Counting certain elements in lists












4












$begingroup$


I have the following data set:



n1 = 1000;

data1 = {#, {RandomInteger[3, 20]}} & /@ Range[n1];



Dimensions@data1

{1000, 2}


I want to count how often data1[[All, 2]] == 1.



My solution, which seems to be wrong:



result1 = Table[

   Count[Flatten@(data1[[i, 2]]), u_ /; u == 1],

   {i, 1, n1}

   ];



Dimensions@result1

{1000}


Now I have 50 appended lists of data1 (= data2).



n3 = 50;



data2 = Array[0 &, n3];



Do[

  data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];

  , {i, 1, n3}

  ];



Dimensions@data2

{50, 1000, 2}


I want to count how often data2[[j, i, 2]] == 1, where {i, 1, n1} and {j, 1, n3}.



My solution for this more complicated list:



result2 = Array[0 &, n3];



Do[

  result2[[j]] =

    Table[

     Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],

     {i, 1, n1}

     ];

  , {j, 1, n3}

  ];



Dimensions@result2

{50, 1000}


How can I replace the Do loops in both cases and improve the performance?






list-manipulation







share|improve this question











$endgroup$

















    4












    $begingroup$


    I have the following data set:



    n1 = 1000;
    
data1 = {#, {RandomInteger[3, 20]}} & /@ Range[n1];
    

    
Dimensions@data1
    
{1000, 2}


    I want to count how often data1[[All, 2]] == 1.



    My solution, which seems to be wrong:



    result1 = Table[
    
   Count[Flatten@(data1[[i, 2]]), u_ /; u == 1],
    
   {i, 1, n1}
    
   ];
    

    
Dimensions@result1
    
{1000}


    Now I have 50 appended lists of data1 (= data2).



    n3 = 50;
    

    
data2 = Array[0 &, n3];
    

    
Do[
    
  data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];
    
  , {i, 1, n3}
    
  ];
    

    
Dimensions@data2
    
{50, 1000, 2}


    I want to count how often data2[[j, i, 2]] == 1, where {i, 1, n1} and {j, 1, n3}.



    My solution for this more complicated list:



    result2 = Array[0 &, n3];
    

    
Do[
    
  result2[[j]] =
    
    Table[
    
     Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
    
     {i, 1, n1}
    
     ];
    
  , {j, 1, n3}
    
  ];
    

    
Dimensions@result2
    
{50, 1000}


    How can I replace the Do loops in both cases and improve the performance?






    list-manipulation







    share|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      I have the following data set:



      n1 = 1000;
      
data1 = {#, {RandomInteger[3, 20]}} & /@ Range[n1];
      

      
Dimensions@data1
      
{1000, 2}


      I want to count how often data1[[All, 2]] == 1.



      My solution, which seems to be wrong:



      result1 = Table[
      
   Count[Flatten@(data1[[i, 2]]), u_ /; u == 1],
      
   {i, 1, n1}
      
   ];
      

      
Dimensions@result1
      
{1000}


      Now I have 50 appended lists of data1 (= data2).



      n3 = 50;
      

      
data2 = Array[0 &, n3];
      

      
Do[
      
  data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];
      
  , {i, 1, n3}
      
  ];
      

      
Dimensions@data2
      
{50, 1000, 2}


      I want to count how often data2[[j, i, 2]] == 1, where {i, 1, n1} and {j, 1, n3}.



      My solution for this more complicated list:



      result2 = Array[0 &, n3];
      

      
Do[
      
  result2[[j]] =
      
    Table[
      
     Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
      
     {i, 1, n1}
      
     ];
      
  , {j, 1, n3}
      
  ];
      

      
Dimensions@result2
      
{50, 1000}


      How can I replace the Do loops in both cases and improve the performance?






      list-manipulation







      share|improve this question











      $endgroup$




      I have the following data set:



      n1 = 1000;
      
data1 = {#, {RandomInteger[3, 20]}} & /@ Range[n1];
      

      
Dimensions@data1
      
{1000, 2}


      I want to count how often data1[[All, 2]] == 1.



      My solution, which seems to be wrong:



      result1 = Table[
      
   Count[Flatten@(data1[[i, 2]]), u_ /; u == 1],
      
   {i, 1, n1}
      
   ];
      

      
Dimensions@result1
      
{1000}


      Now I have 50 appended lists of data1 (= data2).



      n3 = 50;
      

      
data2 = Array[0 &, n3];
      

      
Do[
      
  data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];
      
  , {i, 1, n3}
      
  ];
      

      
Dimensions@data2
      
{50, 1000, 2}


      I want to count how often data2[[j, i, 2]] == 1, where {i, 1, n1} and {j, 1, n3}.



      My solution for this more complicated list:



      result2 = Array[0 &, n3];
      

      
Do[
      
  result2[[j]] =
      
    Table[
      
     Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
      
     {i, 1, n1}
      
     ];
      
  , {j, 1, n3}
      
  ];
      

      
Dimensions@result2
      
{50, 1000}


      How can I replace the Do loops in both cases and improve the performance?






      list-manipulation




      list-manipulation






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 6 hours ago







      lio

















      asked 6 hours ago









      liolio

      1,140217




      1,140217






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          For your result1:



          Count[1] /@ data1[[All, 2, 1]]


          For your result2:



          Map[Count[1], data2[[All, All, 2, 1]], {2}];





          share|improve this answer











          $endgroup$













          • $begingroup$
            This is great ...
            $endgroup$
            – lio
            6 hours ago










          • $begingroup$
            Do you have an idea for result2 improvement?
            $endgroup$
            – lio
            6 hours ago










          • $begingroup$
            @lio Yes, I've added something for result2 as well. You can check that they give the same output as your result1 and result2 respectively.
            $endgroup$
            – MarcoB
            6 hours ago



















          3












          $begingroup$

          r1 = Total[1 - Unitize[data1[[All, 2, 1]] - 1], {2}] 
          

          
Short @ %



          {5,5,4,6,5,8,5,6,5,4,8,6,3,3,2,7,2,9,5,6,4,5,3,8,7,6,8,3,7,5,9,4,7,6,5,4,5,5,<<925>>,7,5,6,6,3,5,4,6,6,9,7,9,5,6,4,3,8,4,4,6,6,9,6,6,3,8,3,4,1,8,4,4,4,5,7,8,5}




          r2 = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], {4}];
          
r2 // Dimensions



          {50, 1000}







          share|improve this answer











          $endgroup$













          • $begingroup$
            Great ... what do you think about result2. Now the Indices in the result2 double loop are correct.
            $endgroup$
            – lio
            6 hours ago










          • $begingroup$
            @lio, please see the update.
            $endgroup$
            – kglr
            5 hours ago











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          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%2f193270%2fcounting-certain-elements-in-lists%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          For your result1:



          Count[1] /@ data1[[All, 2, 1]]


          For your result2:



          Map[Count[1], data2[[All, All, 2, 1]], {2}];





          share|improve this answer











          $endgroup$













          • $begingroup$
            This is great ...
            $endgroup$
            – lio
            6 hours ago










          • $begingroup$
            Do you have an idea for result2 improvement?
            $endgroup$
            – lio
            6 hours ago










          • $begingroup$
            @lio Yes, I've added something for result2 as well. You can check that they give the same output as your result1 and result2 respectively.
            $endgroup$
            – MarcoB
            6 hours ago
















          4












          $begingroup$

          For your result1:



          Count[1] /@ data1[[All, 2, 1]]


          For your result2:



          Map[Count[1], data2[[All, All, 2, 1]], {2}];





          share|improve this answer











          $endgroup$













          • $begingroup$
            This is great ...
            $endgroup$
            – lio
            6 hours ago










          • $begingroup$
            Do you have an idea for result2 improvement?
            $endgroup$
            – lio
            6 hours ago










          • $begingroup$
            @lio Yes, I've added something for result2 as well. You can check that they give the same output as your result1 and result2 respectively.
            $endgroup$
            – MarcoB
            6 hours ago














          4












          4








          4





          $begingroup$

          For your result1:



          Count[1] /@ data1[[All, 2, 1]]


          For your result2:



          Map[Count[1], data2[[All, All, 2, 1]], {2}];





          share|improve this answer











          $endgroup$



          For your result1:



          Count[1] /@ data1[[All, 2, 1]]


          For your result2:



          Map[Count[1], data2[[All, All, 2, 1]], {2}];






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 6 hours ago

























          answered 6 hours ago









          MarcoBMarcoB

          37.5k556113




          37.5k556113












          • $begingroup$
            This is great ...
            $endgroup$
            – lio
            6 hours ago










          • $begingroup$
            Do you have an idea for result2 improvement?
            $endgroup$
            – lio
            6 hours ago










          • $begingroup$
            @lio Yes, I've added something for result2 as well. You can check that they give the same output as your result1 and result2 respectively.
            $endgroup$
            – MarcoB
            6 hours ago


















          • $begingroup$
            This is great ...
            $endgroup$
            – lio
            6 hours ago










          • $begingroup$
            Do you have an idea for result2 improvement?
            $endgroup$
            – lio
            6 hours ago










          • $begingroup$
            @lio Yes, I've added something for result2 as well. You can check that they give the same output as your result1 and result2 respectively.
            $endgroup$
            – MarcoB
            6 hours ago
















          $begingroup$
          This is great ...
          $endgroup$
          – lio
          6 hours ago




          $begingroup$
          This is great ...
          $endgroup$
          – lio
          6 hours ago












          $begingroup$
          Do you have an idea for result2 improvement?
          $endgroup$
          – lio
          6 hours ago




          $begingroup$
          Do you have an idea for result2 improvement?
          $endgroup$
          – lio
          6 hours ago












          $begingroup$
          @lio Yes, I've added something for result2 as well. You can check that they give the same output as your result1 and result2 respectively.
          $endgroup$
          – MarcoB
          6 hours ago




          $begingroup$
          @lio Yes, I've added something for result2 as well. You can check that they give the same output as your result1 and result2 respectively.
          $endgroup$
          – MarcoB
          6 hours ago











          3












          $begingroup$

          r1 = Total[1 - Unitize[data1[[All, 2, 1]] - 1], {2}] 
          

          
Short @ %



          {5,5,4,6,5,8,5,6,5,4,8,6,3,3,2,7,2,9,5,6,4,5,3,8,7,6,8,3,7,5,9,4,7,6,5,4,5,5,<<925>>,7,5,6,6,3,5,4,6,6,9,7,9,5,6,4,3,8,4,4,6,6,9,6,6,3,8,3,4,1,8,4,4,4,5,7,8,5}




          r2 = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], {4}];
          
r2 // Dimensions



          {50, 1000}







          share|improve this answer











          $endgroup$













          • $begingroup$
            Great ... what do you think about result2. Now the Indices in the result2 double loop are correct.
            $endgroup$
            – lio
            6 hours ago










          • $begingroup$
            @lio, please see the update.
            $endgroup$
            – kglr
            5 hours ago
















          3












          $begingroup$

          r1 = Total[1 - Unitize[data1[[All, 2, 1]] - 1], {2}] 
          

          
Short @ %



          {5,5,4,6,5,8,5,6,5,4,8,6,3,3,2,7,2,9,5,6,4,5,3,8,7,6,8,3,7,5,9,4,7,6,5,4,5,5,<<925>>,7,5,6,6,3,5,4,6,6,9,7,9,5,6,4,3,8,4,4,6,6,9,6,6,3,8,3,4,1,8,4,4,4,5,7,8,5}




          r2 = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], {4}];
          
r2 // Dimensions



          {50, 1000}







          share|improve this answer











          $endgroup$













          • $begingroup$
            Great ... what do you think about result2. Now the Indices in the result2 double loop are correct.
            $endgroup$
            – lio
            6 hours ago










          • $begingroup$
            @lio, please see the update.
            $endgroup$
            – kglr
            5 hours ago














          3












          3








          3





          $begingroup$

          r1 = Total[1 - Unitize[data1[[All, 2, 1]] - 1], {2}] 
          

          
Short @ %



          {5,5,4,6,5,8,5,6,5,4,8,6,3,3,2,7,2,9,5,6,4,5,3,8,7,6,8,3,7,5,9,4,7,6,5,4,5,5,<<925>>,7,5,6,6,3,5,4,6,6,9,7,9,5,6,4,3,8,4,4,6,6,9,6,6,3,8,3,4,1,8,4,4,4,5,7,8,5}




          r2 = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], {4}];
          
r2 // Dimensions



          {50, 1000}







          share|improve this answer











          $endgroup$



          r1 = Total[1 - Unitize[data1[[All, 2, 1]] - 1], {2}] 
          

          
Short @ %



          {5,5,4,6,5,8,5,6,5,4,8,6,3,3,2,7,2,9,5,6,4,5,3,8,7,6,8,3,7,5,9,4,7,6,5,4,5,5,<<925>>,7,5,6,6,3,5,4,6,6,9,7,9,5,6,4,3,8,4,4,6,6,9,6,6,3,8,3,4,1,8,4,4,4,5,7,8,5}




          r2 = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], {4}];
          
r2 // Dimensions



          {50, 1000}








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 5 hours ago

























          answered 6 hours ago









          kglrkglr

          189k10205422




          189k10205422












          • $begingroup$
            Great ... what do you think about result2. Now the Indices in the result2 double loop are correct.
            $endgroup$
            – lio
            6 hours ago










          • $begingroup$
            @lio, please see the update.
            $endgroup$
            – kglr
            5 hours ago


















          • $begingroup$
            Great ... what do you think about result2. Now the Indices in the result2 double loop are correct.
            $endgroup$
            – lio
            6 hours ago










          • $begingroup$
            @lio, please see the update.
            $endgroup$
            – kglr
            5 hours ago
















          $begingroup$
          Great ... what do you think about result2. Now the Indices in the result2 double loop are correct.
          $endgroup$
          – lio
          6 hours ago




          $begingroup$
          Great ... what do you think about result2. Now the Indices in the result2 double loop are correct.
          $endgroup$
          – lio
          6 hours ago












          $begingroup$
          @lio, please see the update.
          $endgroup$
          – kglr
          5 hours ago




          $begingroup$
          @lio, please see the update.
          $endgroup$
          – kglr
          5 hours ago


















          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%2f193270%2fcounting-certain-elements-in-lists%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

          How to label and detect the document text images

          Image
          1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C
          Read more

          Vallis Paradisi

          Image
          Coordinata: .mw-parser-output .geo-default,.mw-parser-output .geo-dms,.mw-parser-output .geo-dec{display:inline}.mw-parser-output .geo-nondefault,.mw-parser-output .geo-multi-punct{display:none}.mw-parser-output .longitude,.mw-parser-output .latitude{white-space:nowrap} 33°03′S 71°37′W  /  33.05°S 71.617°W  / -33.05; -71.617 Vallis Paradisi Res apud Vicidata repertae: Civitas : Chilia Numerus incolarum : 296 655 Zona horaria : UTC-4 Situs interretialis Nomen officiale : Valparaíso Gubernium Consilium : municipal council of Valparaíso Geographia Superficies : 401 600 000 Metrum quadratum Coniunctiones urbium Urbes gemellae : Ovetum , Badalona , Santa Fe , Ora longa , Sciamhaevum , Quantunia , Callao , Bat Yam , Melaka , Novorossiysk , Busan , Corduba , Odessa , Rosario , Malacca , Barcino , Soteropolis , Franciscopolis (California) , Guanajuato, Silao Tabula aut despectus Despectus in portum sinumque et urbem Vallis Paradisi Vallis Para
          Read more

          Tabula Rosettana

          Image
          Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a
          Read more