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

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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!

采纳的回答

Davide Masiello
Davide Masiello 2022-10-17
编辑:Davide Masiello 2022-10-17
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
  2 个评论
Daniel Neubauer
Daniel Neubauer 2022-10-17
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
Davide Masiello 2022-10-17
I understand, I have changed my answer.
I believe it should work that way either.

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