使用 Python 从单尾分布生成范围内的随机数

Question

I want to generate a random float in the range [0, 1) from a one-tailed distribution that looks like this

The above is the chi-squared distribution. I can only find resources on drawing from a uniform distribution in a range, however.

Answer 1

You could use a Beta distribution , eg

import numpy as np

np.random.seed(2018)
np.random.beta(2, 5, 10)
#array([ 0.18094173,  0.26192478,  0.14055507,  0.07172968,  0.11830031,
#        0.1027738 ,  0.20499125,  0.23220654,  0.0251325 ,  0.26324832])

Here we draw numbers from a Beta(2, 5) distribution

The Beta distribution is a very versatile and fundamental distribution in statistics; without going into any details, by changing the parameters alpha and beta you can make the distribution left-skewed, right-skewed, uniform, symmetric etc. The distribution is defined on the interval [0, 1] which is consistent with what you're after.

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A more technical comment

While the Kumaraswamy distribution certainly has more benign algebraic properties than the Beta distribution I would argue that the latter is the more fundamental distribution; for example, in Bayesian inference, the Beta distribution often enters as the conjugate prior when dealing with binomial(-like) processes.

Secondly, the mean and variance of the Beta distribution can be expressed quite simply in terms of the parameters alpha , beta ; for example, the mean is simply given by alpha / (alpha + beta) .

Lastly, from a computational and statistical inference point of view, fitting a Beta distribution to data is usually done in a few lines of code in Python (or R), where most Python libraries like numpy and scipy already include methods to deal with the Beta distribution.

Answer 2

I would leaning toward distribution which is naturally bounded on [0...1] interval (or any other [a...b] interval which could be rescaled later), like @MauritsEvers answer. Reason is, you know the distribution and could derive (or read) some interesting facts about it. If you use chi2 adn truncate it, it is unclear how to argue about properties of what you've got.

Personally I prefer Kumaraswamy distribution over Beta distribution, expressions for mean,mode, variance etc are a lot simpler.

Just install it

pip install kumaraswamy

and sample

from kumaraswamy import kumaraswamy

d = kumaraswamy(a=2.0, b=5.0)

q = d.rvs(10)
print(q)

will produce 10 numbers following magenta curve in the Wiki article.

If you don't want Beta or Kumaraswamy, there is fe Logit-normal distribution and quite a few others

Answer 3

Look at the numpy.random.chisquare method library.

numpy.random.chisquare(df, size=None)

>>> np.random.chisquare(2,4)
array([ 1.89920014,  9.00867716,  3.13710533,  5.62318272])

Answer 4

If you want to draw a sample of size N = 5 from a ChiSquare distribution, you can try OpenTURNS library:

import openturns as ot`
# define your distribution. Here, nu = 3. (nu is a float > 0)
distribution = ot.ChiSquare(3) 

# draw a sample of size N from `distribution`
N=5
sample = distribution.getSample(N)

A complete list of distributions is available here

sample has an OpenTURNS format but you can manipulate it as a Numpy array:

s = np.array(Sample)
print(s)
>>>array([[1.65299759],
      [6.78405097],
      [0.88528975],
      [0.87900211],
      [0.25031129]])

You can also easily plot the distribution PDF just by calling : distribution.drawPDF()

Customizations:

from openturns.viewer import View
graph = distribution.drawPDF()
title = str(distribution)[:100].split('\n')[0]
graph.setTitle(title)
View(graph, add_legend=False)