How to fill a zeros 3D array with a random number of ones specified in a 2D matrix?
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
The 3D array A is of size MxNxP.
The 2d matrix B is of size MxN.
For each row of B, there is at maximum one element greater than zero.
Example:
B = [0 3 0; 2 0 0];
A = zeros(2,3,4);
I want to randomly fill A such that it has exactly three ones in the first row and second column and exactly two ones in the second row and first column. For instance:
A(:,:,1) = [0 1 0; 0 0 0]
A(:,:,2) = [0 1 0; 1 0 0]
A(:,:,3) = [0 0 0; 1 0 0]
A(:,:,4) = [0 1 0; 0 0 0]
In other words, the B elements indicate how many ones should be in the third dimension of A with the same row and column indices.
I am currently able to do it with a loop but I'm looking forward to do it in a more efficient way.
This is my code:
B = [0 3 0; 2 0 0];
A = zeros(2,3,4);
for i = 1:size(B,1)
[r,c] = max(B(i,:));
idx = randperm(size(A,3),r);
A(i,c,idx) = 1;
end
0 Kommentare
Antworten (1)
Ameer Hamza
am 2 Okt. 2020
This is a one method
B = [0 3 0; 2 0 0];
n = 4;
A = zeros([size(B) n]);
idx = arrayfun(@(x) {randperm(4, x)}, B);
for i = 1:size(A, 1)
for j = 1:size(A, 2)
A(i, j, idx{i,j}) = 1;
end
end
Siehe auch
Kategorien
Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!