get p*(q*m) matrix from m*n matrix and p*q indexing matrix

hey everyone,

is there an elegant way to get the following:

C=rand(8,n) %given
index=[...
    1 5 1 2 3 4
    2 6 2 3 4 1
    3 7 6 7 8 5
    4 8 5 6 7 8
    1 5 1 2 3 4];
X=[...
    C(1,1) C(5,1) C(1,1) ... ... C(4,1)     C(1,n) C(5,n) C(1,n) ... ... C(4,n)
    C(2,1) C(6,1) C(2,1) ... ... C(1,1)     C(2,n) C(6,n) C(2,n) ... ... C(1,n)
    C(3,1) C(7,1) C(6,1) ... ... C(5,1) ... C(3,n) C(7,n) C(6,n) ... ... C(5,n)
    C(4,1) C(8,1) C(5,1) ... ... C(8,1)     C(4,n) C(8,n) C(5,n) ... ... C(8,n)
    C(1,1) C(5,1) C(1,1) ... ... C(4,1)     C(1,n) C(5,n) C(1,n) ... ... C(4,n)
    ]

X=C(index,:) 

gives the information but not arranged as desired. it is for plotting cubes by edge coordinates in 3d with patch command.

thanks for any help!

Answer 1

[Davide Masiello]

Something like this?

n = 3;
C = rand(8,n)

C = 8×3

0.8036 0.1626 0.1875 0.8851 0.4522 0.1023 0.8075 0.3073 0.0316 0.3548 0.5842 0.8018 0.6769 0.1753 0.1741 0.8239 0.0872 0.1524 0.0985 0.8049 0.5961 0.6590 0.9066 0.9541

index=[...
    1 5 1 2 3 4
    2 6 2 3 4 1
    3 7 6 7 8 5
    4 8 5 6 7 8
    1 5 1 2 3 4];

X = reshape(C(index(:),1:n),size(index,1),size(index,2)*n)

X = 5×18

0.8036 0.6769 0.8036 0.8851 0.8075 0.3548 0.1626 0.1753 0.1626 0.4522 0.3073 0.5842 0.1875 0.1741 0.1875 0.1023 0.0316 0.8018 0.8851 0.8239 0.8851 0.8075 0.3548 0.8036 0.4522 0.0872 0.4522 0.3073 0.5842 0.1626 0.1023 0.1524 0.1023 0.0316 0.8018 0.1875 0.8075 0.0985 0.8239 0.0985 0.6590 0.6769 0.3073 0.8049 0.0872 0.8049 0.9066 0.1753 0.0316 0.5961 0.1524 0.5961 0.9541 0.1741 0.3548 0.6590 0.6769 0.8239 0.0985 0.6590 0.5842 0.9066 0.1753 0.0872 0.8049 0.9066 0.8018 0.9541 0.1741 0.1524 0.5961 0.9541 0.8036 0.6769 0.8036 0.8851 0.8075 0.3548 0.1626 0.1753 0.1626 0.4522 0.3073 0.5842 0.1875 0.1741 0.1875 0.1023 0.0316 0.8018

[Daniel Neubauer]

thanks for the reply.

however, is there a way without looping? my data is rather big so i'd prefer not to loop.

[Davide Masiello]

I understand, I have changed my answer.

I believe it should work that way either.