生成一个随机数列表，总计为1 [英] Generating a list of random numbers, summing to 1

问题描述

我如何列出N个(例如100个)随机数，使它们的和为1?

How would I make a list of N (say 100) random numbers, so that their sum is 1?

我可以列出一个随机数

r = [ran.random() for i in range(1,100)]

我将如何对其进行修改，以使列表的总和为1(这是用于概率模拟).

How would I modify this so that the list sums to 1 (this is for a probability simulation).

推荐答案

最简单的解决方案的确是采用N个随机值并除以和.

The simplest solution is indeed to take N random values and divide by the sum.

更通用的解决方案是使用Dirichlet分布 http://en.wikipedia.org/wiki/Dirichlet_distribution 在numpy中可用.

A more generic solution is to use the Dirichlet distribution http://en.wikipedia.org/wiki/Dirichlet_distribution which is available in numpy.

通过更改分布参数，您可以更改单个数字的随机性"

By changing the parameters of the distribution you can change the "randomness" of individual numbers

>>> import numpy as np, numpy.random
>>> print np.random.dirichlet(np.ones(10),size=1)
[[ 0.01779975  0.14165316  0.01029262  0.168136    0.03061161  0.09046587
   0.19987289  0.13398581  0.03119906  0.17598322]]

>>> print np.random.dirichlet(np.ones(10)/1000.,size=1)
[[  2.63435230e-115   4.31961290e-209   1.41369771e-212   1.42417285e-188
    0.00000000e+000   5.79841280e-143   0.00000000e+000   9.85329725e-005
    9.99901467e-001   8.37460207e-246]]

>>> print np.random.dirichlet(np.ones(10)*1000.,size=1)
[[ 0.09967689  0.10151585  0.10077575  0.09875282  0.09935606  0.10093678
   0.09517132  0.09891358  0.10206595  0.10283501]]

取决于主要参数，狄利克雷分布将给出所有值都接近1./N的矢量，其中N是矢量的长度，或者给出矢量中大多数值约为0的矢量，并且只有一个，或者在这些可能性之间提供一些东西.

Depending on the main parameter the Dirichlet distribution will either give vectors where all the values are close to 1./N where N is the length of the vector, or give vectors where most of the values of the vectors will be ~0 , and there will be a single 1, or give something in between those possibilities.

EDIT(原始答案的5年后):关于Dirichlet分布的另一个有用事实是，如果您生成Gamma分布的一组随机变量，然后将它们除以它们的总和，就自然可以得到它.

EDIT(5 years after the original answer): Another useful fact about the Dirichlet distribution is that you naturally get it, if you generate a Gamma-distributed set of random variables and then divide them by their sum.

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