如何生成随机二进制矩阵，其中每行中只有一位的值为1？

Question

I want to use MATLAB to generate a random binary matrix A (nxm) which satisfies a condition:

Each row contains one position with value 1 . Other positions are value 0 . The position having 1 value is random position.

I tried this code

   n=5;m=10; 
   %% A = randi([0 1], n,m);
   A=zeros(n,m);
   for i=1:n
       rand_pos=randperm(m);
       pos_one=rand_pos(1); % random possition of 1 value
       A(i,pos_one)=1;
   end

Is it correct?

Answer 1

The solution works, but it is inefficient.
You are using randperm to create a vector (array), and then use only the first element of the vector.
You can use randi to create a scalar (single element) instead:

n=5;m=10; 
A=zeros(n,m);
for i=1:m
    %rand_pos gets a random number in range [1, n].
    rand_pos = randi([1, n]);
    A(rand_pos, i)=1;
end

---

You can also use the following "vectorized" solution:

rand_pos_vec = randi([1, n], 1, m);
A(sub2ind([n, m], rand_pos_vec, 1:m)) = 1;

The above solution:

• Creates a vector of random values in range [1, n].
• Use sub2ind to convert "row index" to "matrix index".
• Place 1 in "matrix index".

Answer 2

You can do it in one line using bsxfun and randi :

A = double(bsxfun(@eq, 1:m, randi(n, n, 1)));

This compares the row vector [1 2 ... m] with an n × 1 random vector of values from 1 to n . The comparison is done element-wise with singleton expansion. For each row, exactly one of the values of [1 2 ... m] equals that in the random vector.