matlab - 块化J×2矩阵的行

例如，给定一个J-by-2矩阵

A = [1 2 ; 3 4 ; 5 6]

我想阻止它对角化也就是说，我想要：

B = [1 2 0 0 0 0 ; 0 0 3 4 0 0 ; 0 0 0 0 5 6].

执行此操作的一个命令是：

blkdiag(A(1,:),A(2,:),A(3,:))

如果J很大，这将是一个缓慢而乏味的过程有没有内置的Matlab函数来实现这一点？

最佳答案

下面是一个使用J x 2的linear indexing数组案例的破解方案-

%// Get number of rows
N = size(A,1);

%// Get linear indices of the first column elements positions in output array
idx = 1:2*N+1:(N-1)*(2*N+1)+1;

%// Setup output array
out = zeros(N,N*2);

%// Put first and second column elements into idx and idx+N positions
out([idx(:) idx(:)+N]) = A

只有一个函数调用（忽略size，因为它必须是最小的）开销zeros，甚至可以用this undocumented zeros initialization trick移除-

out(N,N*2) = 0; %// Instead of out = zeros(N,N*2);

样本运行-

A =
     1     2
     3     4
     5     6
     7     8
out =
     1     2     0     0     0     0     0     0
     0     0     3     4     0     0     0     0
     0     0     0     0     5     6     0     0
     0     0     0     0     0     0     7     8

以下是迄今为止发布的解决方案的基准测试。
基准代码

%//Set up some random data
J = 7000;   A = rand(J,2);
%// Warm up tic/toc
for k = 1:100000
    tic(); elapsed = toc();
end
disp('---------------------------------- With @mikkola solution')
tic
temp = mat2cell(A, ones(J,1), 2);
B = blkdiag(temp{:});
toc, clear B temp
disp('---------------------------------- With @Jeff Irwin solution')
tic
m = size(A, 1);
n = size(A, 2);
B = zeros(m, m * n);
for k = 1: n
    B(:, k: n: m * n) = diag(A(:, k));
end
toc, clear B k m n
disp('---------------------------------- With Hacky1 solution')
tic
N = size(A,1);
idx = 1:2*N+1:(N-1)*(2*N+1)+1;
out = zeros(N,N*2);
out([idx(:) idx(:)+N]) = A;
toc, clear out idx N
disp('---------------------------------- With Hacky2 solution')
tic
N = size(A,1);
idx = 1:2*N+1:(N-1)*(2*N+1)+1;
out(N,N*2) = 0;
out([idx(:) idx(:)+N]) = A;
toc, clear out idx N

运行时

---------------------------------- With @mikkola solution
Elapsed time is 0.546584 seconds.
---------------------------------- With @Jeff Irwin solution
Elapsed time is 1.330666 seconds.
---------------------------------- With Hacky1 solution
Elapsed time is 0.455735 seconds.
---------------------------------- With Hacky2 solution
Elapsed time is 0.364227 seconds.