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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

asked 7 hours ago

LittleFingerLittleFinger

864

$endgroup$

add a comment | 

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

asked 7 hours ago

LittleFingerLittleFinger

864

$endgroup$

add a comment | 

$begingroup$

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

asked 7 hours ago

LittleFingerLittleFinger

864

$endgroup$

Timing[For[i = 1;



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

 Total@N@list]

share|improve this question

asked 7 hours ago

LittleFingerLittleFinger

864

share|improve this question

asked 7 hours ago

LittleFingerLittleFinger

864

asked 7 hours ago

LittleFingerLittleFinger

864

asked 7 hours ago

LittleFingerLittleFinger

864

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

answered 6 hours ago

MikeYMikeY

3,208514

$endgroup$

add a comment | 

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

edited 5 hours ago

answered 7 hours ago

MarcoBMarcoB

36.7k556112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 hours ago
  

  
  
  
  

add a comment | 

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

edited 6 hours ago

answered 6 hours ago

Henrik SchumacherHenrik Schumacher

55.4k576154

$endgroup$

• 
  
  
  

  
  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
  6 hours ago
  
  
  

  
  
  
  

add a comment | 

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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

answered 6 hours ago

MikeYMikeY

3,208514

$endgroup$

add a comment | 

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

answered 6 hours ago

MikeYMikeY

3,208514

$endgroup$

add a comment | 

$begingroup$

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

answered 6 hours ago

MikeYMikeY

3,208514

$endgroup$

( For[i = 1; list = {}, i <= 11111, i++, AppendTo[list, i/Pi]];

Total@N@list ) //AbsoluteTiming

(For[i = 1; list = {}, i <= 11111, i++, list = {list, i/Pi}];

Total@N@Flatten@list) // AbsoluteTiming

share|improve this answer

answered 6 hours ago

MikeYMikeY

3,208514

answered 6 hours ago

MikeYMikeY

3,208514

answered 6 hours ago

MikeYMikeY

3,208514

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

edited 5 hours ago

answered 7 hours ago

MarcoBMarcoB

36.7k556112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 hours ago
  

  
  
  
  

add a comment | 

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

edited 5 hours ago

answered 7 hours ago

MarcoBMarcoB

36.7k556112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 5 hours ago

answered 7 hours ago

MarcoBMarcoB

36.7k556112

$endgroup$

Total@Range[11111]/Pi // N



(*Out: 1.96501*10^7 *)

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

edited 5 hours ago

answered 7 hours ago

MarcoBMarcoB

36.7k556112

answered 7 hours ago

MarcoBMarcoB

36.7k556112

answered 7 hours ago

MarcoBMarcoB

36.7k556112

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

edited 6 hours ago

answered 6 hours ago

Henrik SchumacherHenrik Schumacher

55.4k576154

$endgroup$

• 
  
  
  

  
  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
  6 hours ago
  
  
  

  
  
  
  

add a comment | 

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

edited 6 hours ago

answered 6 hours ago

Henrik SchumacherHenrik Schumacher

55.4k576154

$endgroup$

• 
  
  
  

  
  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
  6 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 6 hours ago

answered 6 hours ago

Henrik SchumacherHenrik Schumacher

55.4k576154

$endgroup$

n = 11111;

Total@Range[11111]/Pi // N // RepeatedTiming

n (n + 1)/(2. Pi)// RepeatedTiming

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Henrik SchumacherHenrik Schumacher

55.4k576154

answered 6 hours ago

Henrik SchumacherHenrik Schumacher

55.4k576154

answered 6 hours ago

Henrik SchumacherHenrik Schumacher

55.4k576154