Yhyjyk

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2

$begingroup$

I would like to know what is the most efficient to implement the following computation. Given three lists

 a = a_1,a_2, a_3, …, a_n
 b = b_1,b_2, b_3, …, b_n
 c = c_1,c_2, c_3, …, c_n 

and a function $f(x_1,x_2,x_3)$, obtain

 f(a_1,b_1,c_1) f(a_1,b_1,c_2) ..... f(a_1,b_1,c_n) 
 f(a_2,b_2,c_1) f(a_2,b_2,c_2) ..... f(a_2,b_2,c_n) 
 ..... ..... ..... .....
 f(a_n,b_n,c_1) f(a_n,b_n,c_2) ..... f(a_n,b_n,c_n) 

I cannot find a solution not using For.

list-manipulation

share|improve this question

edited 2 hours ago

corey979

20.9k64382

asked 2 hours ago

SmerdjakovSmerdjakov

1305

$endgroup$

add a comment | 

2

$begingroup$

I would like to know what is the most efficient to implement the following computation. Given three lists

 a = a_1,a_2, a_3, …, a_n
 b = b_1,b_2, b_3, …, b_n
 c = c_1,c_2, c_3, …, c_n 

and a function $f(x_1,x_2,x_3)$, obtain

 f(a_1,b_1,c_1) f(a_1,b_1,c_2) ..... f(a_1,b_1,c_n) 
 f(a_2,b_2,c_1) f(a_2,b_2,c_2) ..... f(a_2,b_2,c_n) 
 ..... ..... ..... .....
 f(a_n,b_n,c_1) f(a_n,b_n,c_2) ..... f(a_n,b_n,c_n) 

I cannot find a solution not using For.

list-manipulation

share|improve this question

edited 2 hours ago

corey979

20.9k64382

asked 2 hours ago

SmerdjakovSmerdjakov

1305

$endgroup$

add a comment | 

$begingroup$

I would like to know what is the most efficient to implement the following computation. Given three lists

 a = a_1,a_2, a_3, …, a_n
 b = b_1,b_2, b_3, …, b_n
 c = c_1,c_2, c_3, …, c_n 

and a function $f(x_1,x_2,x_3)$, obtain

 f(a_1,b_1,c_1) f(a_1,b_1,c_2) ..... f(a_1,b_1,c_n) 
 f(a_2,b_2,c_1) f(a_2,b_2,c_2) ..... f(a_2,b_2,c_n) 
 ..... ..... ..... .....
 f(a_n,b_n,c_1) f(a_n,b_n,c_2) ..... f(a_n,b_n,c_n) 

I cannot find a solution not using For.

list-manipulation

share|improve this question

edited 2 hours ago

corey979

20.9k64382

asked 2 hours ago

SmerdjakovSmerdjakov

1305

$endgroup$

 a = a_1,a_2, a_3, …, a_n
 b = b_1,b_2, b_3, …, b_n
 c = c_1,c_2, c_3, …, c_n 

 f(a_1,b_1,c_1) f(a_1,b_1,c_2) ..... f(a_1,b_1,c_n) 
 f(a_2,b_2,c_1) f(a_2,b_2,c_2) ..... f(a_2,b_2,c_n) 
 ..... ..... ..... .....
 f(a_n,b_n,c_1) f(a_n,b_n,c_2) ..... f(a_n,b_n,c_n) 

share|improve this question

edited 2 hours ago

corey979

20.9k64382

asked 2 hours ago

SmerdjakovSmerdjakov

1305

share|improve this question

edited 2 hours ago

corey979

20.9k64382

asked 2 hours ago

SmerdjakovSmerdjakov

1305

edited 2 hours ago

corey979

20.9k64382

edited 2 hours ago

corey979

20.9k64382

asked 2 hours ago

SmerdjakovSmerdjakov

1305

asked 2 hours ago

SmerdjakovSmerdjakov

1305

3 Answers
3

active

oldest

votes

4

$begingroup$

Here's one way to do it with Outer:

n = 3;
l1 = Array[a, n];
l2 = Array[b, n];
l3 = Array[c, n];

Outer[
 f[#1[[1]], #1[[2]], #2] &,
 Transpose @ l1, l2,
 l3,
 1
]

Out[25]= f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]

share|improve this answer

answered 2 hours ago

Sjoerd SmitSjoerd Smit

4,610817

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or Outer[f[Sequence @@ #1, #2] &, Transpose@l1, l2, l3, 1] so you don't need to unravel #1 manually.
  $endgroup$
  – Roman
  2 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

a = a1, a2, a3, a4, a5;
b = b1, b2, b3, b4, b5;
c = c1, c2, c3, c4, c5;

Table[f[a[[j]], b[[j]], c[[k]]], j, 1, 5, k, 1, 5]

share|improve this answer

answered 2 hours ago

corey979corey979

20.9k64382

$endgroup$

add a comment | 

2

$begingroup$

Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:

n = 3;
l1 = Array[a,n];
l2 = Array[b,n];
l3 = Array[c,n];

Using Thread:

Thread /@ Thread[f[l1, l2, l3], List, 2]

f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]

share|improve this answer

answered 1 hour ago

Carl WollCarl Woll

75.9k3100198

$endgroup$

add a comment | 

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4

$begingroup$

Here's one way to do it with Outer:

n = 3;
l1 = Array[a, n];
l2 = Array[b, n];
l3 = Array[c, n];

Outer[
 f[#1[[1]], #1[[2]], #2] &,
 Transpose @ l1, l2,
 l3,
 1
]

Out[25]= f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]

share|improve this answer

answered 2 hours ago

Sjoerd SmitSjoerd Smit

4,610817

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or Outer[f[Sequence @@ #1, #2] &, Transpose@l1, l2, l3, 1] so you don't need to unravel #1 manually.
  $endgroup$
  – Roman
  2 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Here's one way to do it with Outer:

n = 3;
l1 = Array[a, n];
l2 = Array[b, n];
l3 = Array[c, n];

Outer[
 f[#1[[1]], #1[[2]], #2] &,
 Transpose @ l1, l2,
 l3,
 1
]

Out[25]= f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]

share|improve this answer

answered 2 hours ago

Sjoerd SmitSjoerd Smit

4,610817

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or Outer[f[Sequence @@ #1, #2] &, Transpose@l1, l2, l3, 1] so you don't need to unravel #1 manually.
  $endgroup$
  – Roman
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Here's one way to do it with Outer:

n = 3;
l1 = Array[a, n];
l2 = Array[b, n];
l3 = Array[c, n];

Outer[
 f[#1[[1]], #1[[2]], #2] &,
 Transpose @ l1, l2,
 l3,
 1
]

Out[25]= f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]

share|improve this answer

answered 2 hours ago

Sjoerd SmitSjoerd Smit

4,610817

$endgroup$

n = 3;
l1 = Array[a, n];
l2 = Array[b, n];
l3 = Array[c, n];

Outer[
 f[#1[[1]], #1[[2]], #2] &,
 Transpose @ l1, l2,
 l3,
 1
]

share|improve this answer

answered 2 hours ago

Sjoerd SmitSjoerd Smit

4,610817

answered 2 hours ago

Sjoerd SmitSjoerd Smit

4,610817

answered 2 hours ago

Sjoerd SmitSjoerd Smit

4,610817

2

$begingroup$

a = a1, a2, a3, a4, a5;
b = b1, b2, b3, b4, b5;
c = c1, c2, c3, c4, c5;

Table[f[a[[j]], b[[j]], c[[k]]], j, 1, 5, k, 1, 5]

share|improve this answer

answered 2 hours ago

corey979corey979

20.9k64382

$endgroup$

add a comment | 

2

$begingroup$

a = a1, a2, a3, a4, a5;
b = b1, b2, b3, b4, b5;
c = c1, c2, c3, c4, c5;

Table[f[a[[j]], b[[j]], c[[k]]], j, 1, 5, k, 1, 5]

share|improve this answer

answered 2 hours ago

corey979corey979

20.9k64382

$endgroup$

add a comment | 

$begingroup$

a = a1, a2, a3, a4, a5;
b = b1, b2, b3, b4, b5;
c = c1, c2, c3, c4, c5;

Table[f[a[[j]], b[[j]], c[[k]]], j, 1, 5, k, 1, 5]

share|improve this answer

answered 2 hours ago

corey979corey979

20.9k64382

$endgroup$

a = a1, a2, a3, a4, a5;
b = b1, b2, b3, b4, b5;
c = c1, c2, c3, c4, c5;

Table[f[a[[j]], b[[j]], c[[k]]], j, 1, 5, k, 1, 5]

share|improve this answer

answered 2 hours ago

corey979corey979

20.9k64382

answered 2 hours ago

corey979corey979

20.9k64382

answered 2 hours ago

corey979corey979

20.9k64382

2

$begingroup$

Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:

n = 3;
l1 = Array[a,n];
l2 = Array[b,n];
l3 = Array[c,n];

Using Thread:

Thread /@ Thread[f[l1, l2, l3], List, 2]

f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]

share|improve this answer

answered 1 hour ago

Carl WollCarl Woll

75.9k3100198

$endgroup$

add a comment | 

2

$begingroup$

Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:

n = 3;
l1 = Array[a,n];
l2 = Array[b,n];
l3 = Array[c,n];

Using Thread:

Thread /@ Thread[f[l1, l2, l3], List, 2]

f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]

share|improve this answer

answered 1 hour ago

Carl WollCarl Woll

75.9k3100198

$endgroup$

add a comment | 

$begingroup$

Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:

n = 3;
l1 = Array[a,n];
l2 = Array[b,n];
l3 = Array[c,n];

Using Thread:

Thread /@ Thread[f[l1, l2, l3], List, 2]

f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]

share|improve this answer

answered 1 hour ago

Carl WollCarl Woll

75.9k3100198

$endgroup$

n = 3;
l1 = Array[a,n];
l2 = Array[b,n];
l3 = Array[c,n];

Thread /@ Thread[f[l1, l2, l3], List, 2]

share|improve this answer

answered 1 hour ago

Carl WollCarl Woll

75.9k3100198

answered 1 hour ago

Carl WollCarl Woll

75.9k3100198

answered 1 hour ago

Carl WollCarl Woll

75.9k3100198