Ttdfjt

Word ending in "-ine" for rat-like

Avoiding repetition when using the "snprintf idiom" to write text

Could you fall off a planet if it was being accelerated by engines?

I agreed to cancel a long-planned vacation (with travel costs) due to project deadlines, but now the timeline has all changed again

Making a wall made from glass bricks

How to describe POV characters?

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Can purchasing tickets for high holidays services count as maaser?

What was the point of separating stdout and stderr?

How does mmorpg store data?

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Quantum jump/leap, exist or not, and instantaneous or not (for electrons)?

Origin of the convolution theorem

Is it okay to submit a paper from a master's thesis without informing the advisor?

"I am [the / an] owner of a bookstore"?

A finite 2 group containing the dihedral group of order 16?

When casting a spell with a long casting time, what happens if you don't spend your action on a turn to continue casting?

If two black hole event horizons overlap (touch) can they ever separate again?

Does a lens with a bigger max. aperture focus faster than a lens with a smaller max. aperture?

Does a return economy-class seat between London and San Francisco release 5.28 tonnes of CO2 equivalents?

Adjective for 'made of pus' or 'corrupted by pus' or something of something of pus

/etc/hosts not working

Iterate MapThread with matrices

Equating matrices (or higher order tensors) element-wiseCan I return lists with different dimensions from a compiled function?MapThread multiple functions onto multiple listsFinding elements from a sparse matrixModifying a list of matrices with conditional statementEasy way to perform multiplication of two 2x2 matrices, that contain list elements?Cannot use Part in Pure FunctionFaster position-based duplicate removal in a ragged array?Importing and saving .csv files as variables with the original filebasenamesWhat is the fastest way of combining elemts of lists in rules?

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

4

$begingroup$

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 Jun 20 at 19:30

egwene sedai

asked Jun 20 at 19:08

egwene sedaiegwene sedai

1,8731010 silver badges2121 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What about Outer?
  $endgroup$
  – chuy
  Jun 20 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @chuy Thanks! but Outer may not be a good fit when a,b ... are matrices.
  $endgroup$
  – egwene sedai
  Jun 20 at 19:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just realized that p = a, b, c; Distribute[f[p, p], List] works
  $endgroup$
  – egwene sedai
  Jun 20 at 19:44
  

  
  
  
  

add a comment | 

4

$begingroup$

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 Jun 20 at 19:30

egwene sedai

asked Jun 20 at 19:08

egwene sedaiegwene sedai

1,8731010 silver badges2121 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What about Outer?
  $endgroup$
  – chuy
  Jun 20 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @chuy Thanks! but Outer may not be a good fit when a,b ... are matrices.
  $endgroup$
  – egwene sedai
  Jun 20 at 19:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just realized that p = a, b, c; Distribute[f[p, p], List] works
  $endgroup$
  – egwene sedai
  Jun 20 at 19:44
  

  
  
  
  

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 Jun 20 at 19:30

egwene sedai

asked Jun 20 at 19:08

egwene sedaiegwene sedai

1,8731010 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 Jun 20 at 19:30

egwene sedai

asked Jun 20 at 19:08

egwene sedaiegwene sedai

1,8731010 silver badges2121 bronze badges

share|improve this question

edited Jun 20 at 19:30

egwene sedai

asked Jun 20 at 19:08

egwene sedaiegwene sedai

1,8731010 silver badges2121 bronze badges

asked Jun 20 at 19:08

egwene sedaiegwene sedai

1,8731010 silver badges2121 bronze badges

asked Jun 20 at 19:08

egwene sedaiegwene sedai

1,8731010 silver badges2121 bronze badges

2 Answers
2

active

oldest

votes

9

$begingroup$

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 Jun 20 at 19:18

answered Jun 20 at 19:13

kglrkglr

201k1010 gold badges230230 silver badges458458 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oops, might need to be changed when a,b,c are matrices?
  $endgroup$
  – egwene sedai
  Jun 20 at 19:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Jun 20 at 19:26
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @egwenesedai, does KroneckerProduct@@@Tuples[p, 2] work?
  $endgroup$
  – kglr
  Jun 20 at 19:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Works, thanks a lot!
  $endgroup$
  – egwene sedai
  Jun 20 at 19:29
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  you can also use Tuples[kp[p, p]] /. kp -> KroneckerProduct (to prevent KroneckerProduct evaluating before Tuples does its job)
  $endgroup$
  – kglr
  Jun 20 at 19:31
  

  
  
  
  

add a comment | 

6

$begingroup$

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 Jun 20 at 20:47

Carl WollCarl Woll

86k33 gold badges110110 silver badges220220 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! That's one clever use for Block
  $endgroup$
  – egwene sedai
  Jun 21 at 21:33
  

  
  
  
  

add a comment | 

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9

$begingroup$

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 Jun 20 at 19:18

answered Jun 20 at 19:13

kglrkglr

201k1010 gold badges230230 silver badges458458 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oops, might need to be changed when a,b,c are matrices?
  $endgroup$
  – egwene sedai
  Jun 20 at 19:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Jun 20 at 19:26
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @egwenesedai, does KroneckerProduct@@@Tuples[p, 2] work?
  $endgroup$
  – kglr
  Jun 20 at 19:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Works, thanks a lot!
  $endgroup$
  – egwene sedai
  Jun 20 at 19:29
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  you can also use Tuples[kp[p, p]] /. kp -> KroneckerProduct (to prevent KroneckerProduct evaluating before Tuples does its job)
  $endgroup$
  – kglr
  Jun 20 at 19:31
  

  
  
  
  

add a comment | 

9

$begingroup$

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 Jun 20 at 19:18

answered Jun 20 at 19:13

kglrkglr

201k1010 gold badges230230 silver badges458458 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oops, might need to be changed when a,b,c are matrices?
  $endgroup$
  – egwene sedai
  Jun 20 at 19:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Jun 20 at 19:26
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @egwenesedai, does KroneckerProduct@@@Tuples[p, 2] work?
  $endgroup$
  – kglr
  Jun 20 at 19:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Works, thanks a lot!
  $endgroup$
  – egwene sedai
  Jun 20 at 19:29
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  you can also use Tuples[kp[p, p]] /. kp -> KroneckerProduct (to prevent KroneckerProduct evaluating before Tuples does its job)
  $endgroup$
  – kglr
  Jun 20 at 19:31
  

  
  
  
  

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 Jun 20 at 19:18

answered Jun 20 at 19:13

kglrkglr

201k1010 gold badges230230 silver badges458458 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 Jun 20 at 19:18

answered Jun 20 at 19:13

kglrkglr

201k1010 gold badges230230 silver badges458458 bronze badges

answered Jun 20 at 19:13

kglrkglr

201k1010 gold badges230230 silver badges458458 bronze badges

answered Jun 20 at 19:13

kglrkglr

201k1010 gold badges230230 silver badges458458 bronze badges

6

$begingroup$

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 Jun 20 at 20:47

Carl WollCarl Woll

86k33 gold badges110110 silver badges220220 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! That's one clever use for Block
  $endgroup$
  – egwene sedai
  Jun 21 at 21:33
  

  
  
  
  

add a comment | 

6

$begingroup$

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 Jun 20 at 20:47

Carl WollCarl Woll

86k33 gold badges110110 silver badges220220 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! That's one clever use for Block
  $endgroup$
  – egwene sedai
  Jun 21 at 21:33
  

  
  
  
  

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 Jun 20 at 20:47

Carl WollCarl Woll

86k33 gold badges110110 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 Jun 20 at 20:47

Carl WollCarl Woll

86k33 gold badges110110 silver badges220220 bronze badges

answered Jun 20 at 20:47

Carl WollCarl Woll

86k33 gold badges110110 silver badges220220 bronze badges

answered Jun 20 at 20:47

Carl WollCarl Woll

86k33 gold badges110110 silver badges220220 bronze badges