Dropping list elements from nested list after evaluation












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.






list-manipulation filtering







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.






    list-manipulation filtering







    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.






      list-manipulation filtering







      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




      list-manipulation filtering






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited yesterday









      Roman

      5,20011131




      5,20011131










      asked yesterday









      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 yesterday









                      RomanRoman

                      5,20011131




                      5,20011131























                          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 yesterday









                              Henrik SchumacherHenrik Schumacher

                              60k582168




                              60k582168























                                  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 yesterday









                                      J42161217J42161217

                                      4,468324




                                      4,468324























                                          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 yesterday









                                              Michael E2Michael E2

                                              150k12203482




                                              150k12203482






























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