Filter löschen
Filter löschen

Create random matrix (MATLAB)

4 Ansichten (letzte 30 Tage)
high speed
high speed am 15 Jan. 2022
Kommentiert: high speed am 20 Jan. 2022
Dear,
I have these initial parameters :
numRows = 216;
numCols = 432;
A = zeros(numRows,numCols);
numOnesPerCol = randi(([2,3]),[1,numCols]);
numOnesPerRow = randi(([5,6]),[numRows,1]);
and I want to create a binary matrix with dimensions (numRows*numCols) that has numOnesPerCol and numOnesPerRow.
How can I do that please
  4 Kommentare
high speed
high speed am 15 Jan. 2022
@Tina I will give you a small example to understand the idea
numRows = 5;
numCols = 10;
A = zeros(numRows,numCols);
numOnesPerCol = randi(([2,3]),[1,numCols]); % Number of ones in each column
numOnesPerRow = randi(([5,6]),[numRows,1]); % Number of ones in each Row
I obtain for example this binary matrix
which contain 5 rows and 10 columns, in each column I have 2 or 3 ones. And in each row I have 5 or 6 ones.
high speed
high speed am 15 Jan. 2022
@Torsten I will give you a small example to understand the idea
numRows = 5;
numCols = 10;
A = zeros(numRows,numCols);
numOnesPerCol = randi(([2,3]),[1,numCols]); % Number of ones in each column
numOnesPerRow = randi(([5,6]),[numRows,1]); % Number of ones in each Row
I obtain for example this binary matrix
which contain 5 rows and 10 columns, in each column I have 2 or 3 ones. And in each row I have 5 or 6 ones.

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Torsten
Torsten am 15 Jan. 2022
Bearbeitet: Torsten am 20 Jan. 2022
Use intlinprog for the program
min: sum_{i=1}^{numRows} (e1_i+ + e1_i-) + sum_{j=1}^{numCols} (e2_j+ + e2_j-)
under the constraints
e1_j+ - e1_j- - sum_{i=1}^{numRows} A(i,j) = -numOnesPerCol(j) (j=1,...,numCols)
e2_i+ - e2_i- - sum_{j=1}^{numCols} A(i,j) = -numOnesPerRow(i) (i=1,...,numRows)
e1_j+, e1_j- >= 0 (j=1,...,numCols)
e2_i+, e2_i- >= 0 (i=1,...,numRows)
A(i,j) in {0,1} for all i,j
  6 Kommentare
Torsten
Torsten am 20 Jan. 2022
Bearbeitet: Torsten am 20 Jan. 2022
Test whether this code works for your purpose.
If fopt as output from intlinprog is equal to zero, a matrix with the desired properties exists and is written to console.
If you want to run the program for larger values of numRows and numCols, you should construct Aeq as a sparse matrix.
numRows = 5;
numCols = 10;
%numOnesPerCol = randi(([2,3]),[1,numCols]); % Number of ones in each column
%numOnesPerRow = randi(([5,6]),[numRows,1]); % Number of ones in each Row
numOnesPerCol = [3 2 3 3 2 3 2 3 3 2];
numOnesPerRow = [5;6;5;5;5];
%Usually, nothing needs to be changed after this line
f = [ones(2*numCols,1);ones(2*numRows,1);zeros(numRows*numCols,1)];
intcon = 2*numCols+2*numRows+1:2*numRows+2*numCols+numRows*numCols;
Aeq11 = [eye(numCols,numCols),-eye(numCols,numCols),zeros(numCols,numRows),zeros(numCols,numRows)];
Aeq12 = -repmat(eye(numCols,numCols),1,numRows);
Aeq21 = [zeros(numRows,numCols),zeros(numRows,numCols),eye(numRows,numRows),-eye(numRows,numRows)];
Aeq22 = [];
for i=1:numRows
Aeq22 = [Aeq22;zeros(1,(i-1)*numCols),ones(1,numCols),zeros(1,(numRows-i)*numCols)];
end
Aeq22 = -Aeq22;
Aeq = [Aeq11,Aeq12;Aeq21,Aeq22];
beq = [-numOnesPerCol.';-numOnesPerRow];
Aineq = [];
bineq = [];
lb = [zeros(2*numCols,1);zeros(2*numRows,1);zeros(numCols*numRows,1)];
ub = [Inf(2*numCols,1);Inf(2*numRows,1);ones(numCols*numRows,1)];
[xopt,fopt,status,output] = intlinprog(f,intcon,Aineq,bineq,Aeq,beq,lb,ub);
disp(xopt)
disp(fopt)
disp(status)
disp(output)
if abs(fopt) < eps
for i = 1:numRows
for j = 1:numCols
A(i,j) = xopt(2*numRows+2*numCols+(i-1)*numCols+j);
end
end
disp(A)
end
high speed
high speed am 20 Jan. 2022
@Torsten It works with your code. Thank you so much
I really appreciate your help

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Image Analyst
Image Analyst am 15 Jan. 2022
@high speed I think what you need to do is to first construct a Latin Rectangle:
Then mask it with two numbers like
output = latinRectangle == 1 | latinRectangle == 2;
Sorry I don't have Latin Rectangle code but there is Latin Square, and maybe Latin Rectangle, code in the File Exchange.
  1 Kommentar
high speed
high speed am 15 Jan. 2022
@Image Analyst Thank you for your response, but I think that I don't need to work with Latin Rectangle.
Because the idea here is to obtain binary matrix that contains a variety of ones between 2 and 3 in each column and a variety of ones between 5 and 6 in each row as in this example

Melden Sie sich an, um zu kommentieren.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by