如何增加或减少数组中某项被选中的概率?
How to increase or decrease the probability of an item of an array being picked?
所以,假设我正在制作老虎机之类的东西,要使用我想使用的表情符号,我会在数组中定义它们。
var arr = ["emoji","emoji2","emoji3","emoji4","emoji5"]
假设我希望表情符号 1 - 4 出现的次数多于 5,并说要降低表情符号 5 被选中的概率。
我可以做一些大事,例如:
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)]
但这不是一个非常有效的想法,那么是否可以在不制作非常大的数组的情况下实现上述想法?
我的目标基本上是拥有一个像
这样的数组
var arr = ["emoji","emoji2","emoji3","emoji4","emoji5"]
它会输出一些表情符号 1 - 4 比表情符号 5 更频繁出现的东西,没有大数组。
这样试试
var arr = ["emoji","emoji2","emoji3","emoji4","emoji5"]
var emoji = arr[Math.floor(Math.random() * (Math.random() < 0.75 ? arr.length - 1 : arr.length))]
对于加权概率的一般情况,一种选择是拥有一个其键为累积概率的对象。假设您希望 emoji5 在 4% 的时间内出现 - 那么,累积概率将为 24、48、72、96、100(其中最后一个区间 96 到 100 表示 emoji5 的权重较低)。然后生成一个1-100之间的随机数,并找到第一个大于所选数字的密钥:
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());
}
另一种选择是给每个字符串关联一个weight
数字,并给emoji5
一个低的数字,将权重相加,生成一个介于0和总权重之间的随机数,并找到第一个匹配项:
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());
}
所以,假设我正在制作老虎机之类的东西,要使用我想使用的表情符号,我会在数组中定义它们。
var arr = ["emoji","emoji2","emoji3","emoji4","emoji5"]
假设我希望表情符号 1 - 4 出现的次数多于 5,并说要降低表情符号 5 被选中的概率。
我可以做一些大事,例如:
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)]
但这不是一个非常有效的想法,那么是否可以在不制作非常大的数组的情况下实现上述想法?
我的目标基本上是拥有一个像
这样的数组var arr = ["emoji","emoji2","emoji3","emoji4","emoji5"]
它会输出一些表情符号 1 - 4 比表情符号 5 更频繁出现的东西,没有大数组。
这样试试
var arr = ["emoji","emoji2","emoji3","emoji4","emoji5"]
var emoji = arr[Math.floor(Math.random() * (Math.random() < 0.75 ? arr.length - 1 : arr.length))]
对于加权概率的一般情况,一种选择是拥有一个其键为累积概率的对象。假设您希望 emoji5 在 4% 的时间内出现 - 那么,累积概率将为 24、48、72、96、100(其中最后一个区间 96 到 100 表示 emoji5 的权重较低)。然后生成一个1-100之间的随机数,并找到第一个大于所选数字的密钥:
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());
}
另一种选择是给每个字符串关联一个weight
数字,并给emoji5
一个低的数字,将权重相加,生成一个介于0和总权重之间的随机数,并找到第一个匹配项:
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());
}