# Create random matrix (MATLAB)

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high speed 2022 年 1 月 15 日
コメント済み: high speed 2022 年 1 月 20 日
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
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high speed 2022 年 1 月 15 日
@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.

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### 採用された回答

Torsten 2022 年 1 月 15 日

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
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high speed 2022 年 1 月 20 日
@Torsten It works with your code. Thank you so much

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### その他の回答 (1 件)

Image Analyst 2022 年 1 月 15 日
@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.
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high speed 2022 年 1 月 15 日
@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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