How do I generate multiple binary matrices containing a single one in each row?

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Davide Biaggini
Davide Biaggini 2019-7-31
评论： Davide Biaggini 2019-8-1
I want to generate multiple m x n (size of known matrix A) binary matrices. The rule to respect is that every row has to contain a single 1.
Suppose I want to generate 10 matrices called randmat of size(A). This is the code I've written combining suggestions found on this forum. It should use a for cycle to generate random matrices of 0 and 1, then force each row to sum up to 1. The problem is that I'm using variables of different type, and the code doesn't work (MATLAB error: 'Unable to perform assignment because brace indexing is not supported for variables of this type').
for i = 0:9
[r,c] = size(A);
randmat{i+1} = randi([0 1],r,c);
rowsum = sum(randmat,2);
randmat{i+1} = bsxfun(@rdivide,randmat{i+1},rowsum);
end
How could It be solved?
Thanks in advance for your help.
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Davide Biaggini
Davide Biaggini 2019-7-31
编辑：Davide Biaggini 2019-7-31
Thank you for your comment. Sparse can help saving some memory. However, as you've written it, the syntax means that 1:r stands for the number of rows, and randi(c, q, r) is the number of columns. This gives the sparse error 'Sparse matrix sizes must be non-negative integers less than MAXSIZE as defined by COMPUTER'. Am I mistaken?
By writing:
A=sparse(randi([0 1],r,c))
I get a sparse matrix of size rxc made of zeros and ones, which still doesn't answer my original question.

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Davide Biaggini
Davide Biaggini 2019-7-31
编辑：Davide Biaggini 2019-7-31
For those interested, here is the working solution that I found. The code generates a sequence of values via rand and sets the maximum of each row to 1, while the other values are set to 0. Each random binary array is stored in the initial 3d matrix. As example, in order to generate 5 random 7x3 binary matrices, containing only a single 1 in each row:
k=5; % number of (mxn) matrices
m=7;
n=3;
M = rand(m,n,k);
for i=1:k
M(:,:,i) = double(bsxfun(@eq, M(:,:,i), max(M(:,:,i), [], 2)));
end

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Andrei Bobrov
Andrei Bobrov 2019-7-31
编辑：Andrei Bobrov 2019-8-1
May be so?
m = 8;
n = 6; % [m,n] = size(A);
k = 10;
[~,Anew] = sort(rand(m,n,k),2);% here fixed
M = Anew == 1;
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