快速任意分布随机抽样 [英] Fast arbitrary distribution random sampling

问题描述

random模块( http://docs.python.org/2/library/random.html )具有几个 fixed 函数，可以从中随机采样.例如，random.gauss将从具有给定均值和sigma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，该方法使用我自己的python中的分布尽可能快地在给定间隔内提取许多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，而a, b是抽取样本的限制. 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取根据所需定律分布的随机值.使用这种方法，您可以仅应用倒置函数 到在[0,1]区间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].

找到反函数后，您将以明显的方式根据所需的分布获得1000个数字:

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)]

有关逆变换采样的更多信息:

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

此外，有关该主题的StackOverflow上还有一个很好的问题:

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

• 通过Python方式选择具有不同概率的列表元素

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