MATLAB中的加权随机数 [英] Weighted random numbers in MATLAB

问题描述

如何从权重分配给每个数字的向量a中随机选取N个数字?

How to randomly pick up N numbers from a vector a with weight assigned to each number?

让我们说:

a = 1:3; % possible numbers
weight = [0.3 0.1 0.2]; % corresponding weights

在这种情况下，拾取1的概率应该比拾取2的概率高3倍.

In this case probability to pick up 1 should be 3 times higher than to pick up 2.

所有权重的总和可以是任何值.

Sum of all weights can be anything.

推荐答案

R = randsample([1 2 3], N, true, [0.3 0.1 0.2])

randsample 包含在统计工具箱中

否则，您可以使用某种轮盘选择过程.参见此类似的问题(尽管不是特定于MATLAB的).这是我的单行实现:

Otherwise you can use some kind of roulette-wheel selection process. See this similar question (although not MATLAB specific). Here's my one-line implementation:

a = 1:3;             %# possible numbers
w = [0.3 0.1 0.2];   %# corresponding weights
N = 10;              %# how many numbers to generate

R = a( sum( bsxfun(@ge, rand(N,1), cumsum(w./sum(w))), 2) + 1 )

说明:

考虑间隔[0,1].我们为列表中的每个元素(1:3)分配一个长度的子间隔，该间隔与每个元素的权重成正比；因此1 get和长度0.3/(0.3+0.1+0.2)的间隔，其他都一样.

Consider the interval [0,1]. We assign for each element in the list (1:3) a sub-interval of length proportionate to the weight of each element; therefore 1 get and interval of length 0.3/(0.3+0.1+0.2), same for the others.

现在，如果我们生成一个在[0,1]上具有均匀分布的随机数，则[0,1]中的任何数字都具有相等的被提取概率，因此子间隔的长度决定了该随机数的概率每个间隔中的数字.

Now if we generate a random number with uniform distribution over [0,1], then any number in [0,1] has an equal probability of being picked, thus the sub-intervals' lengths determine the probability of the random number falling in each interval.

这与我在上面所做的操作匹配:选择一个数字X〜U [0,1](更像N数字)，然后以向量化的方式找出它属于哪个间隔..

This matches what I'm doing above: pick a number X~U[0,1] (more like N numbers), then find which interval it falls into in a vectorized way..

您可以通过生成足够大的序列N=1000来检查上述两种技术的结果:

You can check the results of the two techniques above by generating a large enough sequence N=1000:

>> tabulate( R )
  Value    Count   Percent
      1      511     51.10%
      2      160     16.00%
      3      329     32.90%

或多或少与归一化权重匹配的w./sum(w) [0.5 0.16667 0.33333]

which more or less match the normalized weights w./sum(w) [0.5 0.16667 0.33333]

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