selective deletions from list












4














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


















  • 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
















4














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


















  • 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














4












4








4







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 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 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














  • 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








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










2 Answers
2






active

oldest

votes


















9














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





























    2














    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
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      9














      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


























        9














        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
























          9












          9








          9






          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 Dec 11 at 23:50









          kglr

          176k9197402




          176k9197402























              2














              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


























                2














                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
























                  2












                  2








                  2






                  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 Dec 12 at 1:54









                  bill s

                  52.7k375150




                  52.7k375150






























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