Javascript函数生成具有非均匀概率的随机整数 [英] Javascript function to generate random integers with nonuniform probabilities

问题描述

在javascript（或jquery）中有一个简单的函数，它有四个整数及其概率值：1 | 0.41,2 | 0.29,3 | 0.25,4 | 0.05

In javascript (or jquery) is there a simple function to have four integers with their probability values: 1|0.41, 2|0.29, 3|0.25, 4|0.05

如何根据其概率生成这四个数字？

how can I generate these four numbers taking into account their probabilities ?

此问题与此处发布的问题非常类似：生成具有概率的随机整数

This question is very similar to the one posted here: generate random integers with probabilities

然而，解决方案在那里发布：

HOWEVER the solution posted there:

function randomWithProbability() {
  var notRandomNumbers = [1, 1, 1, 1, 2, 2, 2, 3, 3, 4];
  var idx = Math.floor(Math.random() * notRandomNumbers.length);
  return notRandomNumbers[idx];
}

评论中的状态动态创建notRandomNumbers（给定数字及其权重/概率）

states in the comment "create notRandomNumbers dynamically (given the numbers and their weight/probability)"

这不足以满足我的需求。当概率为10％，20％，60％，10％时，这很有效。

This is insufficient for my needs. That works well when the probabilities are say 10%,20%, 60%,10%.

在这种情况下，使用所需的分布构造notRandomNumbers很容易，并且数组大小很小是小。但是在一般情况下概率可能是20.354％，30.254％等，数组大小对于正确建模情况来说是巨大的。

In that case constructing notRandomNumbers with the required distribution is easy and the array size is small. But in the general case where probabilities can be something like 20.354%,30.254% etc , the array size would be huge to correctly model the situation.

是否有一个干净的解决方案这个更普遍的问题？

Is there a clean solution to this more general problem?

编辑：谢谢Georg，解决方案已接受，这是我的最终版本，这可能对其他人有用。我已将累积计算拆分为单独的函数，以避免在每次调用时额外添加以获得新的随机数。

Thanks Georg, solution accepted, here is my final version, which may be useful for others. I have split the calculation of the cumulative into a separate function in order to avoid extra additions at each call to get a new random number.

function getRandomBinFromCumulative(cumulative) {
    var r = Math.random();
    for (var i = 0; i < cumulative.length; i++) {
        if (r <= cumulative[i])
            return i;
    }
}
    function getCummulativeDistribution(probs) {
    var cumulative = [];
    var sum = probs[0];
    probs.forEach(function (p) {
        cumulative.push(sum);
        sum += p;
    });
    // the next 2 lines are optional
    cumulative[cumulative.length - 1] = 1; //force to 1 (if input total was <>1)
    cumulative.shift();  //remove the first 0
    return cumulative;
}
function testRand() {
    var probs = [0.1, 0.3, 0.3, 0.3];
    var c = getCummulativeDistribution(probs);
    console.log(c);
    for (var i = 0; i < 100; i++) {
        console.log(getRandomBinFromCumulative(c));
    }
}

推荐答案

累积概率并返回 current_sum> = random_number 的项目：

Just accumulate the probabilities and return an item for which current_sum >= random_number:

probs = [0.41, 0.29, 0.25, 0.05];

function item() {
    var r = Math.random(), s = 0;
    for(var i = 0; i < probs.length; i++) {
        s += probs[i];
        if(r <= s)
            return i;
    }
}

// generate 100000 randoms

a = [];
c = 0;
while(c++ < 100000) {
    a.push(item());
}

// test actual distibution

c = {}
a.forEach(function(x) {
    c[x] = (c[x] || 0) + 1;
});

probs.forEach(function(_, x) {
    document.write(x + "=" +  c[x] / a.length + "<br>")
});

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