因此,假设我正在制作类似于插槽机器的东西,使用我想要使用的表情符号,我会在数组中定义它们。
var arr = ["emoji","emoji2","emoji3","emoji4","emoji5"]
假设我希望emojis 1-4出现在5以上,并且减少emoji5被选中的概率。
我可以做一些很大的事情,比如:
var arr = [
"emoji","emoji2","emoji3","emoji4",
"emoji","emoji2","emoji3","emoji4",
"emoji","emoji2","emoji3","emoji4",
"emoji","emoji2","emoji3","emoji4",
"emoji","emoji2","emoji3","emoji4",
"emoji","emoji2","emoji3","emoji4","emoji5",
]
var emoji = arr[Math.floor(Math.random() * arr.length)]
但是,这不是一个非常有效的想法,那么,在不制作一个非常大的数组的情况下,是否有可能做到上面的想法呢?
我的目标基本上是有一个数组,比如
var arr = ["emoji","emoji2","emoji3","emoji4","emoji5"]
而且它会输出一些表情符号1-4会比emoji5更频繁出现的东西,而没有一个大数组。
发布于 2020-04-25 08:24:00
对于加权概率的一般情况,一种选择是有一个对象,其键具有累积概率。假设你想要emoji5发生4%的时间-那么,累积概率是24,48,72,96,100 (最后的间隔96到100表示表情符号5的低体重)。然后生成一个介于1-100之间的随机数,并找到第一个键,它大于所选数:
const probs = {
24: "emoji",
48: "emoji2",
72: "emoji3",
96: "emoji4",
100: "emoji5"
};
const keys = Object.keys(probs).map(Number);
const generate = () => {
const rand = Math.floor(Math.random() * 100);
const key = keys.find(key => rand < key);
return probs[key];
};
for (let i = 0; i < 10; i++) {
console.log(generate());
}
另一个选项是将一个weight
数字与每个字符串关联起来,给emoji5
一个低的字符串,将权重相加,在0和总权重之间生成一个随机数,并找到第一个匹配:
const weights = [
[4, 'emoji'],
[4, 'emoji2'],
[4, 'emoji3'],
[4, 'emoji4'],
[1, 'emoji5'],
];
const totalWeight = weights.reduce((a, [weight]) => a + weight, 0);
const weightObj = {};
let weightUsed = 0;
for (const item of weights) {
weightUsed += item[0];
weightObj[weightUsed] = item;
}
const keys = Object.keys(weightObj);
const generate = () => {
const rand = Math.floor(Math.random() * totalWeight);
const key = keys.find(key => rand < key);
return weightObj[key][1];
};
for (let i = 0; i < 10; i++) {
console.log(generate());
}
发布于 2020-04-25 08:22:51
试着这样做
var arr = ["emoji","emoji2","emoji3","emoji4","emoji5"]
var emoji = arr[Math.floor(Math.random() * (Math.random() < 0.75 ? arr.length - 1 : arr.length))]
https://stackoverflow.com/questions/61422806
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