How to leave only the following strings?












3












$begingroup$


Consider a data having the form



data = {{1,7,4,6},{1,6,4,8},{2,4,9,2},{E,...},{1,4,6,3},{4,4,6,2},{E,...},...}


i.e., some number $n_{1}$ of rows followed by row ${E,...}$, then some number $n_{2}$ of rows followed by row ${E,...}$ and so on.



Could you please tell me how to leave only the last rows before ${E,}$, i.e. to obtain



subdata= {{2,4,9,2},{4,4,6,2},...}?





list-manipulation data







share|improve this question









$endgroup$








  • 1




    $begingroup$
    e.g. SequenceCases[data, {x_List, {E, ___}} :> x]
    $endgroup$
    – C. E.
    9 hours ago
















3












$begingroup$


Consider a data having the form



data = {{1,7,4,6},{1,6,4,8},{2,4,9,2},{E,...},{1,4,6,3},{4,4,6,2},{E,...},...}


i.e., some number $n_{1}$ of rows followed by row ${E,...}$, then some number $n_{2}$ of rows followed by row ${E,...}$ and so on.



Could you please tell me how to leave only the last rows before ${E,}$, i.e. to obtain



subdata= {{2,4,9,2},{4,4,6,2},...}?





list-manipulation data







share|improve this question









$endgroup$








  • 1




    $begingroup$
    e.g. SequenceCases[data, {x_List, {E, ___}} :> x]
    $endgroup$
    – C. E.
    9 hours ago














3












3








3





$begingroup$


Consider a data having the form



data = {{1,7,4,6},{1,6,4,8},{2,4,9,2},{E,...},{1,4,6,3},{4,4,6,2},{E,...},...}


i.e., some number $n_{1}$ of rows followed by row ${E,...}$, then some number $n_{2}$ of rows followed by row ${E,...}$ and so on.



Could you please tell me how to leave only the last rows before ${E,}$, i.e. to obtain



subdata= {{2,4,9,2},{4,4,6,2},...}?





list-manipulation data







share|improve this question









$endgroup$




Consider a data having the form



data = {{1,7,4,6},{1,6,4,8},{2,4,9,2},{E,...},{1,4,6,3},{4,4,6,2},{E,...},...}


i.e., some number $n_{1}$ of rows followed by row ${E,...}$, then some number $n_{2}$ of rows followed by row ${E,...}$ and so on.



Could you please tell me how to leave only the last rows before ${E,}$, i.e. to obtain



subdata= {{2,4,9,2},{4,4,6,2},...}?





list-manipulation data




list-manipulation data






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 9 hours ago









John TaylorJohn Taylor

792211




792211








  • 1




    $begingroup$
    e.g. SequenceCases[data, {x_List, {E, ___}} :> x]
    $endgroup$
    – C. E.
    9 hours ago














  • 1




    $begingroup$
    e.g. SequenceCases[data, {x_List, {E, ___}} :> x]
    $endgroup$
    – C. E.
    9 hours ago








1




1




$begingroup$
e.g. SequenceCases[data, {x_List, {E, ___}} :> x]
$endgroup$
– C. E.
9 hours ago




$begingroup$
e.g. SequenceCases[data, {x_List, {E, ___}} :> x]
$endgroup$
– C. E.
9 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

Try SequenceCases:



data = {{1, 7, 4, 6}, {1, 6, 4, 8}, {2, 4, 9, 2}, {E, 1, 2, 3}, 

        {1, 4, 6, 3}, {4, 4, 6, 2}, {E, 4, 5, 6}}

SequenceCases[data, {p_, {E, ___}} :> p]


yields



{{2, 4, 9, 2}, {4, 4, 6, 2}}





share|improve this answer









$endgroup$





















    3












    $begingroup$

    The most idiomatic solution to this problem is, in my opinion, pattern matching (as Sakra has also answered):



    SequenceCases[data, {x_List, {E, ___}} :> x]



    {{2, 4, 9, 2}, {4, 4, 6, 2}}




    But the problem also lends itself to functional solutions, e.g.:



    pairs = Partition[data, 2, 1];
    
If[#[[2, 1]] == E, #[[1]], Nothing] & /@ pairs



    {{2, 4, 9, 2}, {4, 4, 6, 2}}




    Or in one go:



    BlockMap[If[#[[2, 1]] == E, #[[1]], Nothing] &, data, 2, 1]



    {{2, 4, 9, 2}, {4, 4, 6, 2}}







    share|improve this answer











    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Try SequenceCases:



      data = {{1, 7, 4, 6}, {1, 6, 4, 8}, {2, 4, 9, 2}, {E, 1, 2, 3}, 
      
        {1, 4, 6, 3}, {4, 4, 6, 2}, {E, 4, 5, 6}}
      
SequenceCases[data, {p_, {E, ___}} :> p]


      yields



      {{2, 4, 9, 2}, {4, 4, 6, 2}}





      share|improve this answer









      $endgroup$


















        3












        $begingroup$

        Try SequenceCases:



        data = {{1, 7, 4, 6}, {1, 6, 4, 8}, {2, 4, 9, 2}, {E, 1, 2, 3}, 
        
        {1, 4, 6, 3}, {4, 4, 6, 2}, {E, 4, 5, 6}}
        
SequenceCases[data, {p_, {E, ___}} :> p]


        yields



        {{2, 4, 9, 2}, {4, 4, 6, 2}}





        share|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Try SequenceCases:



          data = {{1, 7, 4, 6}, {1, 6, 4, 8}, {2, 4, 9, 2}, {E, 1, 2, 3}, 
          
        {1, 4, 6, 3}, {4, 4, 6, 2}, {E, 4, 5, 6}}
          
SequenceCases[data, {p_, {E, ___}} :> p]


          yields



          {{2, 4, 9, 2}, {4, 4, 6, 2}}





          share|improve this answer









          $endgroup$



          Try SequenceCases:



          data = {{1, 7, 4, 6}, {1, 6, 4, 8}, {2, 4, 9, 2}, {E, 1, 2, 3}, 
          
        {1, 4, 6, 3}, {4, 4, 6, 2}, {E, 4, 5, 6}}
          
SequenceCases[data, {p_, {E, ___}} :> p]


          yields



          {{2, 4, 9, 2}, {4, 4, 6, 2}}






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 9 hours ago









          sakrasakra

          2,8331429




          2,8331429























              3












              $begingroup$

              The most idiomatic solution to this problem is, in my opinion, pattern matching (as Sakra has also answered):



              SequenceCases[data, {x_List, {E, ___}} :> x]



              {{2, 4, 9, 2}, {4, 4, 6, 2}}




              But the problem also lends itself to functional solutions, e.g.:



              pairs = Partition[data, 2, 1];
              
If[#[[2, 1]] == E, #[[1]], Nothing] & /@ pairs



              {{2, 4, 9, 2}, {4, 4, 6, 2}}




              Or in one go:



              BlockMap[If[#[[2, 1]] == E, #[[1]], Nothing] &, data, 2, 1]



              {{2, 4, 9, 2}, {4, 4, 6, 2}}







              share|improve this answer











              $endgroup$


















                3












                $begingroup$

                The most idiomatic solution to this problem is, in my opinion, pattern matching (as Sakra has also answered):



                SequenceCases[data, {x_List, {E, ___}} :> x]



                {{2, 4, 9, 2}, {4, 4, 6, 2}}




                But the problem also lends itself to functional solutions, e.g.:



                pairs = Partition[data, 2, 1];
                
If[#[[2, 1]] == E, #[[1]], Nothing] & /@ pairs



                {{2, 4, 9, 2}, {4, 4, 6, 2}}




                Or in one go:



                BlockMap[If[#[[2, 1]] == E, #[[1]], Nothing] &, data, 2, 1]



                {{2, 4, 9, 2}, {4, 4, 6, 2}}







                share|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  The most idiomatic solution to this problem is, in my opinion, pattern matching (as Sakra has also answered):



                  SequenceCases[data, {x_List, {E, ___}} :> x]



                  {{2, 4, 9, 2}, {4, 4, 6, 2}}




                  But the problem also lends itself to functional solutions, e.g.:



                  pairs = Partition[data, 2, 1];
                  
If[#[[2, 1]] == E, #[[1]], Nothing] & /@ pairs



                  {{2, 4, 9, 2}, {4, 4, 6, 2}}




                  Or in one go:



                  BlockMap[If[#[[2, 1]] == E, #[[1]], Nothing] &, data, 2, 1]



                  {{2, 4, 9, 2}, {4, 4, 6, 2}}







                  share|improve this answer











                  $endgroup$



                  The most idiomatic solution to this problem is, in my opinion, pattern matching (as Sakra has also answered):



                  SequenceCases[data, {x_List, {E, ___}} :> x]



                  {{2, 4, 9, 2}, {4, 4, 6, 2}}




                  But the problem also lends itself to functional solutions, e.g.:



                  pairs = Partition[data, 2, 1];
                  
If[#[[2, 1]] == E, #[[1]], Nothing] & /@ pairs



                  {{2, 4, 9, 2}, {4, 4, 6, 2}}




                  Or in one go:



                  BlockMap[If[#[[2, 1]] == E, #[[1]], Nothing] &, data, 2, 1]



                  {{2, 4, 9, 2}, {4, 4, 6, 2}}








                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 7 hours ago

























                  answered 9 hours ago









                  C. E.C. E.

                  51.4k3101207




                  51.4k3101207






























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