Create random matrix (MATLAB)
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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 个评论
Hi there ! Can you be a liitle clear?Do you want all the rows of first col have the numonesperrow data or do you want the first col of all the row have numonepercoldata ??
You cant merge your parameter in any way but you can try
numRows = 216;
numCols = 432;
A = zeros(numRows,numCols);
numOnesPerCol = randi(([2,3]),[215,numCols]);
numOnesPerRow = randi(([5,6]),[1,numCols]);
a=[numOnesPerRow ;numOnesPerCol ]
or vice versa
Torsten
2022-1-15
numOnesPerCol is a row vector (1x432) with only 2's and 3's.
numOnesPerRow is a row vector (1x216) with only 5's and 6's.
So what do you mean by
"I want to create a matrix with dimensions (numRows*numCols) that has numOnesPerCol and numOnesPerRow"
high speed
2022-1-15
@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
2022-1-15
@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.
采纳的回答
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 个评论
high speed
2022-1-15
Torsten
2022-1-15
Use the optimizer "intlinprog" and solve the system I defined above for the matrix elements A(i,j).
Image Analyst
2022-1-15
编辑：Image Analyst
2022-1-15
A solution may not exist. For example if you wanted, in a 5x10 matrix to have 9 or more in each row but only 1 or 2 in each column. There is no solution for that. Solutions exist only for certain combinations.
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
2022-1-20
更多回答（1 个）
Image Analyst
2022-1-15
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 个评论
high speed
2022-1-15
@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
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