Question

我有以下示例代码生成不同的master矩阵，每次运行时只包含0＆1和1＆＃39;

R=[];
Repeted=50;
Person=10;
for i=1:Repeted
R1=randi([0,1],[Person,21]);
R= [R;R1];
end
sumrows=find(sum(R(1:end,1:end),2)==6); 
if sumrows
B=R(sumrows,:);                            
end
master= unique(B,'rows')

现在我需要从master矩阵中随机选择一些元素并创建一个包含以下行总和的10行和21列的新矩阵：
[10 5 3 6 6 4 5 8 5 9 5 5 5 4 4 4 3 7 4 4 5]

如果master矩阵的元素无法使用规定的总和，那么我想重新创建master矩阵。

例如，如果我的代码生成了以下master矩阵：

master =   [ 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 1 1 1 0 0 0 ;
             0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 0 1 1 1 ;
             0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 1 1 1 ;
             0 0 0 0 0 1 1 0 0 0 1 1 0 0 1 0 1 0 0 0 0 ;
             0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 0 0 1 1 ;
             0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 1 1 0 1 ;
             0 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 1 0 1 1 ;
             0 0 0 1 0 0 1 1 0 0 1 0 0 0 1 0 0 0 0 0 1 ;
             0 0 0 1 0 1 0 1 0 0 0 0 0 1 0 0 1 1 0 0 0 ;
             0 0 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 0 1 0 0 ;
             0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 1 0 1 ;
             0 1 0 0 0 0 1 0 1 1 0 0 0 0 0 0 1 1 0 0 0 ;
             0 1 0 0 1 0 0 1 1 0 0 0 0 0 1 0 0 0 0 1 0 ;
             0 1 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 ;
             0 1 1 0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 0 ;
             0 1 1 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 0 ;
             1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 1 0 0 0 0 ;
             1 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 ;
             1 1 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 ;
             1 1 0 1 0 1 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 ]

由于master矩阵具有足够数量的1和0，因此可以生成新矩阵。所以新矩阵可以如下所示：

          [ 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 ;
            1 1 0 0 0 1 0 0 1 0 1 1 1 0 0 0 1 1 0 0 1 ;
            1 1 0 0 0 0 1 1 0 1 0 0 0 1 1 1 0 1 1 1 1 ;
            1 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 1 0 1 1 1 ;
            1 0 1 1 1 0 1 1 1 1 0 0 0 1 1 1 0 1 0 0 0 ;
            1 1 0 1 1 1 0 1 0 1 1 1 1 0 0 0 1 1 0 0 0 ;
            1 0 0 1 0 0 1 1 0 1 1 1 1 0 0 0 0 1 1 1 1 ;
            1 0 1 1 1 0 0 1 1 1 1 1 1 0 0 0 0 1 0 0 0 ;
            1 1 0 0 1 1 1 1 0 1 1 1 1 0 0 0 0 0 1 1 1 ;
            1 1 0 1 1 0 1 1 1 1 0 0 0 1 1 1 0 1 0 0 0 ]

但如果master矩阵没有足够的1和0，那么我想从我的代码中重新创建master矩阵。

Answer 1

<强>代码： -
我没有对您已编写的代码进行任何更改。完整的代码和解释为注释：

Repeted = 50;   Person = 10;

NewMat = zeros(10,21);  % Initializing 'NewMat' with all zeros 
SRNewMat=[10 5 3 6 6 4 5 8 5 9 5 5 5 4 4 4 3 7 4 4 5]; %Sum of the rows as mentioned by you
SSRNewMat=sum(SRNewMat); % Sum of 'SRNewMat' Matrix

while(1)
    R = [];
    for i = 1:Repeted
        R1 = randi([0,1],[Person,21]);
        R = [R;R1];        
        % In the above step, you are growing the size of 'R' in a loop 
        % which is not a good programming practice. Pre-allocation should
        % be done instead but I am leaving the code, that you wrote, as it is
    end
    sumrows = find(sum(R(1:end,1:end),2)==6);
    if sumrows
        B = R(sumrows,:);
    end
    Master = unique(B,'rows');

    if (sum(Master(:)) >= SSRNewMat) && ((numel(Master)-sum(Master(:)))>= (numel(NewMat)-SSRNewMat))
        % Checking if 'master' has required number of ones & required number of zeros
        for k = 1:21  % 21 is basically size(NewMat,2) i.e. no. of columns of NewMat
          NewMat(1:SRNewMat(k),k) = ones(SRNewMat(k),1);  %Replacing 0's with 1's in NewMat
          NewMat(:,k)= NewMat(randperm(10),k); %Shuffling the rows of kth column  of NewMat
          %In the above line, 10 is basically size(NewMat,1) i.e. no. of rows of NewMat
        end
        break; %No need to run again if mentioned sum is possible
    end
end
NewMat

的输出： -
我运行此代码时生成的输出之一如下所示：

NewMat = 1 0 0 1 1 1 0 1 1 0 1 1 1 1 1 0 1 0 0 1 1 1 1 1 0 1 0 1 1 0 1 0 0 0 1 1 0 0 1 1 0 1 1 1 0 1 0 0 0 1 0 1 0 0 0 0 0 1 0 1 0 0 1 1 0 0 1 1 1 1 1 0 1 1 1 1 0 0 1 0 1 0 1 0 1 0 0 1 0 0 1 0 1 1 1 0 1 0 0 0 0 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 1 1 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 1 0 1 1 0 0 1 0 0 0 0 0 0 0 1 1 1 1 0 0 0 1 0 1 0 1 0 0 0 1 0 1 1 0 0 1 1 0 1 1 1 1 1 1 1 1 0 0 0 1 0 1 0 0 1 0 1 0 1 0 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1

从另一个矩阵的元素创建特定条件的新矩阵

1 个答案: