Mfcttrf

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Iterate MapThread with matrices

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Iterate MapThread with matrices

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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

2

$begingroup$

Consider a function f and a list p=a,b,c. I want to get a list of

f[a,a]
f[a,b]
f[a,c]
f[b,a]
f[b,b]
f[b,c]
f[c,a]
f[c,b]
f[c,c]

In real use, p can have higher dimensions, and f may take more arguments, e.g. I may need to generate a list of f[a,a,a,a] through f[z,z,z,z]. Is there a cleaner way to do this other than making lists that approximately repeat the elements of p then use MapThread? e.g.

p=a,b
p1 = a, a, b, b
p2 = a, b, a, b
MapThread[f, p1, p2]

Here, a, b can be matrices

list-manipulation map

share|improve this question

edited 7 hours ago

egwene sedai

asked 8 hours ago

egwene sedaiegwene sedai

1,8631010 silver badges2121 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What about Outer?
  $endgroup$
  – chuy
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @chuy Thanks! but Outer may not be a good fit when a,b ... are matrices.
  $endgroup$
  – egwene sedai
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just realized that p = a, b, c; Distribute[f[p, p], List] works
  $endgroup$
  – egwene sedai
  7 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Consider a function f and a list p=a,b,c. I want to get a list of

f[a,a]
f[a,b]
f[a,c]
f[b,a]
f[b,b]
f[b,c]
f[c,a]
f[c,b]
f[c,c]

In real use, p can have higher dimensions, and f may take more arguments, e.g. I may need to generate a list of f[a,a,a,a] through f[z,z,z,z]. Is there a cleaner way to do this other than making lists that approximately repeat the elements of p then use MapThread? e.g.

p=a,b
p1 = a, a, b, b
p2 = a, b, a, b
MapThread[f, p1, p2]

Here, a, b can be matrices

list-manipulation map

share|improve this question

edited 7 hours ago

egwene sedai

asked 8 hours ago

egwene sedaiegwene sedai

1,8631010 silver badges2121 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What about Outer?
  $endgroup$
  – chuy
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @chuy Thanks! but Outer may not be a good fit when a,b ... are matrices.
  $endgroup$
  – egwene sedai
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just realized that p = a, b, c; Distribute[f[p, p], List] works
  $endgroup$
  – egwene sedai
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Consider a function f and a list p=a,b,c. I want to get a list of

f[a,a]
f[a,b]
f[a,c]
f[b,a]
f[b,b]
f[b,c]
f[c,a]
f[c,b]
f[c,c]

In real use, p can have higher dimensions, and f may take more arguments, e.g. I may need to generate a list of f[a,a,a,a] through f[z,z,z,z]. Is there a cleaner way to do this other than making lists that approximately repeat the elements of p then use MapThread? e.g.

p=a,b
p1 = a, a, b, b
p2 = a, b, a, b
MapThread[f, p1, p2]

Here, a, b can be matrices

list-manipulation map

share|improve this question

edited 7 hours ago

egwene sedai

asked 8 hours ago

egwene sedaiegwene sedai

1,8631010 silver badges2121 bronze badges

$endgroup$

f[a,a]
f[a,b]
f[a,c]
f[b,a]
f[b,b]
f[b,c]
f[c,a]
f[c,b]
f[c,c]

p=a,b
p1 = a, a, b, b
p2 = a, b, a, b
MapThread[f, p1, p2]

share|improve this question

edited 7 hours ago

egwene sedai

asked 8 hours ago

egwene sedaiegwene sedai

1,8631010 silver badges2121 bronze badges

share|improve this question

edited 7 hours ago

egwene sedai

asked 8 hours ago

egwene sedaiegwene sedai

1,8631010 silver badges2121 bronze badges

asked 8 hours ago

egwene sedaiegwene sedai

1,8631010 silver badges2121 bronze badges

asked 8 hours ago

egwene sedaiegwene sedai

1,8631010 silver badges2121 bronze badges

2 Answers
2

active

oldest

votes

5

$begingroup$

You can use Tuples:

Tuples[f[a, b, c, a, b, c]]

f[a, a], f[a, b], f[a, c], f[b, a], f[b, b], f[b, c], f[c, a],
f[c, b], f[c, c]

f @@@ Tuples[a, b, c, 2]

same result

Tuples[f[a, b, r, s, t, x, y]]

f[a, r, x], f[a, r, y], f[a, s, x], f[a, s, y], f[a, t, x],
f[a, t, y], f[b, r, x], f[b, r, y], f[b, s, x], f[b, s, y],
f[b, t, x], f[b, t, y]

f @@@ Tuples[a, b, r, s, t, x, y]

same result

p = a, b;
p1 = a, a, b, b;
Tuples[f[p, p1]]

f[a, a], f[a, a], f[a, b], f[a, b], f[b, a], f[b, a], f[b, b],
f[b, b]

share|improve this answer

edited 8 hours ago

answered 8 hours ago

kglrkglr

200k1010 gold badges228228 silver badges454454 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oops, might need to be changed when a,b,c are matrices?
  $endgroup$
  – egwene sedai
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! But when I use e.g. 2x2 matrices for a, b, c, p=a,b,c, then Tuples[KroneckerProduct[p, p]] // Dimensions is 262144, 9, 4, and does not give 9 4-by-4 matrices...
  $endgroup$
  – egwene sedai
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @egwenesedai, does KroneckerProduct@@@Tuples[p, 2] work?
  $endgroup$
  – kglr
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Works, thanks a lot!
  $endgroup$
  – egwene sedai
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  you can also use Tuples[kp[p, p]] /. kp -> KroneckerProduct (to prevent KroneckerProduct evaluating before Tuples does its job)
  $endgroup$
  – kglr
  7 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

You could also use Tuples as follows:

Tuples[f[a,b,c], 2]

f[a, a], f[a, b], f[a, c], f[b, a], f[b, b], f[b, c], f[c, a], f[c, b],
f[c, c]

If f evaluates, and you want to do this for matrices, you could do:

p = RandomInteger[1, 3, 2, 2];
Block[KroneckerProduct, Tuples[KroneckerProduct @@ p, 2]] //Dimensions

9, 4, 4

share|improve this answer

answered 6 hours ago

Carl WollCarl Woll

84.9k33 gold badges109109 silver badges220220 bronze badges

$endgroup$

add a comment | 

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5

$begingroup$

You can use Tuples:

Tuples[f[a, b, c, a, b, c]]

f[a, a], f[a, b], f[a, c], f[b, a], f[b, b], f[b, c], f[c, a],
f[c, b], f[c, c]

f @@@ Tuples[a, b, c, 2]

same result

Tuples[f[a, b, r, s, t, x, y]]

f[a, r, x], f[a, r, y], f[a, s, x], f[a, s, y], f[a, t, x],
f[a, t, y], f[b, r, x], f[b, r, y], f[b, s, x], f[b, s, y],
f[b, t, x], f[b, t, y]

f @@@ Tuples[a, b, r, s, t, x, y]

same result

p = a, b;
p1 = a, a, b, b;
Tuples[f[p, p1]]

f[a, a], f[a, a], f[a, b], f[a, b], f[b, a], f[b, a], f[b, b],
f[b, b]

share|improve this answer

edited 8 hours ago

answered 8 hours ago

kglrkglr

200k1010 gold badges228228 silver badges454454 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oops, might need to be changed when a,b,c are matrices?
  $endgroup$
  – egwene sedai
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! But when I use e.g. 2x2 matrices for a, b, c, p=a,b,c, then Tuples[KroneckerProduct[p, p]] // Dimensions is 262144, 9, 4, and does not give 9 4-by-4 matrices...
  $endgroup$
  – egwene sedai
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @egwenesedai, does KroneckerProduct@@@Tuples[p, 2] work?
  $endgroup$
  – kglr
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Works, thanks a lot!
  $endgroup$
  – egwene sedai
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  you can also use Tuples[kp[p, p]] /. kp -> KroneckerProduct (to prevent KroneckerProduct evaluating before Tuples does its job)
  $endgroup$
  – kglr
  7 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

You can use Tuples:

Tuples[f[a, b, c, a, b, c]]

f[a, a], f[a, b], f[a, c], f[b, a], f[b, b], f[b, c], f[c, a],
f[c, b], f[c, c]

f @@@ Tuples[a, b, c, 2]

same result

Tuples[f[a, b, r, s, t, x, y]]

f[a, r, x], f[a, r, y], f[a, s, x], f[a, s, y], f[a, t, x],
f[a, t, y], f[b, r, x], f[b, r, y], f[b, s, x], f[b, s, y],
f[b, t, x], f[b, t, y]

f @@@ Tuples[a, b, r, s, t, x, y]

same result

p = a, b;
p1 = a, a, b, b;
Tuples[f[p, p1]]

f[a, a], f[a, a], f[a, b], f[a, b], f[b, a], f[b, a], f[b, b],
f[b, b]

share|improve this answer

edited 8 hours ago

answered 8 hours ago

kglrkglr

200k1010 gold badges228228 silver badges454454 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oops, might need to be changed when a,b,c are matrices?
  $endgroup$
  – egwene sedai
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! But when I use e.g. 2x2 matrices for a, b, c, p=a,b,c, then Tuples[KroneckerProduct[p, p]] // Dimensions is 262144, 9, 4, and does not give 9 4-by-4 matrices...
  $endgroup$
  – egwene sedai
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @egwenesedai, does KroneckerProduct@@@Tuples[p, 2] work?
  $endgroup$
  – kglr
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Works, thanks a lot!
  $endgroup$
  – egwene sedai
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  you can also use Tuples[kp[p, p]] /. kp -> KroneckerProduct (to prevent KroneckerProduct evaluating before Tuples does its job)
  $endgroup$
  – kglr
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

You can use Tuples:

Tuples[f[a, b, c, a, b, c]]

f[a, a], f[a, b], f[a, c], f[b, a], f[b, b], f[b, c], f[c, a],
f[c, b], f[c, c]

f @@@ Tuples[a, b, c, 2]

same result

Tuples[f[a, b, r, s, t, x, y]]

f[a, r, x], f[a, r, y], f[a, s, x], f[a, s, y], f[a, t, x],
f[a, t, y], f[b, r, x], f[b, r, y], f[b, s, x], f[b, s, y],
f[b, t, x], f[b, t, y]

f @@@ Tuples[a, b, r, s, t, x, y]

same result

p = a, b;
p1 = a, a, b, b;
Tuples[f[p, p1]]

f[a, a], f[a, a], f[a, b], f[a, b], f[b, a], f[b, a], f[b, b],
f[b, b]

share|improve this answer

edited 8 hours ago

answered 8 hours ago

kglrkglr

200k1010 gold badges228228 silver badges454454 bronze badges

$endgroup$

Tuples[f[a, b, c, a, b, c]]

f @@@ Tuples[a, b, c, 2]

Tuples[f[a, b, r, s, t, x, y]]

f @@@ Tuples[a, b, r, s, t, x, y]

p = a, b;
p1 = a, a, b, b;
Tuples[f[p, p1]]

share|improve this answer

edited 8 hours ago

answered 8 hours ago

kglrkglr

200k1010 gold badges228228 silver badges454454 bronze badges

answered 8 hours ago

kglrkglr

200k1010 gold badges228228 silver badges454454 bronze badges

answered 8 hours ago

kglrkglr

200k1010 gold badges228228 silver badges454454 bronze badges

2

$begingroup$

You could also use Tuples as follows:

Tuples[f[a,b,c], 2]

f[a, a], f[a, b], f[a, c], f[b, a], f[b, b], f[b, c], f[c, a], f[c, b],
f[c, c]

If f evaluates, and you want to do this for matrices, you could do:

p = RandomInteger[1, 3, 2, 2];
Block[KroneckerProduct, Tuples[KroneckerProduct @@ p, 2]] //Dimensions

9, 4, 4

share|improve this answer

answered 6 hours ago

Carl WollCarl Woll

84.9k33 gold badges109109 silver badges220220 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

You could also use Tuples as follows:

Tuples[f[a,b,c], 2]

f[a, a], f[a, b], f[a, c], f[b, a], f[b, b], f[b, c], f[c, a], f[c, b],
f[c, c]

If f evaluates, and you want to do this for matrices, you could do:

p = RandomInteger[1, 3, 2, 2];
Block[KroneckerProduct, Tuples[KroneckerProduct @@ p, 2]] //Dimensions

9, 4, 4

share|improve this answer

answered 6 hours ago

Carl WollCarl Woll

84.9k33 gold badges109109 silver badges220220 bronze badges

$endgroup$

add a comment | 

$begingroup$

You could also use Tuples as follows:

Tuples[f[a,b,c], 2]

f[a, a], f[a, b], f[a, c], f[b, a], f[b, b], f[b, c], f[c, a], f[c, b],
f[c, c]

If f evaluates, and you want to do this for matrices, you could do:

p = RandomInteger[1, 3, 2, 2];
Block[KroneckerProduct, Tuples[KroneckerProduct @@ p, 2]] //Dimensions

9, 4, 4

share|improve this answer

answered 6 hours ago

Carl WollCarl Woll

84.9k33 gold badges109109 silver badges220220 bronze badges

$endgroup$

Tuples[f[a,b,c], 2]

p = RandomInteger[1, 3, 2, 2];
Block[KroneckerProduct, Tuples[KroneckerProduct @@ p, 2]] //Dimensions

share|improve this answer

answered 6 hours ago

Carl WollCarl Woll

84.9k33 gold badges109109 silver badges220220 bronze badges

answered 6 hours ago

Carl WollCarl Woll

84.9k33 gold badges109109 silver badges220220 bronze badges

answered 6 hours ago

Carl WollCarl Woll

84.9k33 gold badges109109 silver badges220220 bronze badges