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





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          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


















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