Question

我有N ^ 2个矩阵。每一个都是3x3矩阵。将它们连接到3Nx3N矩阵的一种方法是写入 A(:,:,i)= # 3x3 matrix i=1:N^2

matrix

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

Answer 1

这是一个非常快速的单线程，仅使用RESHAPE和PERMUTE：

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

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

Answer 2

请尝试以下代码：

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 3

涉及mat2cell和reshape

的另一种可能性

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

，那么在一行中也是如此

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 4

为什么不做旧式的预分配和循环？应该很快。

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：对于速度迷们来说，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.

将N ^ 2 3x3矩阵连接成3Nx3N矩阵

4 个答案: