我有2个8x8矩阵，即P0＆amp; P1，
P0 = [a a a a a a a a P1 = [b b b b b b b b
a a a a a a a a b b b b b b b b
a a a a a a a a b b b b b b b b
a a a a a a a a b b b b b b b b
a a a a a a a a b b b b b b b b
a a a a a a a a b b b b b b b b
a a a a a a a a b b b b b b b b
a a a a a a a a] b b b b b b b b]
如何在matlab中创建这样的矩阵H？
H = [P0 P1 0 0 0 0
0 P0 P1 0 0 0
0 0 P0 P1 0 0
0 0 0 P0 P1 0
0 0 0 0 P0 P1]
答案 0 :(得分：1)
由于P0和P1是矩阵，您可以将它们存储在单元格中，H是5x6单元格数组/矩阵。
以下是一个例子。
>> P0 = ones(8)
P0 =
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
>> P1 = ones(8) + 1
P1 =
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
>> H = cell(5,6)
H =
[] [] [] [] [] []
[] [] [] [] [] []
[] [] [] [] [] []
[] [] [] [] [] []
[] [] [] [] [] []
>> [H{:}] = deal(0)
H =
[0] [0] [0] [0] [0] [0]
[0] [0] [0] [0] [0] [0]
[0] [0] [0] [0] [0] [0]
[0] [0] [0] [0] [0] [0]
[0] [0] [0] [0] [0] [0]
>> H(sub2ind(size(H), 1:size(H,1), 1:size(H,1))) = {P0}
H =
[8x8 double] [0] [0] [0] [0] [0]
[0] [8x8 double] [0] [0] [0] [0]
[0] [0] [8x8 double] [0] [0] [0]
[0] [0] [0] [8x8 double] [0] [0]
[0] [0] [0] [0] [8x8 double] [0]
>> H(sub2ind(size(H), 1:size(H,1), 2:size(H,2))) = {P1}
H =
[8x8 double] [8x8 double] [0] [0] [0] [0]
[0] [8x8 double] [8x8 double] [0] [0] [0]
[0] [0] [8x8 double] [8x8 double] [0] [0]
[0] [0] [0] [8x8 double] [8x8 double] [0]
[0] [0] [0] [0] [8x8 double] [8x8 double]
>> H{1,1}
ans =
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
>> H{1,2}
ans =
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
sub2ind函数的用法，如果经常与线性索引的概念紧密相关。这里，sub2ind用于查找将值赋给H的对角线字段所需的线性索引。
您可以在此处找到相关文档：
答案 1 :(得分：0)
我愿意：
a = 1;
b = 2;
N = 8;
H_width = 6;
H_height = 5;
P0 = ones(N)*a;
P1 = ones(N)*b;
P = [P0,P1];
H = zeros(N*H_height, N*H_width);
for hh = 1:min(H_height,H_width-1)
H((hh-1)*N+1:hh*N,(hh-1)*N+1:(hh+1)*N) = P;
end
if H_width < H_height
H((H_width-1)*N+1:H_width*N,end-N+1:end) = P0;
end
N = 2
的示例 1 1 2 2 0 0 0 0 0 0 0 0
1 1 2 2 0 0 0 0 0 0 0 0
0 0 1 1 2 2 0 0 0 0 0 0
0 0 1 1 2 2 0 0 0 0 0 0
0 0 0 0 1 1 2 2 0 0 0 0
0 0 0 0 1 1 2 2 0 0 0 0
0 0 0 0 0 0 1 1 2 2 0 0
0 0 0 0 0 0 1 1 2 2 0 0
0 0 0 0 0 0 0 0 1 1 2 2
0 0 0 0 0 0 0 0 1 1 2 2
答案 2 :(得分：0)
在两个对角线中创建一个包含数字a和b的矩阵，然后使用kron在块中重复这些数字：
m = 2; %// size of P0, P1
n = 4; %// number of blocks
a = 5;
b = 10;
h = diag(a*ones(1,n)) + diag(b*ones(1,n-1),1);
h(end,end+1) = b;
H = kron(h,ones(m));
结果：
H =
5 5 10 10 0 0 0 0 0 0
5 5 10 10 0 0 0 0 0 0
0 0 5 5 10 10 0 0 0 0
0 0 5 5 10 10 0 0 0 0
0 0 0 0 5 5 10 10 0 0
0 0 0 0 5 5 10 10 0 0
0 0 0 0 0 0 5 5 10 10
0 0 0 0 0 0 5 5 10 10