Matlab J-by-2矩阵行的块对角化

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，这将是缓慢而乏味的。是否有一个内置的Matlab函数可以做到这一点？

这里有一个针对

jx2

数组情况的黑客解决方案，使用-

只有一个函数调用（忽略

size

，因为它必须是最小的）开销

zero

，甚至可以使用-

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

基准测试代码

%//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.

---

这是另一个解决方案。不确定它的效率有多高，但它适用于任何大小的矩阵

a

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

这里有一个用于重塑为J-by-1单元数组的元素，其中每个元素包含一行

a

。然后使用

{:}

操作符将内容作为逗号分隔的变量列表推送到

blkdiag

：

%//Set up some random data
J = 100;
A = rand(J,2);
%// Solution for arbitrary J-by-2 A
temp = mat2cell(A, ones(J,1), 2);
B = blkdiag(temp{:})

---

很好的解决方案！我这里有一些计时结果，但通过@Divakar运行基准测试代码重复了它们。下面是我这边的结果

---------------------------------- With @mikkola solution
Elapsed time is 0.100674 seconds.
---------------------------------- With @Jeff Irwin solution
Elapsed time is 0.283275 seconds.
---------------------------------- With @Divakar Hacky1 solution
Elapsed time is 0.079194 seconds.
---------------------------------- With @Divakar Hacky2 solution
Elapsed time is 0.051629 seconds.

---

我找到了一种方法，使用

sparse

在内存和时钟时间方面都优于其他解决方案：

N = size(A,1);
ind_1 = [1:N].';
ind_2 = [1:2:2*N-1].';
A_1   = sparse(ind_1,ind_2,A(:,1),N,2*N);
ind_2 = [2:2:2*N].';
A_2   = sparse(ind_1,ind_2,A(:,2),N,2*N);
out       = A_1 + A_2;

使用与@Divakar相同的基准代码报告以下结果：

---------------------------------- With @mikkola solution
Elapsed time is 0.065136 seconds.
---------------------------------- With @Jeff Irwin solution
Elapsed time is 0.500264 seconds.
---------------------------------- With Hacky1 solution
Elapsed time is 0.200303 seconds.
---------------------------------- With Hacky2 solution
Elapsed time is 0.011991 seconds.
---------------------------------- With @Matt T solution
Elapsed time is 0.000712 seconds.

---

不确定您使用的基准测试代码，但我已经添加了我的代码，以防人们想查看。结果在我这方面是不同的。@Divakar我重复了运行您的代码的基准测试。订购与您的终端相同。无任何函数调用开销的索引显然比FTW更快+1.@rayryeng未记录的MATLAB FTW！；）谢谢设置

out=sparse（N，2*N）

而不是

out（N，N*2）=0

可能是有意义的，这样可以避免

N

很大时的大量开销。@MattT我没有处理过多少sparse，所以我不确定。但是，如果这行得通的话，当然@MattT分配给

稀疏

矩阵的成本非常高。最好是使用非零内容初始化

sparse

矩阵，而不是创建空的

sparse

矩阵并赋值。嵌套调用

diag

会降低性能，但这是解决此问题的最规范的方法之一+1.

N = size(A,1);
ind_1 = [1:N].';
ind_2 = [1:2:2*N-1].';
A_1   = sparse(ind_1,ind_2,A(:,1),N,2*N);
ind_2 = [2:2:2*N].';
A_2   = sparse(ind_1,ind_2,A(:,2),N,2*N);
out       = A_1 + A_2;

---------------------------------- With @mikkola solution
Elapsed time is 0.065136 seconds.
---------------------------------- With @Jeff Irwin solution
Elapsed time is 0.500264 seconds.
---------------------------------- With Hacky1 solution
Elapsed time is 0.200303 seconds.
---------------------------------- With Hacky2 solution
Elapsed time is 0.011991 seconds.
---------------------------------- With @Matt T solution
Elapsed time is 0.000712 seconds.