How to append a matrix element by elementHow to apply a permutation to a symmetric square matrix?How to find the distance of two lists?Partitioned matrix operationsPlotting Matrix versus MatrixMatrix expansion and reorganisationHow to remove empty row-column pairs from an asymmetric square matrixBuilding a matrix function one element at a timeScramble matrix under some conditioncreating a data framework using matrix tensorTake upper triangular part of matrix
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How to append a matrix element by element
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Inverse-quotes-quine
How to append a matrix element by element
How to apply a permutation to a symmetric square matrix?How to find the distance of two lists?Partitioned matrix operationsPlotting Matrix versus MatrixMatrix expansion and reorganisationHow to remove empty row-column pairs from an asymmetric square matrixBuilding a matrix function one element at a timeScramble matrix under some conditioncreating a data framework using matrix tensorTake upper triangular part of matrix
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
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
asked 8 hours ago
KaiKai
5941 silver badge9 bronze badges
$endgroup$
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
asked 8 hours ago
KaiKai
5941 silver badge9 bronze badges
$endgroup$
$begingroup$
You wrote "My matrices are always square and of the same dimensions". That is not true. TryDimensions/@A,B
$endgroup$
– Fortsaint
6 hours ago
$begingroup$
Kai, withAandBas inputs what is the desired result:a, b, c, dor, 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. MatrixAis simply a matrix of empty lists to append to.
$endgroup$
– Kai
1 hour ago
$begingroup$
@kglr see the update
$endgroup$
– Kai
1 hour ago
$begingroup$
withA = , , , ;B = a, b, c, d;cc = e, f, g, h;if you useTranspose[A, B, cc, 3, 1, 2]you get, a, e, , b, f, , c, g, , d, h. But your post says you wanta, 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
asked 8 hours ago
KaiKai
5941 silver badge9 bronze badges
$endgroup$
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
list-manipulation matrix
asked 8 hours ago
KaiKai
5941 silver badge9 bronze badges
asked 8 hours ago
KaiKai
5941 silver badge9 bronze badges
edited 1 hour ago
Kai
asked 8 hours ago
KaiKai
5941 silver badge9 bronze badges
asked 8 hours ago
KaiKai
5941 silver badge9 bronze badges
asked 8 hours ago
KaiKai
5941 silver badge9 bronze badges
5941 silver badge9 bronze badges
$begingroup$
You wrote "My matrices are always square and of the same dimensions". That is not true. TryDimensions/@A,B
$endgroup$
– Fortsaint
6 hours ago
$begingroup$
Kai, withAandBas inputs what is the desired result:a, b, c, dor, 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. MatrixAis simply a matrix of empty lists to append to.
$endgroup$
– Kai
1 hour ago
$begingroup$
@kglr see the update
$endgroup$
– Kai
1 hour ago
$begingroup$
withA = , , , ;B = a, b, c, d;cc = e, f, g, h;if you useTranspose[A, B, cc, 3, 1, 2]you get, a, e, , b, f, , c, g, , d, h. But your post says you wanta, e, b, f, c, g, d, h, no?
$endgroup$
– kglr
27 mins ago
add a comment |
$begingroup$
You wrote "My matrices are always square and of the same dimensions". That is not true. TryDimensions/@A,B
$endgroup$
– Fortsaint
6 hours ago
$begingroup$
Kai, withAandBas inputs what is the desired result:a, b, c, dor, 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. MatrixAis simply a matrix of empty lists to append to.
$endgroup$
– Kai
1 hour ago
$begingroup$
@kglr see the update
$endgroup$
– Kai
1 hour ago
$begingroup$
withA = , , , ;B = a, b, c, d;cc = e, f, g, h;if you useTranspose[A, B, cc, 3, 1, 2]you get, a, e, , b, f, , c, g, , d, h. But your post says you wanta, e, b, f, c, g, d, h, no?
$endgroup$
– kglr
27 mins ago
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
@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
$begingroup$
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 Answers
3
active
oldest
votes
$begingroup$
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
answered 6 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
$endgroup$
$begingroup$
I thinkTranspose[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 |
$begingroup$
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.
answered 7 hours ago
lirtosiastlirtosiast
1785 bronze badges
$endgroup$
$begingroup$
What is the function of the&in the first line?
$endgroup$
– Kai
1 hour ago
add a comment |
$begingroup$
MapThread Flatten at Level 2:
MapThread[Flatten[##] &, A, B, 2]
a, b, c, d
answered 6 hours ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
answered 6 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
$endgroup$
$begingroup$
I thinkTranspose[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
answered 6 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
$endgroup$
$begingroup$
I thinkTranspose[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
answered 6 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
$endgroup$
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
answered 6 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
edited 4 hours ago
answered 6 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
answered 6 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
answered 6 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
85.1k3 gold badges109 silver badges220 bronze badges
$begingroup$
I thinkTranspose[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 |
$begingroup$
I thinkTranspose[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
$begingroup$
I think
Transpose[a, b, c, d, 3, 1, 2] may be enough.$endgroup$
– Roman
4 hours ago
$begingroup$
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
$begingroup$
@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
$begingroup$
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 |
$begingroup$
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.
answered 7 hours ago
lirtosiastlirtosiast
1785 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.
answered 7 hours ago
lirtosiastlirtosiast
1785 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.
answered 7 hours ago
lirtosiastlirtosiast
1785 bronze badges
$endgroup$
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.
answered 7 hours ago
lirtosiastlirtosiast
1785 bronze badges
answered 7 hours ago
lirtosiastlirtosiast
1785 bronze badges
answered 7 hours ago
lirtosiastlirtosiast
1785 bronze badges
answered 7 hours ago
lirtosiastlirtosiast
1785 bronze badges
1785 bronze badges
$begingroup$
What is the function of the&in the first line?
$endgroup$
– Kai
1 hour ago
add a comment |
$begingroup$
What is the function of the&in the first line?
$endgroup$
– Kai
1 hour ago
$begingroup$
What is the function of the
& in the first line?$endgroup$
– Kai
1 hour ago
$begingroup$
What is the function of the
& in the first line?$endgroup$
– Kai
1 hour ago
add a comment |
$begingroup$
MapThread Flatten at Level 2:
MapThread[Flatten[##] &, A, B, 2]
a, b, c, d
answered 6 hours ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
$endgroup$
add a comment |
$begingroup$
MapThread Flatten at Level 2:
MapThread[Flatten[##] &, A, B, 2]
a, b, c, d
answered 6 hours ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
$endgroup$
add a comment |
$begingroup$
MapThread Flatten at Level 2:
MapThread[Flatten[##] &, A, B, 2]
a, b, c, d
answered 6 hours ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
$endgroup$
MapThread Flatten at Level 2:
MapThread[Flatten[##] &, A, B, 2]
a, b, c, d
answered 6 hours ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
edited 6 hours ago
answered 6 hours ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
answered 6 hours ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
answered 6 hours ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
200k10 gold badges229 silver badges455 bronze badges
add a comment |
add a comment |
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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
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Kai, with
AandBas inputs what is the desired result:a, b, c, dor, a, , b, , c, , d?$endgroup$
– kglr
2 hours ago
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@Fortsaint my meaning is that the matrices which are being appended are always the same dimension. Matrix
Ais simply a matrix of empty lists to append to.$endgroup$
– Kai
1 hour ago
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@kglr see the update
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– Kai
1 hour ago
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with
A = , , , ;B = a, b, c, d;cc = e, f, g, h;if you useTranspose[A, B, cc, 3, 1, 2]you get, a, e, , b, f, , c, g, , d, h. But your post says you wanta, e, b, f, c, g, d, h, no?$endgroup$
– kglr
27 mins ago