How to speed up a process












5












$begingroup$


I am wondering how to speed up the following code: Currently it takes more than two sec to be processed.



Timing[For[i = 1;



list = {}, i <= 11111, i++, l = AppendTo[list, i/Pi]];

 Total@N@list]


I am new to Wolfram languages and it's the only thing I could come up with.
Any idea ?



Thanks






performance-tuning timing







share|improve this question









$endgroup$

















    5












    $begingroup$


    I am wondering how to speed up the following code: Currently it takes more than two sec to be processed.



    Timing[For[i = 1;
    

    
list = {}, i <= 11111, i++, l = AppendTo[list, i/Pi]];
    
 Total@N@list]


    I am new to Wolfram languages and it's the only thing I could come up with.
    Any idea ?



    Thanks






    performance-tuning timing







    share|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$


      I am wondering how to speed up the following code: Currently it takes more than two sec to be processed.



      Timing[For[i = 1;
      

      
list = {}, i <= 11111, i++, l = AppendTo[list, i/Pi]];
      
 Total@N@list]


      I am new to Wolfram languages and it's the only thing I could come up with.
      Any idea ?



      Thanks






      performance-tuning timing







      share|improve this question









      $endgroup$




      I am wondering how to speed up the following code: Currently it takes more than two sec to be processed.



      Timing[For[i = 1;
      

      
list = {}, i <= 11111, i++, l = AppendTo[list, i/Pi]];
      
 Total@N@list]


      I am new to Wolfram languages and it's the only thing I could come up with.
      Any idea ?



      Thanks






      performance-tuning timing




      performance-tuning timing






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked yesterday







      user63037





























          3 Answers
          3






          active

          oldest

          votes


















          6












          $begingroup$

          You can see in @MarcoB's answer the enormous speed up possible.



          Even for your code, if you replace the AppendTo with a different structure, you can get orders of magnitude improvement.



          Your code here - note that we don't need l for anything.



          ( For[i = 1; list = {}, i <= 11111, i++, AppendTo[list, i/Pi]];
          
Total@N@list ) //AbsoluteTiming



          {2.12476, 1.96501*10^7}




          Slight modification results in factor of 100 speed up.



          (For[i = 1; list = {}, i <= 11111, i++, list = {list, i/Pi}];
          
Total@N@Flatten@list) // AbsoluteTiming



          {0.0287278, 1.96501*10^7}







          share|improve this answer









          $endgroup$





















            9












            $begingroup$

            Total@Range[11111]/Pi // N
            

            
(*Out: 1.96501*10^7 *)



            In general:




            • avoid For loops: see: Why should I avoid the For loop in Mathematica?

            • avoid Append / AppendTo because they generate a new list every time you add an element; instead, generate lists with Table, Array, Range, Reap / Sow instead;

            • See also: Alternatives to procedural loops and iterating over lists in Mathematica



            Just for some timing context, and to compare For with Do:



            n = 10^6; rpt = RepeatedTiming;
            

            
(For[i = 1; list = {}, i <= n, i++, list = {list, i}]; Total@N@Flatten[list];) // rpt
            
(For[i = 1; list = {}, i <= n, i++, list = {list, i}]; N@Total@Flatten[list];) // rpt
            

            
(list = {}; Do[list = {list, i}, {i, n}]; Total@N@Flatten[list];) // rpt
            
(list = {}; Do[list = {list, i}, {i, n}]; N@Total@Flatten[list];) // rpt
            

            
Total@N@Range[n]; // rpt
            
N@Total@Range[n]; // rpt
            

            
n (n + 1)/2.; // rpt
            

            

            

            
(* Out:
            

            
For loops: 1.35  s
            
           1.40  s
            

            
Do loops:  0.980 s
            
           0.887 s
            

            
Range:     0.01  s
            
           0.007 s
            

            
formula:   1 x 10^-6 s 
            

            
*)





            share|improve this answer











            $endgroup$













            • $begingroup$
              Of course, you don't need to know the formula. Mathematica will work it out for you in fraction of the time it took to do the For loop.
              $endgroup$
              – mikado
              yesterday



















            5












            $begingroup$

            n = 11111;
            
Total@Range[11111]/Pi // N // RepeatedTiming
            
n (n + 1)/(2. Pi)// RepeatedTiming



            {0.000034, 1.96501*10^7}



            {1.2*10^-6, 1.96501*10^7}







            share|improve this answer











            $endgroup$









            • 1




              $begingroup$
              He. Good old Carl Friedrich to the rescue... :-) I'd go for n (n + 1)/2. / Pi then, to avoid N and squeeze the last few microseconds out
              $endgroup$
              – MarcoB
              yesterday













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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6












            $begingroup$

            You can see in @MarcoB's answer the enormous speed up possible.



            Even for your code, if you replace the AppendTo with a different structure, you can get orders of magnitude improvement.



            Your code here - note that we don't need l for anything.



            ( For[i = 1; list = {}, i <= 11111, i++, AppendTo[list, i/Pi]];
            
Total@N@list ) //AbsoluteTiming



            {2.12476, 1.96501*10^7}




            Slight modification results in factor of 100 speed up.



            (For[i = 1; list = {}, i <= 11111, i++, list = {list, i/Pi}];
            
Total@N@Flatten@list) // AbsoluteTiming



            {0.0287278, 1.96501*10^7}







            share|improve this answer









            $endgroup$


















              6












              $begingroup$

              You can see in @MarcoB's answer the enormous speed up possible.



              Even for your code, if you replace the AppendTo with a different structure, you can get orders of magnitude improvement.



              Your code here - note that we don't need l for anything.



              ( For[i = 1; list = {}, i <= 11111, i++, AppendTo[list, i/Pi]];
              
Total@N@list ) //AbsoluteTiming



              {2.12476, 1.96501*10^7}




              Slight modification results in factor of 100 speed up.



              (For[i = 1; list = {}, i <= 11111, i++, list = {list, i/Pi}];
              
Total@N@Flatten@list) // AbsoluteTiming



              {0.0287278, 1.96501*10^7}







              share|improve this answer









              $endgroup$
















                6












                6








                6





                $begingroup$

                You can see in @MarcoB's answer the enormous speed up possible.



                Even for your code, if you replace the AppendTo with a different structure, you can get orders of magnitude improvement.



                Your code here - note that we don't need l for anything.



                ( For[i = 1; list = {}, i <= 11111, i++, AppendTo[list, i/Pi]];
                
Total@N@list ) //AbsoluteTiming



                {2.12476, 1.96501*10^7}




                Slight modification results in factor of 100 speed up.



                (For[i = 1; list = {}, i <= 11111, i++, list = {list, i/Pi}];
                
Total@N@Flatten@list) // AbsoluteTiming



                {0.0287278, 1.96501*10^7}







                share|improve this answer









                $endgroup$



                You can see in @MarcoB's answer the enormous speed up possible.



                Even for your code, if you replace the AppendTo with a different structure, you can get orders of magnitude improvement.



                Your code here - note that we don't need l for anything.



                ( For[i = 1; list = {}, i <= 11111, i++, AppendTo[list, i/Pi]];
                
Total@N@list ) //AbsoluteTiming



                {2.12476, 1.96501*10^7}




                Slight modification results in factor of 100 speed up.



                (For[i = 1; list = {}, i <= 11111, i++, list = {list, i/Pi}];
                
Total@N@Flatten@list) // AbsoluteTiming



                {0.0287278, 1.96501*10^7}








                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered yesterday









                MikeYMikeY

                3,248614




                3,248614























                    9












                    $begingroup$

                    Total@Range[11111]/Pi // N
                    

                    
(*Out: 1.96501*10^7 *)



                    In general:




                    • avoid For loops: see: Why should I avoid the For loop in Mathematica?

                    • avoid Append / AppendTo because they generate a new list every time you add an element; instead, generate lists with Table, Array, Range, Reap / Sow instead;

                    • See also: Alternatives to procedural loops and iterating over lists in Mathematica



                    Just for some timing context, and to compare For with Do:



                    n = 10^6; rpt = RepeatedTiming;
                    

                    
(For[i = 1; list = {}, i <= n, i++, list = {list, i}]; Total@N@Flatten[list];) // rpt
                    
(For[i = 1; list = {}, i <= n, i++, list = {list, i}]; N@Total@Flatten[list];) // rpt
                    

                    
(list = {}; Do[list = {list, i}, {i, n}]; Total@N@Flatten[list];) // rpt
                    
(list = {}; Do[list = {list, i}, {i, n}]; N@Total@Flatten[list];) // rpt
                    

                    
Total@N@Range[n]; // rpt
                    
N@Total@Range[n]; // rpt
                    

                    
n (n + 1)/2.; // rpt
                    

                    

                    

                    
(* Out:
                    

                    
For loops: 1.35  s
                    
           1.40  s
                    

                    
Do loops:  0.980 s
                    
           0.887 s
                    

                    
Range:     0.01  s
                    
           0.007 s
                    

                    
formula:   1 x 10^-6 s 
                    

                    
*)





                    share|improve this answer











                    $endgroup$













                    • $begingroup$
                      Of course, you don't need to know the formula. Mathematica will work it out for you in fraction of the time it took to do the For loop.
                      $endgroup$
                      – mikado
                      yesterday
















                    9












                    $begingroup$

                    Total@Range[11111]/Pi // N
                    

                    
(*Out: 1.96501*10^7 *)



                    In general:




                    • avoid For loops: see: Why should I avoid the For loop in Mathematica?

                    • avoid Append / AppendTo because they generate a new list every time you add an element; instead, generate lists with Table, Array, Range, Reap / Sow instead;

                    • See also: Alternatives to procedural loops and iterating over lists in Mathematica



                    Just for some timing context, and to compare For with Do:



                    n = 10^6; rpt = RepeatedTiming;
                    

                    
(For[i = 1; list = {}, i <= n, i++, list = {list, i}]; Total@N@Flatten[list];) // rpt
                    
(For[i = 1; list = {}, i <= n, i++, list = {list, i}]; N@Total@Flatten[list];) // rpt
                    

                    
(list = {}; Do[list = {list, i}, {i, n}]; Total@N@Flatten[list];) // rpt
                    
(list = {}; Do[list = {list, i}, {i, n}]; N@Total@Flatten[list];) // rpt
                    

                    
Total@N@Range[n]; // rpt
                    
N@Total@Range[n]; // rpt
                    

                    
n (n + 1)/2.; // rpt
                    

                    

                    

                    
(* Out:
                    

                    
For loops: 1.35  s
                    
           1.40  s
                    

                    
Do loops:  0.980 s
                    
           0.887 s
                    

                    
Range:     0.01  s
                    
           0.007 s
                    

                    
formula:   1 x 10^-6 s 
                    

                    
*)





                    share|improve this answer











                    $endgroup$













                    • $begingroup$
                      Of course, you don't need to know the formula. Mathematica will work it out for you in fraction of the time it took to do the For loop.
                      $endgroup$
                      – mikado
                      yesterday














                    9












                    9








                    9





                    $begingroup$

                    Total@Range[11111]/Pi // N
                    

                    
(*Out: 1.96501*10^7 *)



                    In general:




                    • avoid For loops: see: Why should I avoid the For loop in Mathematica?

                    • avoid Append / AppendTo because they generate a new list every time you add an element; instead, generate lists with Table, Array, Range, Reap / Sow instead;

                    • See also: Alternatives to procedural loops and iterating over lists in Mathematica



                    Just for some timing context, and to compare For with Do:



                    n = 10^6; rpt = RepeatedTiming;
                    

                    
(For[i = 1; list = {}, i <= n, i++, list = {list, i}]; Total@N@Flatten[list];) // rpt
                    
(For[i = 1; list = {}, i <= n, i++, list = {list, i}]; N@Total@Flatten[list];) // rpt
                    

                    
(list = {}; Do[list = {list, i}, {i, n}]; Total@N@Flatten[list];) // rpt
                    
(list = {}; Do[list = {list, i}, {i, n}]; N@Total@Flatten[list];) // rpt
                    

                    
Total@N@Range[n]; // rpt
                    
N@Total@Range[n]; // rpt
                    

                    
n (n + 1)/2.; // rpt
                    

                    

                    

                    
(* Out:
                    

                    
For loops: 1.35  s
                    
           1.40  s
                    

                    
Do loops:  0.980 s
                    
           0.887 s
                    

                    
Range:     0.01  s
                    
           0.007 s
                    

                    
formula:   1 x 10^-6 s 
                    

                    
*)





                    share|improve this answer











                    $endgroup$



                    Total@Range[11111]/Pi // N
                    

                    
(*Out: 1.96501*10^7 *)



                    In general:




                    • avoid For loops: see: Why should I avoid the For loop in Mathematica?

                    • avoid Append / AppendTo because they generate a new list every time you add an element; instead, generate lists with Table, Array, Range, Reap / Sow instead;

                    • See also: Alternatives to procedural loops and iterating over lists in Mathematica



                    Just for some timing context, and to compare For with Do:



                    n = 10^6; rpt = RepeatedTiming;
                    

                    
(For[i = 1; list = {}, i <= n, i++, list = {list, i}]; Total@N@Flatten[list];) // rpt
                    
(For[i = 1; list = {}, i <= n, i++, list = {list, i}]; N@Total@Flatten[list];) // rpt
                    

                    
(list = {}; Do[list = {list, i}, {i, n}]; Total@N@Flatten[list];) // rpt
                    
(list = {}; Do[list = {list, i}, {i, n}]; N@Total@Flatten[list];) // rpt
                    

                    
Total@N@Range[n]; // rpt
                    
N@Total@Range[n]; // rpt
                    

                    
n (n + 1)/2.; // rpt
                    

                    

                    

                    
(* Out:
                    

                    
For loops: 1.35  s
                    
           1.40  s
                    

                    
Do loops:  0.980 s
                    
           0.887 s
                    

                    
Range:     0.01  s
                    
           0.007 s
                    

                    
formula:   1 x 10^-6 s 
                    

                    
*)






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited yesterday

























                    answered yesterday









                    MarcoBMarcoB

                    36.7k556113




                    36.7k556113












                    • $begingroup$
                      Of course, you don't need to know the formula. Mathematica will work it out for you in fraction of the time it took to do the For loop.
                      $endgroup$
                      – mikado
                      yesterday


















                    • $begingroup$
                      Of course, you don't need to know the formula. Mathematica will work it out for you in fraction of the time it took to do the For loop.
                      $endgroup$
                      – mikado
                      yesterday
















                    $begingroup$
                    Of course, you don't need to know the formula. Mathematica will work it out for you in fraction of the time it took to do the For loop.
                    $endgroup$
                    – mikado
                    yesterday




                    $begingroup$
                    Of course, you don't need to know the formula. Mathematica will work it out for you in fraction of the time it took to do the For loop.
                    $endgroup$
                    – mikado
                    yesterday











                    5












                    $begingroup$

                    n = 11111;
                    
Total@Range[11111]/Pi // N // RepeatedTiming
                    
n (n + 1)/(2. Pi)// RepeatedTiming



                    {0.000034, 1.96501*10^7}



                    {1.2*10^-6, 1.96501*10^7}







                    share|improve this answer











                    $endgroup$









                    • 1




                      $begingroup$
                      He. Good old Carl Friedrich to the rescue... :-) I'd go for n (n + 1)/2. / Pi then, to avoid N and squeeze the last few microseconds out
                      $endgroup$
                      – MarcoB
                      yesterday


















                    5












                    $begingroup$

                    n = 11111;
                    
Total@Range[11111]/Pi // N // RepeatedTiming
                    
n (n + 1)/(2. Pi)// RepeatedTiming



                    {0.000034, 1.96501*10^7}



                    {1.2*10^-6, 1.96501*10^7}







                    share|improve this answer











                    $endgroup$









                    • 1




                      $begingroup$
                      He. Good old Carl Friedrich to the rescue... :-) I'd go for n (n + 1)/2. / Pi then, to avoid N and squeeze the last few microseconds out
                      $endgroup$
                      – MarcoB
                      yesterday
















                    5












                    5








                    5





                    $begingroup$

                    n = 11111;
                    
Total@Range[11111]/Pi // N // RepeatedTiming
                    
n (n + 1)/(2. Pi)// RepeatedTiming



                    {0.000034, 1.96501*10^7}



                    {1.2*10^-6, 1.96501*10^7}







                    share|improve this answer











                    $endgroup$



                    n = 11111;
                    
Total@Range[11111]/Pi // N // RepeatedTiming
                    
n (n + 1)/(2. Pi)// RepeatedTiming



                    {0.000034, 1.96501*10^7}



                    {1.2*10^-6, 1.96501*10^7}








                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited yesterday

























                    answered yesterday









                    Henrik SchumacherHenrik Schumacher

                    55.4k576154




                    55.4k576154








                    • 1




                      $begingroup$
                      He. Good old Carl Friedrich to the rescue... :-) I'd go for n (n + 1)/2. / Pi then, to avoid N and squeeze the last few microseconds out
                      $endgroup$
                      – MarcoB
                      yesterday
















                    • 1




                      $begingroup$
                      He. Good old Carl Friedrich to the rescue... :-) I'd go for n (n + 1)/2. / Pi then, to avoid N and squeeze the last few microseconds out
                      $endgroup$
                      – MarcoB
                      yesterday










                    1




                    1




                    $begingroup$
                    He. Good old Carl Friedrich to the rescue... :-) I'd go for n (n + 1)/2. / Pi then, to avoid N and squeeze the last few microseconds out
                    $endgroup$
                    – MarcoB
                    yesterday






                    $begingroup$
                    He. Good old Carl Friedrich to the rescue... :-) I'd go for n (n + 1)/2. / Pi then, to avoid N and squeeze the last few microseconds out
                    $endgroup$
                    – MarcoB
                    yesterday




















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