朱莉娅（Julia）中有固定行总和的整数的随机矩阵

发布于 2025-01-18 08:10:21 字数 360 浏览 0 评论 0原文

我想创建一个带有以下约束的随机矩阵：

所有值均应是整数值
固定行总和（用第100、50、10、5等行减小）
随机值从0到6，

我尝试创建一个随机矩阵并将其扩展到我想要的固定行总和，但是后来我得到了我获得浮点值的问题（圆形也无济于事）

A = rand(0:6, 5, 10)

sum_vec = [10,5,3,2,2]

for i = 1:length(A[:,1])
    # row sum 
    s = sum(A[i,:])
    A[i,:]  = (A[i,:] /s ) * sum_vec[i]
end

非常感谢您的帮助！

原文

I would like to create a random matrix with the following constraints:

all values should be Integer values
fixed row sums (decreasing with row e.g. 100, 50, 10, 5 etc.)
random values from 0 to 6

I tried to create a random matrix and scale it to my fixed row sums I want, but then I get the problem that I get Float values (rounding does not help either)

A = rand(0:6, 5, 10)

sum_vec = [10,5,3,2,2]

for i = 1:length(A[:,1])
    # row sum 
    s = sum(A[i,:])
    A[i,:]  = (A[i,:] /s ) * sum_vec[i]
end

Thanks a lot for your help!

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影子的影子 2025-01-25 08:10:21

你还没有说你的随机数应该遵循什么分布，
所以我做出了一些武断的选择。

这是一个示例函数，您可以使用它来生成您想要的内容，其中 n 是必须是 0 到 6 之间的整数的元素数量，s 是它们所需的和：

function fixed_sum_sample(n, s)
    s > 6*n || s < 0 && throw(ArgumentError("wrong arguments"))
    x = zeros(Int, n)
    options = Set(1:n)
    for _ in 1:s
        i = rand(options) # you might want something else than uniform sampling
        x[i] += 1
        x[i] == 6 && pop!(options, i)
    end
    return x
end

You have not said what distribution your random numbers should follow,
so I have made some arbitrary choice.

Here is an example function you can use to generate what you want, where n is the number of elements that have to be integers from 0 to 6, and s is their desired sum:

function fixed_sum_sample(n, s)
    s > 6*n || s < 0 && throw(ArgumentError("wrong arguments"))
    x = zeros(Int, n)
    options = Set(1:n)
    for _ in 1:s
        i = rand(options) # you might want something else than uniform sampling
        x[i] += 1
        x[i] == 6 && pop!(options, i)
    end
    return x
end

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