用给定的概率矩阵生成随机数 [英] Generate random number with given probability matlab

问题描述

我想以给定的概率生成一个随机数，但不确定如何:

I want to generate a random number with a given probability but I'm not sure how to:

我需要一个介于1到3之间的数字

I need a number between 1 and 3

num = ceil(rand*3);

但是我需要不同的值才能具有不同的生成概率.

but I need different values to have different probabilities of generating eg.

0.5 chance of 1
0.1 chance of 2
0.4 chance of 3

我敢肯定这很简单，但是我想不起来怎么做.

I'm sure this is straightforward but I can't think of how to do it.

推荐答案

简单的解决方案是生成具有均匀分布的数字(使用

The simple solution is to generate a number with a uniform distribution (using rand), and manipulate it a bit:

r = rand;
prob = [0.5, 0.1, 0.4];
x = sum(r >= cumsum([0, prob]));

或单线:

x = sum(rand >= cumsum([0, 0.5, 0.1, 0.4]));

说明

此处r是0到1之间的均匀分布的随机数.要生成1到3之间的整数，诀窍是将[0，1]范围划分为3个段，其中每个段的长度与其相应的概率成正比.就您而言，您将:

Explanation

Here r is a uniformly distributed random number between 0 and 1. To generate an integer number between 1 and 3, the trick is to divide the [0, 1] range into 3 segments, where the length of each segment is proportional to its corresponding probability. In your case, you would have:

• 段[0，0.5)，对应于数字1.
• 段[0.5，0.6)，对应于数字2.
• 段[0.6，1]，对应于数字3.

r落在任何细分中的概率与您希望每个数字的概率成正比. sum(r >= cumsum([0, prob]))只是将整数映射到段之一的一种理想方法.

The probability of r falling within any of the segments is proportional to the probabilities you want for each number. sum(r >= cumsum([0, prob])) is just a fancy way of mapping an integer number to one of the segments.

如果您对创建随机数矢量/矩阵感兴趣，可以使用循环或 arrayfun :

If you're interested in creating a vector/matrix of random numbers, you can use a loop or arrayfun:

r = rand(3); % # Any size you want
x = arrayfun(@(z)sum(z >= cumsum([0, prob])), r);

当然，还有矢量化解决方案，我太懒了，无法编写.

Of course, there's also a vectorized solution, I'm just too lazy to write it.

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