selective deletions from list











up vote
2
down vote

favorite












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







share|improve this question


















  • 1




    elements 1 and 5 also have identical first and last members...Shouldn't you erase the first element, too?
    – J42161217
    2 hours ago










  • Take a look at DeleteDuplicatesBy.
    – Kuba♦
    2 hours ago












  • 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
    2 hours ago















up vote
2
down vote

favorite












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







share|improve this question


















  • 1




    elements 1 and 5 also have identical first and last members...Shouldn't you erase the first element, too?
    – J42161217
    2 hours ago










  • Take a look at DeleteDuplicatesBy.
    – Kuba♦
    2 hours ago












  • 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
    2 hours ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











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







share|improve this question













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






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 2 hours ago









Suite401

969312




969312








  • 1




    elements 1 and 5 also have identical first and last members...Shouldn't you erase the first element, too?
    – J42161217
    2 hours ago










  • Take a look at DeleteDuplicatesBy.
    – Kuba♦
    2 hours ago












  • 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
    2 hours ago














  • 1




    elements 1 and 5 also have identical first and last members...Shouldn't you erase the first element, too?
    – J42161217
    2 hours ago










  • Take a look at DeleteDuplicatesBy.
    – Kuba♦
    2 hours ago












  • 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
    2 hours ago








1




1




elements 1 and 5 also have identical first and last members...Shouldn't you erase the first element, too?
– J42161217
2 hours ago




elements 1 and 5 also have identical first and last members...Shouldn't you erase the first element, too?
– J42161217
2 hours ago












Take a look at DeleteDuplicatesBy.
– Kuba♦
2 hours ago






Take a look at DeleteDuplicatesBy.
– Kuba♦
2 hours ago














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
2 hours ago




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
2 hours ago










2 Answers
2






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up vote
6
down vote













SequenceReplace[lis, {OrderlessPatternSequence[{a_, "x", c_}, {a_, b_, c_}]} :> {a, b, c}]



{{"b", "x", "d"}, {"a", "z", "b"}, {"a", "x", "c"}, {"b", "z", "d"}}







share|improve this answer




























    up vote
    0
    down vote













    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"}}





    share|improve this answer





















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      2 Answers
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      2 Answers
      2






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      up vote
      6
      down vote













      SequenceReplace[lis, {OrderlessPatternSequence[{a_, "x", c_}, {a_, b_, c_}]} :> {a, b, c}]



      {{"b", "x", "d"}, {"a", "z", "b"}, {"a", "x", "c"}, {"b", "z", "d"}}







      share|improve this answer

























        up vote
        6
        down vote













        SequenceReplace[lis, {OrderlessPatternSequence[{a_, "x", c_}, {a_, b_, c_}]} :> {a, b, c}]



        {{"b", "x", "d"}, {"a", "z", "b"}, {"a", "x", "c"}, {"b", "z", "d"}}







        share|improve this answer























          up vote
          6
          down vote










          up vote
          6
          down vote









          SequenceReplace[lis, {OrderlessPatternSequence[{a_, "x", c_}, {a_, b_, c_}]} :> {a, b, c}]



          {{"b", "x", "d"}, {"a", "z", "b"}, {"a", "x", "c"}, {"b", "z", "d"}}







          share|improve this answer












          SequenceReplace[lis, {OrderlessPatternSequence[{a_, "x", c_}, {a_, b_, c_}]} :> {a, b, c}]



          {{"b", "x", "d"}, {"a", "z", "b"}, {"a", "x", "c"}, {"b", "z", "d"}}








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 2 hours ago









          kglr

          175k9197402




          175k9197402






















              up vote
              0
              down vote













              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"}}





              share|improve this answer

























                up vote
                0
                down vote













                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"}}





                share|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  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"}}





                  share|improve this answer












                  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"}}






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 19 mins ago









                  bill s

                  52.5k375149




                  52.5k375149






























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