快速任意分布随机采样(逆变换采样) [英] Fast arbitrary distribution random sampling (inverse transform sampling)

问题描述

random 模块(http://docs.python.org/2/library/random.html) 有几个固定函数可以随机抽样.例如，random.gauss 将从具有给定均值和西格玛值的正态分布中采样随机点.

The random module (http://docs.python.org/2/library/random.html) has several fixed functions to randomly sample from. For example random.gauss will sample random point from a normal distribution with a given mean and sigma values.

我正在寻找一种方法，在 python<中使用我自己的分布尽可能快地 在给定间隔之间提取多个 N 随机样本.这就是我的意思:

I'm looking for a way to extract a number N of random samples between a given interval using my own distribution as fast as possible in python. This is what I mean:

def my_dist(x):
    # Some distribution, assume c1,c2,c3 and c4 are known.
    f = c1*exp(-((x-c2)**c3)/c4)
    return f

# Draw N random samples from my distribution between given limits a,b.
N = 1000
N_rand_samples = ran_func_sample(my_dist, a, b, N)

其中 ran_func_sample 是我所追求的，a, b 是从中抽取样本的限制.python 中有没有类似的东西?

where ran_func_sample is what I'm after and a, b are the limits from which to draw the samples. Is there anything of that sort in python?

推荐答案

您需要使用逆变换采样方法来获得根据您想要的规律分布的随机值.使用这种方法，您只需应用反函数到区间[0,1]内具有标准均匀分布的随机数.

You need to use Inverse transform sampling method to get random values distributed according to a law you want. Using this method you can just apply inverted function to random numbers having standard uniform distribution in the interval [0,1].

找到倒置函数后，您可以通过这种明显的方式根据所需的分布得到 1000 个数字分布:

After you find the inverted function, you get 1000 numbers distributed according to the needed distribution this obvious way:

[inverted_function(random.random()) for x in range(1000)]

更多关于逆变换采样:

• http://en.wikipedia.org/wiki/Inverse_transform_sampling

此外，StackOverflow 上有一个与该主题相关的好问题:

Also, there is a good question on StackOverflow related to the topic:

• Pythonic 选择不同概率列表元素的方法

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