Creating random integer row array each element different upper limit and sum of elements add up to a number

1 次查看（过去 30 天）

Biswanath Mahanty 2020-9-24

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/599269-creating-random-integer-row-array-each-element-different-upper-limit-and-sum-of-elements-add-up-to-a

编辑： Ameer Hamza 2020-9-28

I am looking for a intutuve code to generate 1x6 array where each element is random integer (with in specific upper limit) and sums to a given number.

For example a random array where element lower limit [ 0 0 0 0 0 0] and upper limit [ 9 12 25 6 14 7]. The sum of the random integer number adds up to 24, e.g.

[4 3 8 1 3 4 1]

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Ameer Hamza 2020-9-24

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/599269-creating-random-integer-row-array-each-element-different-upper-limit-and-sum-of-elements-add-up-to-a#answer_499900

编辑：Ameer Hamza 2020-9-28

在 MATLAB Online 中打开

See this excellent FEX package by John: https://www.mathworks.com/matlabcentral/fileexchange/49795-randfixedlinearcombination. However, It generates fractional numbers. Following code proposes one modification. I don't know how it will change the probability distribution.

lb = [0 0 0 0 0 0];
ub = [9 12 25 6 14 7];
n = numel(lb);
s = 24;
x = round(randFixedLinearCombination(1, s, [1 1 1 1 1 1], lb, ub));
k = sum(x) - s;
if k > 0
    idx = find(x >= (lb+1));
    idx = idx(randperm(numel(idx), k));
    x(idx) = x(idx) - 1;
else
    idx = find(x <= (ub-1));
    idx = idx(randperm(numel(idx), -k));
    x(idx) = x(idx) + 1;
end

8 个评论
显示 6更早的评论隐藏 6更早的评论

Bruno Luong 2020-9-27

编辑：Bruno Luong 2020-9-27

I think you can do better if you replace the first filtering with <= >= tests rather than strict inequality.

It still not perfect though.

Ameer Hamza 2020-9-28

Yes, I guess this will fail when the bounds are very tight in accordance with the sum constraints. The point about inequality is also valid. For the given bounds, the method does not seem to fail.

请先登录，再进行评论。