Performance-tuning a simple 2D Table construct

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I've recently been playing with some performance tuning over a seemingly simple construction where I build a 2D array of Sin[x*y] for x, y∈Interval[0, 2π, 0, 2π].

I think it's also a good example of how vastly different results one can get depending on how efficiently the operation is performed.

Here's the operation in the simplest/least efficient way I could think up:

Table[N@Sin[x*y], 
 x, 0, 2π, π/250, 
 y, 0, 2π, π/250
 ]; // RepeatedTiming // First

2.01

Now, how can we speed this up? (or slow it down in non-intuitive ways)

For reference, the best I managed to get was 0.0011 but all answers are good answers, especially if they provide analysis of how they speed up/slow down the problem.

Also feel free to cut down on the number of points 250 was simply most evocative.

performance-tuning

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edited 35 mins ago

asked 41 mins ago

b3m2a1

25.2k254148

add a comment | 

up vote
1
down vote

favorite

I've recently been playing with some performance tuning over a seemingly simple construction where I build a 2D array of Sin[x*y] for x, y∈Interval[0, 2π, 0, 2π].

I think it's also a good example of how vastly different results one can get depending on how efficiently the operation is performed.

Here's the operation in the simplest/least efficient way I could think up:

Table[N@Sin[x*y], 
 x, 0, 2π, π/250, 
 y, 0, 2π, π/250
 ]; // RepeatedTiming // First

2.01

Now, how can we speed this up? (or slow it down in non-intuitive ways)

For reference, the best I managed to get was 0.0011 but all answers are good answers, especially if they provide analysis of how they speed up/slow down the problem.

Also feel free to cut down on the number of points 250 was simply most evocative.

performance-tuning

share|improve this question

edited 35 mins ago

asked 41 mins ago

b3m2a1

25.2k254148

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I've recently been playing with some performance tuning over a seemingly simple construction where I build a 2D array of Sin[x*y] for x, y∈Interval[0, 2π, 0, 2π].

I think it's also a good example of how vastly different results one can get depending on how efficiently the operation is performed.

Here's the operation in the simplest/least efficient way I could think up:

Table[N@Sin[x*y], 
 x, 0, 2π, π/250, 
 y, 0, 2π, π/250
 ]; // RepeatedTiming // First

2.01

Now, how can we speed this up? (or slow it down in non-intuitive ways)

For reference, the best I managed to get was 0.0011 but all answers are good answers, especially if they provide analysis of how they speed up/slow down the problem.

Also feel free to cut down on the number of points 250 was simply most evocative.

performance-tuning

share|improve this question

edited 35 mins ago

asked 41 mins ago

b3m2a1

25.2k254148

I've recently been playing with some performance tuning over a seemingly simple construction where I build a 2D array of Sin[x*y] for x, y∈Interval[0, 2π, 0, 2π].

I think it's also a good example of how vastly different results one can get depending on how efficiently the operation is performed.

Here's the operation in the simplest/least efficient way I could think up:

Table[N@Sin[x*y], 
 x, 0, 2π, π/250, 
 y, 0, 2π, π/250
 ]; // RepeatedTiming // First

2.01

Now, how can we speed this up? (or slow it down in non-intuitive ways)

For reference, the best I managed to get was 0.0011 but all answers are good answers, especially if they provide analysis of how they speed up/slow down the problem.

Also feel free to cut down on the number of points 250 was simply most evocative.

performance-tuning

performance-tuning

share|improve this question

edited 35 mins ago

asked 41 mins ago

b3m2a1

25.2k254148

share|improve this question

edited 35 mins ago

asked 41 mins ago

b3m2a1

25.2k254148

share|improve this question

share|improve this question

edited 35 mins ago

edited 35 mins ago

edited 35 mins ago

asked 41 mins ago

b3m2a1

25.2k254148

asked 41 mins ago

b3m2a1

25.2k254148

asked 41 mins ago

b3m2a1

25.2k254148

25.2k254148

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote

Use packed arrays, and vector operations. Here are 2 possibilities:

Table[N@Sin[x*y],x,0,2π,π/12,y,0,2π,π/12]; //RepeatedTiming//First

With[pa=Developer`ToPackedArray @ N @ Range[0,2π,π/12],
 Sin @ Outer[Times, pa, pa]
]; //RepeatedTiming//First

With[pa=Developer`ToPackedArray @ N @ Range[0,2π,π/12],
 Sin @ KroneckerProduct[pa, pa]
]; //RepeatedTiming//First

0.0025

0.0000322

0.0000314

Update

It seems that converting to a packed array takes the most time, so here is a version that avoids packing:

With[pa = Range[0, 24] N[Pi/12],
 Sin @ KroneckerProduct[pa, pa]
]; //RepeatedTiming //First

8.3*10^-6

share|improve this answer

edited 21 mins ago

answered 35 mins ago

Carl Woll

63.4k282163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I like the use of KroneckerProduct. I hadn't thought to use that.
  – b3m2a1
  33 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Range[0., 2 [Pi], 2 [Pi]/ 500] or Subdivide[0., 2 Pi, 500] are even faster because they generate packed arrays right from the start.
  – Henrik Schumacher
  19 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenrikSchumacher oh good thinking with Subdivide. My brain just immediately jumps to Range but there's really no reason for that. It also looks like Subdivide blows Range out of the water here...
  – b3m2a1
  16 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenrikSchumacher another odd subtle point... Range[0., 2. [Pi], [Pi]/250] is much slower than Range[0., 2. [Pi], [Pi]/250.]
  – b3m2a1
  12 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

With[r = Range[0., 2Pi, Pi/12.] , Sin @ Outer[Times, r, r]];// RepeatedTiming// First

0.000014

share|improve this answer

answered 28 mins ago

kglr

169k8192395

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

Use packed arrays, and vector operations. Here are 2 possibilities:

Table[N@Sin[x*y],x,0,2π,π/12,y,0,2π,π/12]; //RepeatedTiming//First

With[pa=Developer`ToPackedArray @ N @ Range[0,2π,π/12],
 Sin @ Outer[Times, pa, pa]
]; //RepeatedTiming//First

With[pa=Developer`ToPackedArray @ N @ Range[0,2π,π/12],
 Sin @ KroneckerProduct[pa, pa]
]; //RepeatedTiming//First

0.0025

0.0000322

0.0000314

Update

It seems that converting to a packed array takes the most time, so here is a version that avoids packing:

With[pa = Range[0, 24] N[Pi/12],
 Sin @ KroneckerProduct[pa, pa]
]; //RepeatedTiming //First

8.3*10^-6

share|improve this answer

edited 21 mins ago

answered 35 mins ago

Carl Woll

63.4k282163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I like the use of KroneckerProduct. I hadn't thought to use that.
  – b3m2a1
  33 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Range[0., 2 [Pi], 2 [Pi]/ 500] or Subdivide[0., 2 Pi, 500] are even faster because they generate packed arrays right from the start.
  – Henrik Schumacher
  19 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenrikSchumacher oh good thinking with Subdivide. My brain just immediately jumps to Range but there's really no reason for that. It also looks like Subdivide blows Range out of the water here...
  – b3m2a1
  16 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenrikSchumacher another odd subtle point... Range[0., 2. [Pi], [Pi]/250] is much slower than Range[0., 2. [Pi], [Pi]/250.]
  – b3m2a1
  12 mins ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

Use packed arrays, and vector operations. Here are 2 possibilities:

Table[N@Sin[x*y],x,0,2π,π/12,y,0,2π,π/12]; //RepeatedTiming//First

With[pa=Developer`ToPackedArray @ N @ Range[0,2π,π/12],
 Sin @ Outer[Times, pa, pa]
]; //RepeatedTiming//First

With[pa=Developer`ToPackedArray @ N @ Range[0,2π,π/12],
 Sin @ KroneckerProduct[pa, pa]
]; //RepeatedTiming//First

0.0025

0.0000322

0.0000314

Update

It seems that converting to a packed array takes the most time, so here is a version that avoids packing:

With[pa = Range[0, 24] N[Pi/12],
 Sin @ KroneckerProduct[pa, pa]
]; //RepeatedTiming //First

8.3*10^-6

share|improve this answer

edited 21 mins ago

answered 35 mins ago

Carl Woll

63.4k282163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I like the use of KroneckerProduct. I hadn't thought to use that.
  – b3m2a1
  33 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Range[0., 2 [Pi], 2 [Pi]/ 500] or Subdivide[0., 2 Pi, 500] are even faster because they generate packed arrays right from the start.
  – Henrik Schumacher
  19 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenrikSchumacher oh good thinking with Subdivide. My brain just immediately jumps to Range but there's really no reason for that. It also looks like Subdivide blows Range out of the water here...
  – b3m2a1
  16 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenrikSchumacher another odd subtle point... Range[0., 2. [Pi], [Pi]/250] is much slower than Range[0., 2. [Pi], [Pi]/250.]
  – b3m2a1
  12 mins ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Use packed arrays, and vector operations. Here are 2 possibilities:

Table[N@Sin[x*y],x,0,2π,π/12,y,0,2π,π/12]; //RepeatedTiming//First

With[pa=Developer`ToPackedArray @ N @ Range[0,2π,π/12],
 Sin @ Outer[Times, pa, pa]
]; //RepeatedTiming//First

With[pa=Developer`ToPackedArray @ N @ Range[0,2π,π/12],
 Sin @ KroneckerProduct[pa, pa]
]; //RepeatedTiming//First

0.0025

0.0000322

0.0000314

Update

It seems that converting to a packed array takes the most time, so here is a version that avoids packing:

With[pa = Range[0, 24] N[Pi/12],
 Sin @ KroneckerProduct[pa, pa]
]; //RepeatedTiming //First

8.3*10^-6

share|improve this answer

edited 21 mins ago

answered 35 mins ago

Carl Woll

63.4k282163

Use packed arrays, and vector operations. Here are 2 possibilities:

Table[N@Sin[x*y],x,0,2π,π/12,y,0,2π,π/12]; //RepeatedTiming//First

With[pa=Developer`ToPackedArray @ N @ Range[0,2π,π/12],
 Sin @ Outer[Times, pa, pa]
]; //RepeatedTiming//First

With[pa=Developer`ToPackedArray @ N @ Range[0,2π,π/12],
 Sin @ KroneckerProduct[pa, pa]
]; //RepeatedTiming//First

0.0025

0.0000322

0.0000314

Update

It seems that converting to a packed array takes the most time, so here is a version that avoids packing:

With[pa = Range[0, 24] N[Pi/12],
 Sin @ KroneckerProduct[pa, pa]
]; //RepeatedTiming //First

8.3*10^-6

share|improve this answer

edited 21 mins ago

answered 35 mins ago

Carl Woll

63.4k282163

share|improve this answer

share|improve this answer

edited 21 mins ago

edited 21 mins ago

edited 21 mins ago

answered 35 mins ago

Carl Woll

63.4k282163

answered 35 mins ago

Carl Woll

63.4k282163

answered 35 mins ago

Carl Woll

63.4k282163

63.4k282163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I like the use of KroneckerProduct. I hadn't thought to use that.
  – b3m2a1
  33 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Range[0., 2 [Pi], 2 [Pi]/ 500] or Subdivide[0., 2 Pi, 500] are even faster because they generate packed arrays right from the start.
  – Henrik Schumacher
  19 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenrikSchumacher oh good thinking with Subdivide. My brain just immediately jumps to Range but there's really no reason for that. It also looks like Subdivide blows Range out of the water here...
  – b3m2a1
  16 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenrikSchumacher another odd subtle point... Range[0., 2. [Pi], [Pi]/250] is much slower than Range[0., 2. [Pi], [Pi]/250.]
  – b3m2a1
  12 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I like the use of KroneckerProduct. I hadn't thought to use that.
  – b3m2a1
  33 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Range[0., 2 [Pi], 2 [Pi]/ 500] or Subdivide[0., 2 Pi, 500] are even faster because they generate packed arrays right from the start.
  – Henrik Schumacher
  19 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenrikSchumacher oh good thinking with Subdivide. My brain just immediately jumps to Range but there's really no reason for that. It also looks like Subdivide blows Range out of the water here...
  – b3m2a1
  16 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenrikSchumacher another odd subtle point... Range[0., 2. [Pi], [Pi]/250] is much slower than Range[0., 2. [Pi], [Pi]/250.]
  – b3m2a1
  12 mins ago
  

  
  
  
  

I like the use of KroneckerProduct. I hadn't thought to use that.
– b3m2a1
33 mins ago

I like the use of KroneckerProduct. I hadn't thought to use that.
– b3m2a1
33 mins ago

1

1

Range[0., 2 [Pi], 2 [Pi]/ 500] or Subdivide[0., 2 Pi, 500] are even faster because they generate packed arrays right from the start.
– Henrik Schumacher
19 mins ago

Range[0., 2 [Pi], 2 [Pi]/ 500] or Subdivide[0., 2 Pi, 500] are even faster because they generate packed arrays right from the start.
– Henrik Schumacher
19 mins ago

@HenrikSchumacher oh good thinking with Subdivide. My brain just immediately jumps to Range but there's really no reason for that. It also looks like Subdivide blows Range out of the water here...
– b3m2a1
16 mins ago

@HenrikSchumacher oh good thinking with Subdivide. My brain just immediately jumps to Range but there's really no reason for that. It also looks like Subdivide blows Range out of the water here...
– b3m2a1
16 mins ago

@HenrikSchumacher another odd subtle point... Range[0., 2. [Pi], [Pi]/250] is much slower than Range[0., 2. [Pi], [Pi]/250.]
– b3m2a1
12 mins ago

@HenrikSchumacher another odd subtle point... Range[0., 2. [Pi], [Pi]/250] is much slower than Range[0., 2. [Pi], [Pi]/250.]
– b3m2a1
12 mins ago

add a comment | 

up vote
2
down vote

With[r = Range[0., 2Pi, Pi/12.] , Sin @ Outer[Times, r, r]];// RepeatedTiming// First

0.000014

share|improve this answer

answered 28 mins ago

kglr

169k8192395

add a comment | 

up vote
2
down vote

With[r = Range[0., 2Pi, Pi/12.] , Sin @ Outer[Times, r, r]];// RepeatedTiming// First

0.000014

share|improve this answer

answered 28 mins ago

kglr

169k8192395

add a comment | 

up vote
2
down vote

up vote
2
down vote

With[r = Range[0., 2Pi, Pi/12.] , Sin @ Outer[Times, r, r]];// RepeatedTiming// First

0.000014

share|improve this answer

answered 28 mins ago

kglr

169k8192395

With[r = Range[0., 2Pi, Pi/12.] , Sin @ Outer[Times, r, r]];// RepeatedTiming// First

0.000014

share|improve this answer

answered 28 mins ago

kglr

169k8192395

share|improve this answer

share|improve this answer

answered 28 mins ago

kglr

169k8192395

answered 28 mins ago

kglr

169k8192395

answered 28 mins ago

kglr

169k8192395

169k8192395

add a comment | 

add a comment | 

 

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