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How to leave only the following strings?



Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?How can I create a class of matrices programmatically?Manipulate excel file and plotHow do I partition a matrix?How to create a table of tables?Turn the following values into percentageHow to select the data in a given way?Reading CSV data from a streamHow to remove from the data the rows of with fixed number of elements?Padding lists for accurate plottingHow to remove the given columns and strings?










2












$begingroup$


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.
    7 hours ago















2












$begingroup$


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.
    7 hours ago













2












2








2





$begingroup$


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 7 hours ago









John TaylorJohn Taylor

787211




787211







  • 1




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












  • 1




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







1




1




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




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










2 Answers
2






active

oldest

votes


















2












$begingroup$

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









$endgroup$




















    2












    $begingroup$

    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









      2












      $begingroup$

      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









      $endgroup$

















        2












        $begingroup$

        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









        $endgroup$















          2












          2








          2





          $begingroup$

          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 7 hours ago









          sakrasakra

          2,8231429




          2,8231429





















              2












              $begingroup$

              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











              $endgroup$

















                2












                $begingroup$

                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











                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  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 5 hours ago

























                  answered 7 hours ago









                  C. E.C. E.

                  51.4k3101207




                  51.4k3101207



























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