Matlab：随机数取决于另一个

Question

我的 Matlab 代码需要您的帮助...

我想在循环中设置一个随机数（0 到 1 之间）：

 for i=1:20
m1(i)= rand;

第二个随机数（n1(i) 应根据第一个最大值设置大于/小于 0.2，但至少大于/小于 m1(i) 0.1，没有负值。

所以最后我想要两个数字在 0 和 1 之间的向量，但是 m1 中的元素 1 和 n1 中的元素 1 不应该相差太大，但也不能太接近另一个...

非常感谢您的提示，我不明白...

Answer 1

这应该可行

for i=1:20
    m1(i)=rand;
    a=.2;% desired range
    n1(i) = (m1(i)-a).*rand(1,1) + m1(i)/2; % creating the second random vector 
  end

Answer 2

N = 100;
m = rand(N,1); %random numbers in [0 1]
n1 = rand(N,1); %random numbers in [0 1]
%now we want to change the range of n from [0 1] to [m+0.1 m+0.2]
n1 = n1*0.1; %random numbers in [0 0.1]
n1 = n1+m+0.1; %random numbers in [m+0.1 m+0.2]

%to get a range [m-0.2 m-0.1]
n2 = rand(N,1); %random numbers in [0 1]
n2 = n2*0.1; %random numbers in [0 0.1]
n2 = n2-0.2+m; 

%but if n2 is negative we should have n1
n2(n2<0) = n1(n2<0);

%random put half the indexes as n1
[~,idx] = sort(rand(N,1));
n2(idx(1:floor(N/2)))=n1(idx(1:floor(N/2)));

n2 是你的向量。

Answer 3

尝试了几个解决方案，这是我想出的第一个完整解决方案。符合以下所有约束

• .1 < |m-n| < .2
• 0 < m < 1
• 0 < n < 1

我的 while 循环效率不高，它比将 n(i) 约束为 [0,1] 更简洁、更容易实现；它还保持随机分布。请注意，0 < n < 1 约束将使n 相对于m 向0.5 倾斜

ub = .2;
lb = .1;

for i = 1:1000
  m(i) = rand;


  n(i) = -1;
  % Determine n with [0,1] constraints
  while(n(i)<0 || n(i)>1)    
    n(i) = 2*(ub-lb)*rand + m(i)-(ub-lb);     % Determine the upper bound shell
    n(i) = n(i) + sign(n(i)-m(i))*lb;      % Add the lower bound
  end
  y(i) = m(i) - n(i);                    % Check the difference

end


figure;
hist(y,100);

Answer 4

这是一个线性化的解决方案：

len = 10000;                  %length of your vector
m1  = rand(len,1);            %input vector
g = (randi([0,1],len,1)*2-1); %random [-1 or 1] generator
a = 0.1.*m1.*g+m1;            %use a = 0.1.*g+m1 for an absolute [0.1,0.2] interval
b = 0.2.*m1.*g+m1;            %use b = 0.2.*g+m1 for an absolute [0.1,0.2] interval
n1 = a + (b-a).*rand(len,1);  %output vector