elements.– That Gravity Guy
Dec 11 at 23:53
add a comment |
2 Answers
2
active
oldest
votes
SequenceReplace[lis, {OrderlessPatternSequence[{a_, "x", c_}, {a_, b_, c_}]} :> {a, b, c}]
{{"b", "x", "d"}, {"a", "z", "b"}, {"a", "x", "c"}, {"b", "z", "d"}}
answered Dec 11 at 23:50
kglr
176k9197402
add a comment |
One approach is to look at the differences between adjacent elements, find those which equal {0,0,x_}, and remove them from the list.
lis[[Complement[Range[Length[lis]],Flatten@Position[Differences[lis], {0, 0, x_}]]]]
{{"b", "x", "d"}, {"a", "z", "b"}, {"a", "x", "c"}, {"b", "z", "d"}}
answered Dec 12 at 1:54
bill s
52.7k375150
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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SequenceReplace[lis, {OrderlessPatternSequence[{a_, "x", c_}, {a_, b_, c_}]} :> {a, b, c}]
{{"b", "x", "d"}, {"a", "z", "b"}, {"a", "x", "c"}, {"b", "z", "d"}}
answered Dec 11 at 23:50
kglr
176k9197402
add a comment |
SequenceReplace[lis, {OrderlessPatternSequence[{a_, "x", c_}, {a_, b_, c_}]} :> {a, b, c}]
{{"b", "x", "d"}, {"a", "z", "b"}, {"a", "x", "c"}, {"b", "z", "d"}}
answered Dec 11 at 23:50
kglr
176k9197402
add a comment |
SequenceReplace[lis, {OrderlessPatternSequence[{a_, "x", c_}, {a_, b_, c_}]} :> {a, b, c}]
{{"b", "x", "d"}, {"a", "z", "b"}, {"a", "x", "c"}, {"b", "z", "d"}}
answered Dec 11 at 23:50
kglr
176k9197402
SequenceReplace[lis, {OrderlessPatternSequence[{a_, "x", c_}, {a_, b_, c_}]} :> {a, b, c}]
{{"b", "x", "d"}, {"a", "z", "b"}, {"a", "x", "c"}, {"b", "z", "d"}}
answered Dec 11 at 23:50
kglr
176k9197402
answered Dec 11 at 23:50
kglr
176k9197402
answered Dec 11 at 23:50
kglr
176k9197402
answered Dec 11 at 23:50
kglr
176k9197402
176k9197402
add a comment |
add a comment |
One approach is to look at the differences between adjacent elements, find those which equal {0,0,x_}, and remove them from the list.
lis[[Complement[Range[Length[lis]],Flatten@Position[Differences[lis], {0, 0, x_}]]]]
{{"b", "x", "d"}, {"a", "z", "b"}, {"a", "x", "c"}, {"b", "z", "d"}}
answered Dec 12 at 1:54
bill s
52.7k375150
add a comment |
One approach is to look at the differences between adjacent elements, find those which equal {0,0,x_}, and remove them from the list.
lis[[Complement[Range[Length[lis]],Flatten@Position[Differences[lis], {0, 0, x_}]]]]
{{"b", "x", "d"}, {"a", "z", "b"}, {"a", "x", "c"}, {"b", "z", "d"}}
answered Dec 12 at 1:54
bill s
52.7k375150
add a comment |
One approach is to look at the differences between adjacent elements, find those which equal {0,0,x_}, and remove them from the list.
lis[[Complement[Range[Length[lis]],Flatten@Position[Differences[lis], {0, 0, x_}]]]]
{{"b", "x", "d"}, {"a", "z", "b"}, {"a", "x", "c"}, {"b", "z", "d"}}
answered Dec 12 at 1:54
bill s
52.7k375150
One approach is to look at the differences between adjacent elements, find those which equal {0,0,x_}, and remove them from the list.
lis[[Complement[Range[Length[lis]],Flatten@Position[Differences[lis], {0, 0, x_}]]]]
{{"b", "x", "d"}, {"a", "z", "b"}, {"a", "x", "c"}, {"b", "z", "d"}}
answered Dec 12 at 1:54
bill s
52.7k375150
answered Dec 12 at 1:54
bill s
52.7k375150
answered Dec 12 at 1:54
bill s
52.7k375150
answered Dec 12 at 1:54
bill s
52.7k375150
52.7k375150
add a comment |
add a comment |
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2
elements 1 and 5 also have identical first and last members...Shouldn't you erase the first element, too?
– J42161217
Dec 11 at 23:35
Take a look at
DeleteDuplicatesBy.– Kuba♦
Dec 11 at 23:47
Another approach could be to start with
Split[lis, #1[[1]] == #2[[1]] && #1[[3]] == #2[[3]] &]to group the consecutive elements.– That Gravity Guy
Dec 11 at 23:53