Hea*_*ota 23 javascript random probability
我对如何使用概率生成整数值感到困惑.
作为一个例子,我有四个整数的概率值:1 | 0.4,2 | 0.3,3 | 0.2,4 | 0.1
如何在考虑其概率的情况下生成这四个数字?
Ser*_*sev 36
这是一个有用的技巧:-)
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];
}
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bhu*_*ups 27
一个简单的天真方法可以是:
function getRandom(){
var num=Math.random();
if(num < 0.3) return 1; //probability 0.3
else if(num < 0.6) return 2; // probability 0.3
else if(num < 0.9) return 3; //probability 0.3
else return 4; //probability 0.1
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Rom*_*098 14
基于@bhups答案的更灵活的解决方案.这使用概率值数组(权重).'权重'元素的总和应该等于1.
var weights = [0.3, 0.3, 0.3, 0.1]; // probabilities
var results = [1, 2, 3, 4]; // values to return
function getRandom () {
var num = Math.random(),
s = 0,
lastIndex = weights.length - 1;
for (var i = 0; i < lastIndex; ++i) {
s += weights[i];
if (num < s) {
return results[i];
}
}
return results[lastIndex];
};
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我建议对概率和其余随机数进行连续检查。
该函数首先将返回值设置为最后一个可能的索引,然后迭代,直到随机值的其余部分小于实际概率。
概率之和必须为一。
function getRandomIndexByProbability(probabilities) {
var r = Math.random(),
index = probabilities.length - 1;
probabilities.some(function (probability, i) {
if (r < probability) {
index = i;
return true;
}
r -= probability;
});
return index;
}
var i,
probabilities = [0.4, 0.3, 0.2, 0.09, 0.01 ],
count = {},
index;
probabilities.forEach(function (a) { count[a] = 0; });
for (i = 0; i < 1e6; i++) {
index = getRandomIndexByProbability(probabilities);
count[probabilities[index]]++
}
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这是我发现最灵活的解决方案,可以在任何具有概率的对象集中进行挑选:
// set of object with probabilities:
const set = {1:0.4,2:0.3,3:0.2,4:0.1};
// get probabilities sum:
var sum = 0;
for(let j in set){
sum += set[j];
}
// choose random integers:
console.log(pick_random());
function pick_random(){
var pick = Math.random()*sum;
for(let j in set){
pick -= set[j];
if(pick <= 0){
return j;
}
}
}Run Code Online (Sandbox Code Playgroud)