Question

我需要生成3个随机浮点数（点后恰好有2位数），特定范围之间的特定平均值。

例如：

我希望平均值为37.14。我的最小值是31，而我的最大值是42。所以我想在这个范围内产生3个随机数，它给出了我想要的平均值：

41.35， 37.31， 32.77

我遇到了以下代码，但确实生成了整数。

n = 3;
xmean = 37;
xmin = 31;
xmax = 42;
xeps = 0.01;
x = randi([xmin xmax],n,1);
while abs(xmean - mean(x)) >= xeps
    if xmean > mean(x)
        x(find(x < xmean,1)) = randi([xmean xmax]);
    elseif xmean < mean(x)
        x(find(x > xmean,1)) = randi([xmin xmean]);
    end
end

非常感谢任何帮助。

Answer 1

只需将randi替换为rand：

n = 3;
xmean = 37;
xmin = 31;
xmax = 42;
xeps = 0.01;
x = 0.01*round(100*(xmin+(xmax-xmin)*rand(n,1)));
while abs(xmean - mean(x)) >= xeps
    if xmean > mean(x)
        x(find(x < xmean,1)) = 0.01*round(100*(xmean+(xmax-xmean)*rand(1)));
    elseif xmean < mean(x)
        x(find(x > xmean,1)) = 0.01*round(100*(xmin+(xmean-xmin)*rand(1)));
    end
end

Answer 2

另一个解决方案是将xmin xmean xmax乘以100.然后将结果除以100：

n = 3;
xmean = 37;
xmin = 31;
xmax = 42;
xeps = 0.01;

xmin = xmin*100;
xmax = xmax*100;
xmean = xmean*100;

x = randi([xmin xmax],n,1);
while abs(xmean - mean(x)) >= xeps
    if xmean > mean(x)
        x(find(x < xmean,1)) = randi([xmean xmax]);
    elseif xmean < mean(x)
        x(find(x > xmean,1)) = randi([xmin xmean]);
    end
end
x = x .* 0.01

Answer 3

考虑到约束，可以自由选择三个随机数中的两个。因此，您可以在范围内绘制两个数字并计算第三个数字。由于约束是均值，因此您知道至少有一个数字必须大于均值，并且必须小于均值才能减少迭代次数，直到找到满足范围的解决方案。

xmean = 37;
xmin = 31;
xmax = 42;

% scale for easier rounding to two digits
xmin = xmin*100;
xmax = xmax*100;
xmean = xmean*100;

foundSolution = false;
x = Nan(1,3);


while ~foundSolution
x(1) = randi([xmin,xmean]);
x(2) = randi([xmean,xmax]);
x(3) = 3*xmean-sum(x(1:2));

if x(3) >= xmin && x(3) <= xmax
   foundSolution = true;
end
end

%# rescale back
x = x/100;

在MATLAB中使用特定浮点均值在min和max之间生成3个随机浮点数

3 个答案: