期间的随机数（使用逻辑）

Question

使用下面的代码：

t=10;
period=9;
A=zeros(36,36,t);
for j=1:t
   A(7:period:end,7:period:end,j)=rand(1,1);
   A(8:period:end,8:period:end,j)=rand(1,1);
   A(9:period:end,9:period:end,j)=rand(1,1);
end

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编辑：这是 A。我在视觉上将它们分成 9x9，以便更容易看到模式。

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 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 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   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 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 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
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 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0   0 0 0 0 0 0 1 0 0   0 0 0 0 0 0 1 0 0   0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0   0 0 0 0 0 0 0 1 0   0 0 0 0 0 0 0 1 0   0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1   0 0 0 0 0 0 0 0 1   0 0 0 0 0 0 0 0 1   0 0 0 0 0 0 0 0 1

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 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 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   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 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 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
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 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0   0 0 0 0 0 0 1 0 0   0 0 0 0 0 0 1 0 0   0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0   0 0 0 0 0 0 0 1 0   0 0 0 0 0 0 0 1 0   0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1   0 0 0 0 0 0 0 0 1   0 0 0 0 0 0 0 0 1   0 0 0 0 0 0 0 0 1

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 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 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   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 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 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
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 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0   0 0 0 0 0 0 1 0 0   0 0 0 0 0 0 1 0 0   0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0   0 0 0 0 0 0 0 1 0   0 0 0 0 0 0 0 1 0   0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1   0 0 0 0 0 0 0 0 1   0 0 0 0 0 0 0 0 1   0 0 0 0 0 0 0 0 1

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 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 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   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 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 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
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 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0   0 0 0 0 0 0 1 0 0   0 0 0 0 0 0 1 0 0   0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0   0 0 0 0 0 0 0 1 0   0 0 0 0 0 0 0 1 0   0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1   0 0 0 0 0 0 0 0 1   0 0 0 0 0 0 0 0 1   0 0 0 0 0 0 0 0 1

我想将每个标记为 1 的数字设为 0 到 1 之间的随机数。是否有必要创建一个向量（48,1），然后插入 1 所在的值？

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当然，这些数字不是完全随机的，每个“周期”都会复制相同的三个随机数字。

有没有一种简单的方法可以创建随机数，使每个随机生成的值都是随机的？

非常感谢。

Answer 1

这比我一开始想的要复杂一些。这是我的解决方案：

A = zeros(36,36,t);
idx = sub2ind([36 36 t],repmat(34:36,1,t),repmat(34:36,1,t),repmat(1:t,1,3));
A(idx) = rand(length(idx),1);

逻辑索引的解决方案是：

A(logical(A)) = rand(nnz(A),1);

但你必须准备A作为问题中给出的二进制矩阵。