JAVA - 如何生成对某些数字加权的随机数? [英] JAVA - How do I generate a random number that is weighted towards certain numbers?

问题描述

如何在 Java 中(使用 Eclipse)生成一个随机数，该数对某些数字进行加权?

How do I generate a random number in Java, (using Eclipse) that is weighted towards certain numbers?

例如，我有 100 个数字，每个数字都有 1% 的几率被平均选择.但是，如果某个数字，比如说 5，在 100 中显示了 8 次，那么它现在有 8% 的机会被随机选择，而其他数字仍然是 1%.

Example, I have 100 numbers, each can be equally chosen 1% of the time. However, if a certain number, lets say 5, is displayed 8 times in that 100, it now has an 8% chance to be randomly selected, while the other numbers are still at 1%.

如何编写一个随机生成器，给我 10 个数字并给数字 5 显示 8% 的优势?

How can I write a random generator that gives me 10 numbers and gives the number 5 that 8% advantage to show up?

如果我没有很好地解释，谢谢和抱歉.

Thanks and sorry if I did not explain that well.

推荐答案

public static void main(String[] args){
    ArrayList<Integer> list = new ArrayList<Integer>();
    list.add(5);list.add(2);// to be continued ...        
    int randomInt = list.get((int) (Math.random() * list.size()));
    System.out.println(randomInt);
}

你会得到你的值列表，然后你只需要随机选择一个索引并取回值

You'll get your List of values, and then you just need to pick randomly an index and get back the value

Math.random() 将在 [0;1[ 中选择一个值，如果乘以 10 个元素的 List 的大小，您将得到一个[0;10[ 中的值，当您将其转换为 int 时，您将获得 10 种可能的索引来查看 List

Math.random() will pick a value in [0;1[, if you multiply by the size of a List of 10 elements, you'll a value in [0;10[ , when you cast as an int you'll get 10 possibilities of index to look into the List

ArrayList(List 的实现)允许重复元素，所以:如果您添加 1,2,2,3,4,5,6,7,8,9(顺序无关紧要)，您将有 10 个元素，每个元素都可以找到 1/10 = 10/100 = 10% 的概率(又名机会)，但是出现两次的元素 2 将有 20% 机会在方法中被选中，出现的数字越多，它就越有机会被选中被选择增加，正如你所说的(数字出现的次数)/(元素的总数)(这是基础数学)

The ArrayList (implementation of List) allows duplicate element so : if you add 1,2,2,3,4,5,6,7,8,9 (the order does not matter) you'll have 10 elements with each one to be find with 1/10 = 10/100 = 10% of probability (aka chance), but the element 2 which appears twice will have 20% chance to be choosen in the method, more the number appears more it's chance to be choosen increase, with as you said (number of time the number appears)/(total number of element) (this is basic maths)

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