在指定概率分布范围内随机 select 个数字

Question

我想 select 一个范围为 3-10 或 4-6 的随机数。数字应该 selected，数字越小，它被 selected 的机会就越大。我下面的代码只有 selects 每个数字的概率相等。

private int bonusPoints;
private double randomBonusPoints = Math.Random() * 100;

bonusPoints = (int)randomBonusPoints;

如何从 P(3,4,5)=85%, P(6,7,8)=10%, P(9,10)=5% 等发行版中 select？

Answer 1

这是最简单的（从 "understanding the code" 的角度来看）的方法：

int choice;
double r = Math.random();

if(r < .5){  //50% chance to choose 4
   choice = 4;
}
else if(r < .9){  //40% chance to choose 5
   choice = 5;
}
else{  //10% chance to choose 6
   choice = 6;
}

显然，您可以针对其他数字和选择这些数字的其他机会进行调整，但这只是为了演示基础知识。

另请注意，谷歌搜索 "weighted random number generator java" returns 大量结果，包括来自 Whosebug 的大量答案。

Answer 2

如果您的分布可能会发生变化，所以您不想像 Kevin 的回答那样对您的分布进行硬编码，您可以使用 NavigableMap with the ceilingEntry 方法。这允许您为 selection 选项分配权重。然后，您可以生成一个从 0 到权重总和的随机数。然后根据您的加权分布统计输出。

select 条目的代码在创建地图后非常短：

Random rand = new Random();
...
double rnd = rand.nextDouble() * totalWeight;
int elem = map.ceilingEntry(rnd).getValue();

要创建任意分布，您可以执行以下操作：

int[] options = new int[]{3,4,5,6,7,8,9,10};
double[] weights = new double[]{ 0.85/3d, 0.85/3d, 0.85/3d,
                                 0.10/3d, 0.10/3d, 0.10/3d,
                                 0.05/2d, 0.05/2d };

NavigableMap<Double, Integer> map = new TreeMap<Double, Integer>();
double totalWeight = 0d;

for (int i = 0; i < weights.length; i++) {
    totalWeight += weights[i];
    map.put(totalWeight, options[i]);
}

测试超过 10000 个 select 离子给出了以下权重：

03: 28.99% 
04: 28.10% 
05: 28.06% 
06: 3.27% 
07: 3.62% 
08: 3.08% 
09: 2.40% 
10: 2.48%

测试代码：

//  select from the weighted elements
Random rand = new Random();
HashMap<Integer, Double> freqs = new HashMap<Integer, Double>();
int iterations = 10000;
for(int i = 0; i < iterations; i++) {
    double rnd = rand.nextDouble() * totalWeight;
    int elem = map.ceilingEntry(rnd).getValue();
    freqs.put(elem, (freqs.containsKey(elem) ? freqs.get(elem) : 0) + (1d/iterations));
}
Map<Integer, Double> sortedFreqs = new TreeMap<Integer, Double>(freqs);

for(Map.Entry<Integer,Double> entry : sortedFreqs.entrySet()) {
    System.out.printf("%02d: %.2f%% %n", entry.getKey(), entry.getValue() * 100d);
}

Answer 3

对初始随机数使用 Math.pow 可提供平滑的标度，而不会破坏随机性。选择你的体重是有争议的，但结果看起来不错。您也可以清楚地放大或缩小它。

public long weightedRandom(long lowest, long highest, double weight) {
    // Even distribution r >= 0 and < 1.
    double r = Math.random();
    // Add the weight while we are still between 0 and 1.
    r = Math.pow(r, weight);
    // Scale it - r >= 0 and <= highest - lowest.
    r = r * (highest - lowest + 1);
    // Translate to lowest.
    r += lowest;
    // Floor to long.
    return (long) r;
}

private void test(double weight) {
    List<Integer> results = new ArrayList<>(10);
    for (int i = 0; i < 10; i++) {
        results.add(0);
    }
    for (int i = 0; i < 1000; i++) {
        int r = (int) weightedRandom(0, results.size() - 1, weight);
        results.set(r, results.get(r) + 1);
    }
    System.out.println("Weight: " + weight + " Results: " + results);
}

public void test() {
    test(1);
    test(10);
    test(.1);
    test(2);
}

结果如下：

Weight: 1.0 Results: [119, 91, 112, 84, 96, 95, 86, 112, 93, 112]
Weight: 10.0 Results: [773, 57, 44, 37, 23, 15, 16, 9, 18, 8]
Weight: 0.1 Results: [0, 0, 0, 0, 0, 8, 14, 76, 243, 659]
Weight: 2.0 Results: [331, 119, 100, 85, 87, 71, 53, 59, 48, 47]

这对我来说像是一个比例因子。

在指定概率分布范围内随机 select 个数字

Randomly select numbers in a range from a specified probability distribution

java

random

range