How do I convert a 2d matrix to a 3d matrix and vice versa efficiently?

Hello,

In this post, I need converting 3d matrix A to 2d matrix C1 and revese case.

When I try to convert 2d matrix C1 back to 3d matrix A, it failed I am not sure the reasons. (sum(A-h3,'all'))

In reshape I should do [2,3,4] not [2,4,3], but I am wondering why I should change the order in size.

So many thanks in advance.

>> size(A) % ans = 2 3 4

>> size(C1) % ans = 8 3

%%%% code begin

A(:,:,1) = [1 2;3 4;5 6]';

A(:,:,2) = [10 20;30 40;50 60]';

A(:,:,3) = [11 21;31 41;51 61]';

A(:,:,4) = [100 200;300 400;500 600]';

C1 = [A(:,:,1);A(:,:,2);A(:,:,3);A(:,:,4)];

permute(reshape(C1,size(A)), [1 3 2])

s1=size(A,1); s2=size(A,2); s3=size(A,3);

permute(reshape(C1,[s1,s3,s2]), [1 3 2]) % need a comment for reverseing order [2,4,3] not [2,3,4]

%permute(reshape(C1,[2,4,3]), [1 3 2])

h1=permute(A,[1 3 2]); % c/d2, v1 (1/2)

h2=reshape(h1,[],size(A,2),1);

%h3=permute(reshape(h2,size(A)), [1 3 2])

h3=permute(reshape(h2,[s1,s3,s2]), [1 3 2]) % need a comment for reverseing order

sum(A-h3,'all')

%%%% code end

Answer 1

[Stephen23]

format compact
A(:,:,1) = [1,2;3,4;5,6]';
A(:,:,2) = [10,20;30,40;50,60]';
A(:,:,3) = [11,21;31,41;51,61]';
A(:,:,4) = [100,200;300,400;500,600]'

A = 
A(:,:,1) =
     1     3     5
     2     4     6
A(:,:,2) =
    10    30    50
    20    40    60
A(:,:,3) =
    11    31    51
    21    41    61
A(:,:,4) =
   100   300   500
   200   400   600

C1 = reshape(permute(A,[2,1,3]),size(A,2),[]).'

C1 = 8×3

1 3 5 2 4 6 10 30 50 20 40 60 11 31 51 21 41 61 100 300 500 200 400 600

[A(:,:,1);A(:,:,2);A(:,:,3);A(:,:,4)] % your desired C1 matrix

ans = 8×3

1 3 5 2 4 6 10 30 50 20 40 60 11 31 51 21 41 61 100 300 500 200 400 600

B = permute(reshape(C1.',[3,2,4]),[2,1,3])

B = 
B(:,:,1) =
     1     3     5
     2     4     6
B(:,:,2) =
    10    30    50
    20    40    60
B(:,:,3) =
    11    31    51
    21    41    61
B(:,:,4) =
   100   300   500
   200   400   600

[Yeon-Mo Yang]

Appreciate your help. Thanks a lot!

Answer 2

[William Rose]

@Yeon-Mo Yang,

A(:,:,1) = [1 2;3 4;5 6]';
A(:,:,2) = [10 20;30 40;50 60]';
A(:,:,3) = [11 21;31 41;51 61]';
A(:,:,4) = [100 200;300 400;500 600]';
C1 = [A(:,:,1);A(:,:,2);A(:,:,3);A(:,:,4)];
D=permute(reshape(C1,[2 2 2 3]),[1 4 3 2]);
E=reshape(D,[2 3 4]);
fprintf('size(A)=%d by %d by %d, size(E)=%d by %d by %d, sum(A-E)=%.1f.\n',...
    size(A),size(E),sum(A-E,'all'));

size(A)=2 by 3 by 4, size(E)=2 by 3 by 4, sum(A-E)=0.0.

Try it.

[Yeon-Mo Yang]

Thanks the time.