Selecting elements from lists based on string pattern












1












$begingroup$


I have a list:



lis = {{"abc", 1, 2}, {"cde", 3, 4}, {"fgbc", 5, 6}}


I would like to select list items that have first elements that end in the string "bc" to give:



res = {{1,2},{5,6}}


This is a variant on this question which has to do with strings only. Thank you as always for suggestions.










share|improve this question









$endgroup$

















    1












    $begingroup$


    I have a list:



    lis = {{"abc", 1, 2}, {"cde", 3, 4}, {"fgbc", 5, 6}}


    I would like to select list items that have first elements that end in the string "bc" to give:



    res = {{1,2},{5,6}}


    This is a variant on this question which has to do with strings only. Thank you as always for suggestions.










    share|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I have a list:



      lis = {{"abc", 1, 2}, {"cde", 3, 4}, {"fgbc", 5, 6}}


      I would like to select list items that have first elements that end in the string "bc" to give:



      res = {{1,2},{5,6}}


      This is a variant on this question which has to do with strings only. Thank you as always for suggestions.










      share|improve this question









      $endgroup$




      I have a list:



      lis = {{"abc", 1, 2}, {"cde", 3, 4}, {"fgbc", 5, 6}}


      I would like to select list items that have first elements that end in the string "bc" to give:



      res = {{1,2},{5,6}}


      This is a variant on this question which has to do with strings only. Thank you as always for suggestions.







      list-manipulation string-manipulation






      share|improve this question













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      asked 3 hours ago









      Suite401Suite401

      988312




      988312






















          3 Answers
          3






          active

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          2












          $begingroup$

          Pick[lis[[All, 2 ;;]], StringEndsQ[lis[[All, 1]], "bc"]]



          {{1, 2}, {5, 6}}




          Also



          f[{_String?(StringEndsQ["bc"]), x__}] := {x}
          f[_] := Sequence
          f /@ lis



          {{1, 2}, {5, 6}}







          share|improve this answer









          $endgroup$





















            1












            $begingroup$

            This uses string matching on the first element of each sublist and then discards the first element from the matches.



            lis = {{"abc", 1, 2}, {"cde", 3, 4}, {"fgbc", 5, 6}};
            Rest/@Select[lis,StringMatchQ[First[#],RegularExpression["\w*bc"]]&]

            (* {{1, 2}, {5, 6}} *)


            You may need to adjust the pattern given to RegularExpression if your strings do not all begin with word characters.






            share|improve this answer









            $endgroup$





















              1












              $begingroup$

              lis = {{"abc", 1, 2}, {"cde", 3, 4}, {"fgbc", 5, 6}};


              Using Cases



              Cases[lis, {_?(StringTake[#, -2] == "bc" &), x_, y_} :> {x, y}, 
              Infinity]

              (* {{1, 2}, {5, 6}} *)


              or



              Rest /@ Cases[lis, {_?(StringTake[#, -2] == "bc" &), __}, Infinity]

              (* {{1, 2}, {5, 6}} *)


              Or using Select



              Rest /@ Select[lis, StringTake[#[[1]], -2] == "bc" &]

              (* {{1, 2}, {5, 6}} *)


              Or using DeleteCases



              Rest /@ DeleteCases[lis, {_?(StringTake[#, -2] != "bc" &), __}, 
              Infinity]

              (* {{1, 2}, {5, 6}} *)





              share|improve this answer









              $endgroup$













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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

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                active

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                2












                $begingroup$

                Pick[lis[[All, 2 ;;]], StringEndsQ[lis[[All, 1]], "bc"]]



                {{1, 2}, {5, 6}}




                Also



                f[{_String?(StringEndsQ["bc"]), x__}] := {x}
                f[_] := Sequence
                f /@ lis



                {{1, 2}, {5, 6}}







                share|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Pick[lis[[All, 2 ;;]], StringEndsQ[lis[[All, 1]], "bc"]]



                  {{1, 2}, {5, 6}}




                  Also



                  f[{_String?(StringEndsQ["bc"]), x__}] := {x}
                  f[_] := Sequence
                  f /@ lis



                  {{1, 2}, {5, 6}}







                  share|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Pick[lis[[All, 2 ;;]], StringEndsQ[lis[[All, 1]], "bc"]]



                    {{1, 2}, {5, 6}}




                    Also



                    f[{_String?(StringEndsQ["bc"]), x__}] := {x}
                    f[_] := Sequence
                    f /@ lis



                    {{1, 2}, {5, 6}}







                    share|improve this answer









                    $endgroup$



                    Pick[lis[[All, 2 ;;]], StringEndsQ[lis[[All, 1]], "bc"]]



                    {{1, 2}, {5, 6}}




                    Also



                    f[{_String?(StringEndsQ["bc"]), x__}] := {x}
                    f[_] := Sequence
                    f /@ lis



                    {{1, 2}, {5, 6}}








                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 30 mins ago









                    kglrkglr

                    181k10200413




                    181k10200413























                        1












                        $begingroup$

                        This uses string matching on the first element of each sublist and then discards the first element from the matches.



                        lis = {{"abc", 1, 2}, {"cde", 3, 4}, {"fgbc", 5, 6}};
                        Rest/@Select[lis,StringMatchQ[First[#],RegularExpression["\w*bc"]]&]

                        (* {{1, 2}, {5, 6}} *)


                        You may need to adjust the pattern given to RegularExpression if your strings do not all begin with word characters.






                        share|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          This uses string matching on the first element of each sublist and then discards the first element from the matches.



                          lis = {{"abc", 1, 2}, {"cde", 3, 4}, {"fgbc", 5, 6}};
                          Rest/@Select[lis,StringMatchQ[First[#],RegularExpression["\w*bc"]]&]

                          (* {{1, 2}, {5, 6}} *)


                          You may need to adjust the pattern given to RegularExpression if your strings do not all begin with word characters.






                          share|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            This uses string matching on the first element of each sublist and then discards the first element from the matches.



                            lis = {{"abc", 1, 2}, {"cde", 3, 4}, {"fgbc", 5, 6}};
                            Rest/@Select[lis,StringMatchQ[First[#],RegularExpression["\w*bc"]]&]

                            (* {{1, 2}, {5, 6}} *)


                            You may need to adjust the pattern given to RegularExpression if your strings do not all begin with word characters.






                            share|improve this answer









                            $endgroup$



                            This uses string matching on the first element of each sublist and then discards the first element from the matches.



                            lis = {{"abc", 1, 2}, {"cde", 3, 4}, {"fgbc", 5, 6}};
                            Rest/@Select[lis,StringMatchQ[First[#],RegularExpression["\w*bc"]]&]

                            (* {{1, 2}, {5, 6}} *)


                            You may need to adjust the pattern given to RegularExpression if your strings do not all begin with word characters.







                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered 2 hours ago









                            BillBill

                            5,62569




                            5,62569























                                1












                                $begingroup$

                                lis = {{"abc", 1, 2}, {"cde", 3, 4}, {"fgbc", 5, 6}};


                                Using Cases



                                Cases[lis, {_?(StringTake[#, -2] == "bc" &), x_, y_} :> {x, y}, 
                                Infinity]

                                (* {{1, 2}, {5, 6}} *)


                                or



                                Rest /@ Cases[lis, {_?(StringTake[#, -2] == "bc" &), __}, Infinity]

                                (* {{1, 2}, {5, 6}} *)


                                Or using Select



                                Rest /@ Select[lis, StringTake[#[[1]], -2] == "bc" &]

                                (* {{1, 2}, {5, 6}} *)


                                Or using DeleteCases



                                Rest /@ DeleteCases[lis, {_?(StringTake[#, -2] != "bc" &), __}, 
                                Infinity]

                                (* {{1, 2}, {5, 6}} *)





                                share|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  lis = {{"abc", 1, 2}, {"cde", 3, 4}, {"fgbc", 5, 6}};


                                  Using Cases



                                  Cases[lis, {_?(StringTake[#, -2] == "bc" &), x_, y_} :> {x, y}, 
                                  Infinity]

                                  (* {{1, 2}, {5, 6}} *)


                                  or



                                  Rest /@ Cases[lis, {_?(StringTake[#, -2] == "bc" &), __}, Infinity]

                                  (* {{1, 2}, {5, 6}} *)


                                  Or using Select



                                  Rest /@ Select[lis, StringTake[#[[1]], -2] == "bc" &]

                                  (* {{1, 2}, {5, 6}} *)


                                  Or using DeleteCases



                                  Rest /@ DeleteCases[lis, {_?(StringTake[#, -2] != "bc" &), __}, 
                                  Infinity]

                                  (* {{1, 2}, {5, 6}} *)





                                  share|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    lis = {{"abc", 1, 2}, {"cde", 3, 4}, {"fgbc", 5, 6}};


                                    Using Cases



                                    Cases[lis, {_?(StringTake[#, -2] == "bc" &), x_, y_} :> {x, y}, 
                                    Infinity]

                                    (* {{1, 2}, {5, 6}} *)


                                    or



                                    Rest /@ Cases[lis, {_?(StringTake[#, -2] == "bc" &), __}, Infinity]

                                    (* {{1, 2}, {5, 6}} *)


                                    Or using Select



                                    Rest /@ Select[lis, StringTake[#[[1]], -2] == "bc" &]

                                    (* {{1, 2}, {5, 6}} *)


                                    Or using DeleteCases



                                    Rest /@ DeleteCases[lis, {_?(StringTake[#, -2] != "bc" &), __}, 
                                    Infinity]

                                    (* {{1, 2}, {5, 6}} *)





                                    share|improve this answer









                                    $endgroup$



                                    lis = {{"abc", 1, 2}, {"cde", 3, 4}, {"fgbc", 5, 6}};


                                    Using Cases



                                    Cases[lis, {_?(StringTake[#, -2] == "bc" &), x_, y_} :> {x, y}, 
                                    Infinity]

                                    (* {{1, 2}, {5, 6}} *)


                                    or



                                    Rest /@ Cases[lis, {_?(StringTake[#, -2] == "bc" &), __}, Infinity]

                                    (* {{1, 2}, {5, 6}} *)


                                    Or using Select



                                    Rest /@ Select[lis, StringTake[#[[1]], -2] == "bc" &]

                                    (* {{1, 2}, {5, 6}} *)


                                    Or using DeleteCases



                                    Rest /@ DeleteCases[lis, {_?(StringTake[#, -2] != "bc" &), __}, 
                                    Infinity]

                                    (* {{1, 2}, {5, 6}} *)






                                    share|improve this answer












                                    share|improve this answer



                                    share|improve this answer










                                    answered 2 hours ago









                                    Bob HanlonBob Hanlon

                                    59.5k33596




                                    59.5k33596






























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