Generating a list of random numbers, summing to 1

Question

• This question is not a duplicate of Getting N random numbers whose sum is M because:
  1. Most answers there are about theory, not a specific coding solution in python to answer this question
  2. The accepted answer here is 5 years older than the one answer in the duplicate that answers this question.
  3. The duplicate accepted answer does not answer this question

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

I can make a list of random numbers with

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

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

Answer 1

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

A more generic solution is to use the Dirichlet distribution which is available in numpy.

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]]

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 (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.

Answer 2

The best way to do this is to simply make a list of as many numbers as you wish, then divide them all by the sum. They are totally random this way.

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

or, as suggested by @TomKealy, keep the sum and creation in one loop:

rs = []
s = 0
for i in range(100):
    r = ran.random()
    s += r
    rs.append(r)

For the fastest performance, use numpy:

import numpy as np
a = np.random.random(100)
a /= a.sum()

And you can give the random numbers any distribution you want, for a probability distribution:

a = np.random.normal(size=100)
a /= a.sum()

---- Timing ----

In [52]: %%timeit
    ...: r = [ran.random() for i in range(1,100)]
    ...: s = sum(r)
    ...: r = [ i/s for i in r ]
   ....: 
1000 loops, best of 3: 231 µs per loop

In [53]: %%timeit
   ....: rs = []
   ....: s = 0
   ....: for i in range(100):
   ....:     r = ran.random()
   ....:     s += r
   ....:     rs.append(r)
   ....: 
10000 loops, best of 3: 39.9 µs per loop

In [54]: %%timeit
   ....: a = np.random.random(100)
   ....: a /= a.sum()
   ....: 
10000 loops, best of 3: 21.8 µs per loop

Answer 3

Dividing each number by the total may not give you the distribution you want. For example, with two numbers, the pair x,y = random.random(), random.random() picks a point uniformly on the square 0<=x<1, 0<=y<1. Dividing by the sum "projects" that point (x,y) onto the line x+y=1 along the line from (x,y) to the origin. Points near (0.5,0.5) will be much more likely than points near (0.1,0.9).

For two variables, then, x = random.random(), y=1-x gives a uniform distribution along the geometrical line segment.

With 3 variables, you are picking a random point in a cube and projecting (radially, through the origin), but points near the center of the triangle will be more likely than points near the vertices. The resulting points are on a triangle in the x+y+z plane. If you need unbiased choice of points in that triangle, scaling is no good.

The problem gets complicated in n-dimensions, but you can get a low-precision (but high accuracy, for all you laboratory science fans!) estimate by picking uniformly from the set of all n-tuples of non-negative integers adding up to N, and then dividing each of them by N.

I recently came up with an algorithm to do that for modest-sized n, N. It should work for n=100 and N = 1,000,000 to give you 6-digit randoms. See my answer at:

Create constrained random numbers?