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How to append a matrix element by element

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How to append a matrix element by element

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3

$begingroup$

Consider if I have some matrices constructed as such:

A = 
 , ,
 , 
 
B = 
 a , b ,
 c , d 
 
C = 
 e , f ,
 g , h 
 
...

A serves as a matrix of empty lists. I want to be able to append B, C, etc. to A, to obtain a final matrix like

A = 
 a, e, ... , b, f, ... ,
 c, g, ... , d, h, ... 
 

How can I do this? My matrices are always square and of the same dimensions. An alternative which could also be useful is to Join two matrices element by element, if the matrix elements of both are already lists.

list-manipulation matrix

share|improve this question

edited 1 hour ago

Kai

asked 8 hours ago

KaiKai

59411 silver badge99 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You wrote "My matrices are always square and of the same dimensions". That is not true. Try Dimensions/@A,B
  $endgroup$
  – Fortsaint
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Kai, with A and B as inputs what is the desired result: a, b, c, d or , a, , b, , c, , d?
  $endgroup$
  – kglr
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Fortsaint my meaning is that the matrices which are being appended are always the same dimension. Matrix A is simply a matrix of empty lists to append to.
  $endgroup$
  – Kai
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kglr see the update
  $endgroup$
  – Kai
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  with A = , , , ;B = a, b, c, d;cc = e, f, g, h; if you use Transpose[A, B, cc, 3, 1, 2] you get , a, e, , b, f, , c, g, , d, h. But your post says you want a, e, b, f, c, g, d, h, no?
  $endgroup$
  – kglr
  27 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Consider if I have some matrices constructed as such:

A = 
 , ,
 , 
 
B = 
 a , b ,
 c , d 
 
C = 
 e , f ,
 g , h 
 
...

A serves as a matrix of empty lists. I want to be able to append B, C, etc. to A, to obtain a final matrix like

A = 
 a, e, ... , b, f, ... ,
 c, g, ... , d, h, ... 
 

How can I do this? My matrices are always square and of the same dimensions. An alternative which could also be useful is to Join two matrices element by element, if the matrix elements of both are already lists.

list-manipulation matrix

share|improve this question

edited 1 hour ago

Kai

asked 8 hours ago

KaiKai

59411 silver badge99 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You wrote "My matrices are always square and of the same dimensions". That is not true. Try Dimensions/@A,B
  $endgroup$
  – Fortsaint
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Kai, with A and B as inputs what is the desired result: a, b, c, d or , a, , b, , c, , d?
  $endgroup$
  – kglr
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Fortsaint my meaning is that the matrices which are being appended are always the same dimension. Matrix A is simply a matrix of empty lists to append to.
  $endgroup$
  – Kai
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kglr see the update
  $endgroup$
  – Kai
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  with A = , , , ;B = a, b, c, d;cc = e, f, g, h; if you use Transpose[A, B, cc, 3, 1, 2] you get , a, e, , b, f, , c, g, , d, h. But your post says you want a, e, b, f, c, g, d, h, no?
  $endgroup$
  – kglr
  27 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Consider if I have some matrices constructed as such:

A = 
 , ,
 , 
 
B = 
 a , b ,
 c , d 
 
C = 
 e , f ,
 g , h 
 
...

A serves as a matrix of empty lists. I want to be able to append B, C, etc. to A, to obtain a final matrix like

A = 
 a, e, ... , b, f, ... ,
 c, g, ... , d, h, ... 
 

How can I do this? My matrices are always square and of the same dimensions. An alternative which could also be useful is to Join two matrices element by element, if the matrix elements of both are already lists.

list-manipulation matrix

share|improve this question

edited 1 hour ago

Kai

asked 8 hours ago

KaiKai

59411 silver badge99 bronze badges

$endgroup$

A = 
 , ,
 , 
 
B = 
 a , b ,
 c , d 
 
C = 
 e , f ,
 g , h 
 
...

A = 
 a, e, ... , b, f, ... ,
 c, g, ... , d, h, ... 
 

share|improve this question

edited 1 hour ago

Kai

asked 8 hours ago

KaiKai

59411 silver badge99 bronze badges

share|improve this question

edited 1 hour ago

Kai

asked 8 hours ago

KaiKai

59411 silver badge99 bronze badges

asked 8 hours ago

KaiKai

59411 silver badge99 bronze badges

asked 8 hours ago

KaiKai

59411 silver badge99 bronze badges

3 Answers
3

active

oldest

votes

3

$begingroup$

Suppose you have 4 matrices:

SeedRandom[1]
a, b, c, d = RandomInteger[1, 4, 2, 2];

Then, you can use Transpose to construct the desired matrix:

Transpose[a, b, c, d, 3, 1, 2]

1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0

share|improve this answer

edited 4 hours ago

answered 6 hours ago

Carl WollCarl Woll

85.1k33 gold badges109109 silver badges220220 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think Transpose[a, b, c, d, 3, 1, 2] may be enough.
  $endgroup$
  – Roman
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Roman Thanks, much simpler!
  $endgroup$
  – Carl Woll
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm still wrapping my head around the second argument of transpose, but this looks like the simplest way for me to achieve what I want.
  $endgroup$
  – Kai
  1 hour ago
  

  
  
  
  

add a comment | 

5

$begingroup$

To join nested arrays elementwise, use the third argument of Join. For matrices, depth 1 (default) is rows, 2 is columns, and 3 is elements.

a = Array[&, 2,2];

b = 
b1, b2,
b3, b4
;

listEach = Map[List, #, 2]&

appendEach[x_,y_] := Join[x, listEach[y], 3]

Print[appendEach[appendEach[a, b], b+5]]
(* b1, 5 + b1, b2, 5 + b2, b3, 5 + b3, b4, 5 + b4 *)

Try it online!

However, if you are appending many matrices it may be more efficient to create a list of matrices using Sow / Reap, then Transpose into a matrix of lists.

share|improve this answer

answered 7 hours ago

lirtosiastlirtosiast

17855 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the function of the & in the first line?
  $endgroup$
  – Kai
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

MapThread Flatten at Level 2:

MapThread[Flatten[##] &, A, B, 2]

a, b, c, d

share|improve this answer

edited 6 hours ago

answered 6 hours ago

kglrkglr

200k1010 gold badges229229 silver badges455455 bronze badges

$endgroup$

add a comment | 

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3

$begingroup$

Suppose you have 4 matrices:

SeedRandom[1]
a, b, c, d = RandomInteger[1, 4, 2, 2];

Then, you can use Transpose to construct the desired matrix:

Transpose[a, b, c, d, 3, 1, 2]

1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0

share|improve this answer

edited 4 hours ago

answered 6 hours ago

Carl WollCarl Woll

85.1k33 gold badges109109 silver badges220220 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think Transpose[a, b, c, d, 3, 1, 2] may be enough.
  $endgroup$
  – Roman
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Roman Thanks, much simpler!
  $endgroup$
  – Carl Woll
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm still wrapping my head around the second argument of transpose, but this looks like the simplest way for me to achieve what I want.
  $endgroup$
  – Kai
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Suppose you have 4 matrices:

SeedRandom[1]
a, b, c, d = RandomInteger[1, 4, 2, 2];

Then, you can use Transpose to construct the desired matrix:

Transpose[a, b, c, d, 3, 1, 2]

1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0

share|improve this answer

edited 4 hours ago

answered 6 hours ago

Carl WollCarl Woll

85.1k33 gold badges109109 silver badges220220 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think Transpose[a, b, c, d, 3, 1, 2] may be enough.
  $endgroup$
  – Roman
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Roman Thanks, much simpler!
  $endgroup$
  – Carl Woll
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm still wrapping my head around the second argument of transpose, but this looks like the simplest way for me to achieve what I want.
  $endgroup$
  – Kai
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

Suppose you have 4 matrices:

SeedRandom[1]
a, b, c, d = RandomInteger[1, 4, 2, 2];

Then, you can use Transpose to construct the desired matrix:

Transpose[a, b, c, d, 3, 1, 2]

1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0

share|improve this answer

edited 4 hours ago

answered 6 hours ago

Carl WollCarl Woll

85.1k33 gold badges109109 silver badges220220 bronze badges

$endgroup$

SeedRandom[1]
a, b, c, d = RandomInteger[1, 4, 2, 2];

Transpose[a, b, c, d, 3, 1, 2]

share|improve this answer

edited 4 hours ago

answered 6 hours ago

Carl WollCarl Woll

85.1k33 gold badges109109 silver badges220220 bronze badges

answered 6 hours ago

Carl WollCarl Woll

85.1k33 gold badges109109 silver badges220220 bronze badges

answered 6 hours ago

Carl WollCarl Woll

85.1k33 gold badges109109 silver badges220220 bronze badges

5

$begingroup$

To join nested arrays elementwise, use the third argument of Join. For matrices, depth 1 (default) is rows, 2 is columns, and 3 is elements.

a = Array[&, 2,2];

b = 
b1, b2,
b3, b4
;

listEach = Map[List, #, 2]&

appendEach[x_,y_] := Join[x, listEach[y], 3]

Print[appendEach[appendEach[a, b], b+5]]
(* b1, 5 + b1, b2, 5 + b2, b3, 5 + b3, b4, 5 + b4 *)

Try it online!

However, if you are appending many matrices it may be more efficient to create a list of matrices using Sow / Reap, then Transpose into a matrix of lists.

share|improve this answer

answered 7 hours ago

lirtosiastlirtosiast

17855 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the function of the & in the first line?
  $endgroup$
  – Kai
  1 hour ago
  

  
  
  
  

add a comment | 

5

$begingroup$

To join nested arrays elementwise, use the third argument of Join. For matrices, depth 1 (default) is rows, 2 is columns, and 3 is elements.

a = Array[&, 2,2];

b = 
b1, b2,
b3, b4
;

listEach = Map[List, #, 2]&

appendEach[x_,y_] := Join[x, listEach[y], 3]

Print[appendEach[appendEach[a, b], b+5]]
(* b1, 5 + b1, b2, 5 + b2, b3, 5 + b3, b4, 5 + b4 *)

Try it online!

However, if you are appending many matrices it may be more efficient to create a list of matrices using Sow / Reap, then Transpose into a matrix of lists.

share|improve this answer

answered 7 hours ago

lirtosiastlirtosiast

17855 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the function of the & in the first line?
  $endgroup$
  – Kai
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

To join nested arrays elementwise, use the third argument of Join. For matrices, depth 1 (default) is rows, 2 is columns, and 3 is elements.

a = Array[&, 2,2];

b = 
b1, b2,
b3, b4
;

listEach = Map[List, #, 2]&

appendEach[x_,y_] := Join[x, listEach[y], 3]

Print[appendEach[appendEach[a, b], b+5]]
(* b1, 5 + b1, b2, 5 + b2, b3, 5 + b3, b4, 5 + b4 *)

Try it online!

However, if you are appending many matrices it may be more efficient to create a list of matrices using Sow / Reap, then Transpose into a matrix of lists.

share|improve this answer

answered 7 hours ago

lirtosiastlirtosiast

17855 bronze badges

$endgroup$

a = Array[&, 2,2];

b = 
b1, b2,
b3, b4
;

listEach = Map[List, #, 2]&

appendEach[x_,y_] := Join[x, listEach[y], 3]

Print[appendEach[appendEach[a, b], b+5]]
(* b1, 5 + b1, b2, 5 + b2, b3, 5 + b3, b4, 5 + b4 *)

share|improve this answer

answered 7 hours ago

lirtosiastlirtosiast

17855 bronze badges

answered 7 hours ago

lirtosiastlirtosiast

17855 bronze badges

answered 7 hours ago

lirtosiastlirtosiast

17855 bronze badges

3

$begingroup$

MapThread Flatten at Level 2:

MapThread[Flatten[##] &, A, B, 2]

a, b, c, d

share|improve this answer

edited 6 hours ago

answered 6 hours ago

kglrkglr

200k1010 gold badges229229 silver badges455455 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

MapThread Flatten at Level 2:

MapThread[Flatten[##] &, A, B, 2]

a, b, c, d

share|improve this answer

edited 6 hours ago

answered 6 hours ago

kglrkglr

200k1010 gold badges229229 silver badges455455 bronze badges

$endgroup$

add a comment | 

$begingroup$

MapThread Flatten at Level 2:

MapThread[Flatten[##] &, A, B, 2]

a, b, c, d

share|improve this answer

edited 6 hours ago

answered 6 hours ago

kglrkglr

200k1010 gold badges229229 silver badges455455 bronze badges

$endgroup$

MapThread[Flatten[##] &, A, B, 2]

share|improve this answer

edited 6 hours ago

answered 6 hours ago

kglrkglr

200k1010 gold badges229229 silver badges455455 bronze badges

answered 6 hours ago

kglrkglr

200k1010 gold badges229229 silver badges455455 bronze badges

answered 6 hours ago

kglrkglr

200k1010 gold badges229229 silver badges455455 bronze badges