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Dropping list elements from nested list after evaluation



Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?How do I delete items at the same position from every sub-list within a list?Using and replacing sequential elements of a listFinding the main parent after sorting in a multiplication processDeleting certain elements from a nested listData selection by comparing elements from different sublists in a nested listSelect from nested list and dropping non matching elementsSelecting elements in nested listRemove elements from deep nested listTake a specific eigenvalue from a list in its unevaluated formEliminate empty elements from a list with a specific pattern










4












$begingroup$


I need to create a new list from a nested list but using the evaluation as criteria to drop the elements. For example let's say that that I have the following list:



list1=1,1,-(-1)^3,x,2*x,1,1,(-1)^3,x,2*x,
1,1,x,2*x,3*x,1,1,-x,-2*x,-3*x


and I need to eliminate the elements of list1 that the absolute value of the third element give $1$, i.d. $-(-1)^3$ and $(-1)^3$, to obtain



 list2=1,1,x,2*x,3*x,1,1,-x,-2*x,-3*x


In this case, list1 was created with the code



For[i = 1, i < 4, i++,
For[j = 1, j < 4, j++,
list1[i, j, p_] = Sort[Eigenvalues[mat[i, j, p]]];
] ]


I have been trying to use Select but until now I am not been able to create list2 to plot it with



list2=ParallelTable[Select[Abs[eigval[i, j, p][[3]]], 
Abs[#] != 1 &] , i, 1, 4, j,1,4]


I am still learning to uses cases in Mathematica so I am not sure how to do it. Do you know if there is wise way to do it? Thanks in advance.










share|improve this question











$endgroup$
















    4












    $begingroup$


    I need to create a new list from a nested list but using the evaluation as criteria to drop the elements. For example let's say that that I have the following list:



    list1=1,1,-(-1)^3,x,2*x,1,1,(-1)^3,x,2*x,
    1,1,x,2*x,3*x,1,1,-x,-2*x,-3*x


    and I need to eliminate the elements of list1 that the absolute value of the third element give $1$, i.d. $-(-1)^3$ and $(-1)^3$, to obtain



     list2=1,1,x,2*x,3*x,1,1,-x,-2*x,-3*x


    In this case, list1 was created with the code



    For[i = 1, i < 4, i++,
    For[j = 1, j < 4, j++,
    list1[i, j, p_] = Sort[Eigenvalues[mat[i, j, p]]];
    ] ]


    I have been trying to use Select but until now I am not been able to create list2 to plot it with



    list2=ParallelTable[Select[Abs[eigval[i, j, p][[3]]], 
    Abs[#] != 1 &] , i, 1, 4, j,1,4]


    I am still learning to uses cases in Mathematica so I am not sure how to do it. Do you know if there is wise way to do it? Thanks in advance.










    share|improve this question











    $endgroup$














      4












      4








      4





      $begingroup$


      I need to create a new list from a nested list but using the evaluation as criteria to drop the elements. For example let's say that that I have the following list:



      list1=1,1,-(-1)^3,x,2*x,1,1,(-1)^3,x,2*x,
      1,1,x,2*x,3*x,1,1,-x,-2*x,-3*x


      and I need to eliminate the elements of list1 that the absolute value of the third element give $1$, i.d. $-(-1)^3$ and $(-1)^3$, to obtain



       list2=1,1,x,2*x,3*x,1,1,-x,-2*x,-3*x


      In this case, list1 was created with the code



      For[i = 1, i < 4, i++,
      For[j = 1, j < 4, j++,
      list1[i, j, p_] = Sort[Eigenvalues[mat[i, j, p]]];
      ] ]


      I have been trying to use Select but until now I am not been able to create list2 to plot it with



      list2=ParallelTable[Select[Abs[eigval[i, j, p][[3]]], 
      Abs[#] != 1 &] , i, 1, 4, j,1,4]


      I am still learning to uses cases in Mathematica so I am not sure how to do it. Do you know if there is wise way to do it? Thanks in advance.










      share|improve this question











      $endgroup$




      I need to create a new list from a nested list but using the evaluation as criteria to drop the elements. For example let's say that that I have the following list:



      list1=1,1,-(-1)^3,x,2*x,1,1,(-1)^3,x,2*x,
      1,1,x,2*x,3*x,1,1,-x,-2*x,-3*x


      and I need to eliminate the elements of list1 that the absolute value of the third element give $1$, i.d. $-(-1)^3$ and $(-1)^3$, to obtain



       list2=1,1,x,2*x,3*x,1,1,-x,-2*x,-3*x


      In this case, list1 was created with the code



      For[i = 1, i < 4, i++,
      For[j = 1, j < 4, j++,
      list1[i, j, p_] = Sort[Eigenvalues[mat[i, j, p]]];
      ] ]


      I have been trying to use Select but until now I am not been able to create list2 to plot it with



      list2=ParallelTable[Select[Abs[eigval[i, j, p][[3]]], 
      Abs[#] != 1 &] , i, 1, 4, j,1,4]


      I am still learning to uses cases in Mathematica so I am not sure how to do it. Do you know if there is wise way to do it? Thanks in advance.







      list-manipulation filtering






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Apr 11 at 16:30









      Roman

      5,40311131




      5,40311131










      asked Apr 11 at 13:50









      morsmors

      716




      716




















          4 Answers
          4






          active

          oldest

          votes


















          7












          $begingroup$

          If you prefer using DeleteCases,



          list2 = DeleteCases[list1, _?(Abs[#[[3]]] == 1 &)]



          1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x







          share|improve this answer









          $endgroup$




















            5












            $begingroup$

            Delete[
            list1,
            Position[Abs[list1[[All, 3]]], 1]
            ]



            1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x







            share|improve this answer









            $endgroup$




















              5












              $begingroup$

              if you want to use Select, try this



              Select[list1,!NumberQ@#[[3]]||Abs[#[[3]]]!=1&] 



              1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x







              share|improve this answer









              $endgroup$




















                1












                $begingroup$

                This is pretty efficient on unpacked arrays (the Listable attribute assumes list1[[All, 3]] is a flat list, as it is in the OP's example):



                Block[signal,
                SetAttributes[signal, Listable];
                signal[1] = 1; signal[_] = 0;
                Pick[list1, signal@Abs[list1[[All, 3]]], 0]
                ]
                (* 1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x *)


                (For packed arrays, one would probably want to use Unitize[x-1] instead of signal.)






                share|improve this answer









                $endgroup$













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                  4 Answers
                  4






                  active

                  oldest

                  votes








                  4 Answers
                  4






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  7












                  $begingroup$

                  If you prefer using DeleteCases,



                  list2 = DeleteCases[list1, _?(Abs[#[[3]]] == 1 &)]



                  1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x







                  share|improve this answer









                  $endgroup$

















                    7












                    $begingroup$

                    If you prefer using DeleteCases,



                    list2 = DeleteCases[list1, _?(Abs[#[[3]]] == 1 &)]



                    1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x







                    share|improve this answer









                    $endgroup$















                      7












                      7








                      7





                      $begingroup$

                      If you prefer using DeleteCases,



                      list2 = DeleteCases[list1, _?(Abs[#[[3]]] == 1 &)]



                      1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x







                      share|improve this answer









                      $endgroup$



                      If you prefer using DeleteCases,



                      list2 = DeleteCases[list1, _?(Abs[#[[3]]] == 1 &)]



                      1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x








                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered Apr 11 at 16:29









                      RomanRoman

                      5,40311131




                      5,40311131





















                          5












                          $begingroup$

                          Delete[
                          list1,
                          Position[Abs[list1[[All, 3]]], 1]
                          ]



                          1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x







                          share|improve this answer









                          $endgroup$

















                            5












                            $begingroup$

                            Delete[
                            list1,
                            Position[Abs[list1[[All, 3]]], 1]
                            ]



                            1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x







                            share|improve this answer









                            $endgroup$















                              5












                              5








                              5





                              $begingroup$

                              Delete[
                              list1,
                              Position[Abs[list1[[All, 3]]], 1]
                              ]



                              1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x







                              share|improve this answer









                              $endgroup$



                              Delete[
                              list1,
                              Position[Abs[list1[[All, 3]]], 1]
                              ]



                              1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x








                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered Apr 11 at 14:01









                              Henrik SchumacherHenrik Schumacher

                              60.2k583169




                              60.2k583169





















                                  5












                                  $begingroup$

                                  if you want to use Select, try this



                                  Select[list1,!NumberQ@#[[3]]||Abs[#[[3]]]!=1&] 



                                  1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x







                                  share|improve this answer









                                  $endgroup$

















                                    5












                                    $begingroup$

                                    if you want to use Select, try this



                                    Select[list1,!NumberQ@#[[3]]||Abs[#[[3]]]!=1&] 



                                    1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x







                                    share|improve this answer









                                    $endgroup$















                                      5












                                      5








                                      5





                                      $begingroup$

                                      if you want to use Select, try this



                                      Select[list1,!NumberQ@#[[3]]||Abs[#[[3]]]!=1&] 



                                      1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x







                                      share|improve this answer









                                      $endgroup$



                                      if you want to use Select, try this



                                      Select[list1,!NumberQ@#[[3]]||Abs[#[[3]]]!=1&] 



                                      1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x








                                      share|improve this answer












                                      share|improve this answer



                                      share|improve this answer










                                      answered Apr 11 at 14:16









                                      J42161217J42161217

                                      4,498324




                                      4,498324





















                                          1












                                          $begingroup$

                                          This is pretty efficient on unpacked arrays (the Listable attribute assumes list1[[All, 3]] is a flat list, as it is in the OP's example):



                                          Block[signal,
                                          SetAttributes[signal, Listable];
                                          signal[1] = 1; signal[_] = 0;
                                          Pick[list1, signal@Abs[list1[[All, 3]]], 0]
                                          ]
                                          (* 1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x *)


                                          (For packed arrays, one would probably want to use Unitize[x-1] instead of signal.)






                                          share|improve this answer









                                          $endgroup$

















                                            1












                                            $begingroup$

                                            This is pretty efficient on unpacked arrays (the Listable attribute assumes list1[[All, 3]] is a flat list, as it is in the OP's example):



                                            Block[signal,
                                            SetAttributes[signal, Listable];
                                            signal[1] = 1; signal[_] = 0;
                                            Pick[list1, signal@Abs[list1[[All, 3]]], 0]
                                            ]
                                            (* 1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x *)


                                            (For packed arrays, one would probably want to use Unitize[x-1] instead of signal.)






                                            share|improve this answer









                                            $endgroup$















                                              1












                                              1








                                              1





                                              $begingroup$

                                              This is pretty efficient on unpacked arrays (the Listable attribute assumes list1[[All, 3]] is a flat list, as it is in the OP's example):



                                              Block[signal,
                                              SetAttributes[signal, Listable];
                                              signal[1] = 1; signal[_] = 0;
                                              Pick[list1, signal@Abs[list1[[All, 3]]], 0]
                                              ]
                                              (* 1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x *)


                                              (For packed arrays, one would probably want to use Unitize[x-1] instead of signal.)






                                              share|improve this answer









                                              $endgroup$



                                              This is pretty efficient on unpacked arrays (the Listable attribute assumes list1[[All, 3]] is a flat list, as it is in the OP's example):



                                              Block[signal,
                                              SetAttributes[signal, Listable];
                                              signal[1] = 1; signal[_] = 0;
                                              Pick[list1, signal@Abs[list1[[All, 3]]], 0]
                                              ]
                                              (* 1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x *)


                                              (For packed arrays, one would probably want to use Unitize[x-1] instead of signal.)







                                              share|improve this answer












                                              share|improve this answer



                                              share|improve this answer










                                              answered Apr 12 at 0:11









                                              Michael E2Michael E2

                                              151k12203482




                                              151k12203482



























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