make an r by m permutation matrix

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fadams18
fadams18 2021-4-6
回答: Parag 2025-3-5
I have the following code :
A = eye(r);
L = [];
for i = 1:10
A = A(randperm(r),:);
L = [L A]; % 10 x 100 matrix with only one nonzero entry per column
end
The variable L is a 10x 100 size matrix. however i want to make a more intelligent code such that,
the dimensions will depend on the dimension of an input data matrix. for example when i input X, i.e.
% r is my rank. m and n are the dimensions.
X = rand(m,r)* rand(r,n);
% I want L to have dimension of [m by r] or [r by n] .
% i.e. it should pick on of the sides of X and the other side r

回答(1 个)

Parag
Parag 2025-3-5
Hi, the below code generates the dimensions m, n, and r randomly, ensuring that r is always less than or equal to the minimum of m and n. It then constructs a matrix X as the product of two random matrices of sizes (m×r) and (r×n). Based on the dimensions of X, the function generate_L creates a matrix L of size either (m×r) or (r×n), using randomly permuted identity matrices. This ensures L has only one nonzero entry per column while adapting to the input matrix X.
Finally, the code outputs the randomly chosen values of m, n, and r, along with the size of L, making the entire process fully automated and flexible.
function L = generate_L(X, r)
[m, n] = size(X);
% Determine the size of L based on X
if m >= n
L = generate_permutation_matrix(m, r); % Size [m x r]
else
L = generate_permutation_matrix(r, n); % Size [r x n]
end
end
function L = generate_permutation_matrix(rows, cols)
L = [];
I = eye(rows);
for i = 1:cols
permuted_I = I(randperm(rows), :); % Random permutation of identity
L = [L, permuted_I]; % Construct the final matrix
end
end
m = randi([5, 15]);
n = randi([5, 15]);
r = randi([min(m, n)-2, min(m, n)]); % Random r ensuring 1 <= r <= min(m, n)
X = rand(m, r) * rand(r, n);
L = generate_L(X, r);
disp(['m = ', num2str(m), ', n = ', num2str(n), ', r = ', num2str(r)]);
disp('Size of L:');
disp(L)
disp(size(L)); % Will output either [m x r] or [r x n]

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