生成总数为1的N个数字 [英] Generating N numbers that sum to 1

问题描述

给出一个大小为 n 的数组，我想为每个索引生成随机概率，以使 Sigma(a [0] .. a [n-1])= 1

Given an array of size n I want to generate random probabilities for each index such that Sigma(a[0]..a[n-1])=1

一个可能的结果可能是:

One possible result might be:

0     1     2     3     4
0.15  0.2   0.18  0.22  0.25

另一个完全合法的结果可能是:

Another perfectly legal result can be:

0     1     2     3     4
0.01  0.01  0.96  0.01  0.01

如何才能轻松，快速地生成这些内容?任何语言的答案都可以，最好是Java.

How can I generate these easily and quickly? Answers in any language are fine, Java preferred.

推荐答案

您要完成的任务无异于从N维单位单纯形中绘制随机点.

The task you are trying to accomplish is tantamount to drawing a random point from the N-dimensional unit simplex.

http://en.wikipedia.org/wiki/Simplex#Random_sampling 可能帮助您.

天真的解决方案可能如下:

A naive solution might go as following:

public static double[] getArray(int n)
    {
        double a[] = new double[n];
        double s = 0.0d;
        Random random = new Random();
        for (int i = 0; i < n; i++)
        {
           a [i] = 1.0d - random.nextDouble();
           a [i] = -1 * Math.log(a[i]);
           s += a[i];
        }
        for (int i = 0; i < n; i++)
        {
           a [i] /= s;
        }
        return a;
    }

要从N维单位单纯形中均匀地绘制一个点，我们必须获取一个指数分布的分布随机变量的矢量，然后通过这些变量的总和对其进行归一化.要获得指数分布的值，我们对均匀分布的值取负 log .

To draw a point uniformly from the N-dimensional unit simplex, we must take a vector of exponentially distributed random variables, then normalize it by the sum of those variables. To get an exponentially distributed value, we take a negative log of uniformly distributed value.

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