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 2 days ago









      Roman

      5,24511131




      5,24511131










      asked 2 days ago









      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









                $endgroup$














                  Your Answer








                  StackExchange.ready(function() {
                  var channelOptions = {
                  tags: "".split(" "),
                  id: "387"
                  };
                  initTagRenderer("".split(" "), "".split(" "), channelOptions);

                  StackExchange.using("externalEditor", function() {
                  // Have to fire editor after snippets, if snippets enabled
                  if (StackExchange.settings.snippets.snippetsEnabled) {
                  StackExchange.using("snippets", function() {
                  createEditor();
                  });
                  }
                  else {
                  createEditor();
                  }
                  });

                  function createEditor() {
                  StackExchange.prepareEditor({
                  heartbeatType: 'answer',
                  autoActivateHeartbeat: false,
                  convertImagesToLinks: false,
                  noModals: true,
                  showLowRepImageUploadWarning: true,
                  reputationToPostImages: null,
                  bindNavPrevention: true,
                  postfix: "",
                  imageUploader: {
                  brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
                  contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
                  allowUrls: true
                  },
                  onDemand: true,
                  discardSelector: ".discard-answer"
                  ,immediatelyShowMarkdownHelp:true
                  });


                  }
                  });














                  draft saved

                  draft discarded


















                  StackExchange.ready(
                  function () {
                  StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f195019%2fdropping-list-elements-from-nested-list-after-evaluation%23new-answer', 'question_page');
                  }
                  );

                  Post as a guest















                  Required, but never shown

























                  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 2 days ago









                      RomanRoman

                      5,24511131




                      5,24511131























                          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 2 days ago









                              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 2 days ago









                                      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 2 days ago









                                              Michael E2Michael E2

                                              150k12203482




                                              150k12203482






























                                                  draft saved

                                                  draft discarded




















































                                                  Thanks for contributing an answer to Mathematica Stack Exchange!


                                                  • Please be sure to answer the question. Provide details and share your research!

                                                  But avoid …



                                                  • Asking for help, clarification, or responding to other answers.

                                                  • Making statements based on opinion; back them up with references or personal experience.


                                                  Use MathJax to format equations. MathJax reference.


                                                  To learn more, see our tips on writing great answers.




                                                  draft saved


                                                  draft discarded














                                                  StackExchange.ready(
                                                  function () {
                                                  StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f195019%2fdropping-list-elements-from-nested-list-after-evaluation%23new-answer', 'question_page');
                                                  }
                                                  );

                                                  Post as a guest















                                                  Required, but never shown





















































                                                  Required, but never shown














                                                  Required, but never shown












                                                  Required, but never shown







                                                  Required, but never shown

































                                                  Required, but never shown














                                                  Required, but never shown












                                                  Required, but never shown







                                                  Required, but never shown







                                                  -

                                                  Popular posts from this blog

                                                  數位音樂下載

                                                  Image
                                                  數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音
                                                  Read more

                                                  格利澤436b

                                                  Image
                                                  格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP
                                                  Read more

                                                  When can things happen in Etherscan, such as the picture below?

                                                  Image
                                                  2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  | 
                                                  Read more