将 N^2 3x3 矩阵串联成 3Nx3N 矩阵

Question

我有 N^2 矩阵。 每一个都是一个 3x3 矩阵。 将它们连接成 3Nx3N 矩阵的一种方法是编写 A(:,:,i)= # 3x3 矩阵 i=1:N^2

B=[A11 A12 ..A1N;A21 ...A2N;...] 但是当N很大时是一项繁琐的工作。 你们提供什么？

Answer 1

尝试以下代码：

N = 4;
A = rand(3,3,N^2);                     %# 3-by-3-by-N^2

c1 = squeeze( num2cell(A,[1 2]) );
c2 = cell(N,1);
for i=0:N-1
    c2{i+1} = cat(2, c1{i*N+1:(i+1)*N});
end

B = cat(1, c2{:});                     %# 3N-by-3N

Answer 2

另一种可能性涉及 mat2cell 和 reshape

N = 2;
A = rand(3,3,N^2);  

C = mat2cell(A,3,3,ones(N^2,1));
C = reshape(C,N,N)'; %'# make a N-by-N cell array and transpose

%# catenate into 3N-by-3N cell array
B = cell2mat(C);

如果您更喜欢的话，这里在一行中相同

B = cell2mat(reshape(mat2cell(A,2,2,ones(N^2,1)),N,N)');

对于 N=2

>> A = rand(3,3,N^2)
A(:,:,1) =
      0.40181      0.12332      0.41727
     0.075967      0.18391     0.049654
      0.23992      0.23995      0.90272
A(:,:,2) =
      0.94479      0.33772       0.1112
      0.49086      0.90005      0.78025
      0.48925      0.36925      0.38974
A(:,:,3) =
      0.24169      0.13197      0.57521
      0.40391      0.94205      0.05978
     0.096455      0.95613      0.23478
A(:,:,4) =
      0.35316     0.043024      0.73172
      0.82119      0.16899      0.64775
     0.015403      0.64912      0.45092

B =
      0.40181      0.12332      0.41727      0.94479      0.33772       0.1112
     0.075967      0.18391     0.049654      0.49086      0.90005      0.78025
      0.23992      0.23995      0.90272      0.48925      0.36925      0.38974
      0.24169      0.13197      0.57521      0.35316     0.043024      0.73172
      0.40391      0.94205      0.05978      0.82119      0.16899      0.64775
     0.096455      0.95613      0.23478     0.015403      0.64912      0.45092

Answer 3

为什么不进行老式的预分配和循环呢？应该相当快。

N = 4;
A = rand(3,3,N^2);  % Assuming column major order for Aij
8
B = zeros(3*N, 3*N);
for j = 1:N^2
    ix = mod(j-1, N)*3 + 1;
    iy = floor((j-1)/N)*3 + 1;
    fprintf('%02d - %02d\n', ix, iy);
    B(ix:ix+2, iy:iy+2) = A(:,:,j);
end

编辑：对于速度迷来说，这里是排名：

N = 200;
A = rand(3,3,N^2);  % test set

@gnovice solution: Elapsed time is 0.013069 seconds.
@Amro    solution: Elapsed time is 0.203308 seconds.
@Rich C  solution: Elapsed time is 0.887077 seconds.
@Jonas   solution: Elapsed time is 7.065174 seconds.

Answer 4

这是一个非常快速的单行代码，仅使用 RESHAPE 和 PERMUTE：

B = reshape(permute(reshape(A,3,3*N,N),[2 1 3]),3*N,3*N).';

还有一个测试：

>> N=2;
>> A = rand(3,3,N^2)
A(:,:,1) =
    0.5909    0.6571    0.8082
    0.7118    0.6090    0.7183
    0.4694    0.9588    0.5582
A(:,:,2) =
    0.1791    0.6844    0.6286
    0.4164    0.4140    0.5833
    0.1380    0.1099    0.8970
A(:,:,3) =
    0.2232    0.2355    0.1214
    0.1782    0.6873    0.3394
    0.5645    0.4745    0.9763
A(:,:,4) =
    0.5334    0.7559    0.9984
    0.8454    0.7618    0.1065
    0.0549    0.5029    0.3226

>> B = reshape(permute(reshape(A,3,3*N,N),[2 1 3]),3*N,3*N).'
B =
    0.5909    0.6571    0.8082    0.1791    0.6844    0.6286
    0.7118    0.6090    0.7183    0.4164    0.4140    0.5833
    0.4694    0.9588    0.5582    0.1380    0.1099    0.8970
    0.2232    0.2355    0.1214    0.5334    0.7559    0.9984
    0.1782    0.6873    0.3394    0.8454    0.7618    0.1065
    0.5645    0.4745    0.9763    0.0549    0.5029    0.3226