How to generate a list of random numbers so their sum would be equal to a randomly chosen number

Question

I want to generate a list of random distribution of numbers so their sum would be equal to a randomly chosen number. For example, if randomly chosen number is 5, the distribution would be [1 2 2] or [2 3] or [1 1 1 2] and so on. Any suggestions are welcome!

Answer 1

Let n be the number you want values to add up to. Generate a random sample of random size (less than n), consisting of values in the range 1 to n exclusive of n. Now add the endpoints 0 and n, and sort. Successive differences of the sorted values will sum to n.

import random as r

def random_sum_to(n):
    a = r.sample(range(1, n), r.randint(1, n-1)) + [0, n]
    list.sort(a)
    return [a[i+1] - a[i] for i in range(len(a) - 1)]

print(random_sum_to(20))  # yields, e.g., [4, 1, 1, 2, 4, 2, 2, 4]

If you'd like to be able to specify the number of terms in the sum explicitly, or have it be random if unspecified, add an optional argument:

import random as r

def random_sum_to(n, num_terms = None):
    num_terms = (num_terms or r.randint(2, n)) - 1
    a = r.sample(range(1, n), num_terms) + [0, n]
    list.sort(a)
    return [a[i+1] - a[i] for i in range(len(a) - 1)]

print(random_sum_to(20, 3))   # [9, 7, 4] for example
print(random_sum_to(5))       # [1, 1, 2, 1] for example

Answer 2

In a loop, you could keep drawing a random number between 1 and the remaining sum until you've reached your total

from random import randint
def generate_values(n):
    values = []
    while n > 0:
        value = randint(1, n)
        values.append(value)
        n -= value
    return values

A few samples of such a function

>>> generate_values(20)
[17, 1, 1, 1]
>>> generate_values(20)
[10, 4, 4, 1, 1]
>>> generate_values(20)
[14, 4, 1, 1]
>>> generate_values(20)
[5, 2, 4, 1, 5, 1, 1, 1]
>>> generate_values(20)
[2, 13, 5]
>>> generate_values(20)
[14, 3, 2, 1]

Answer 3

Consider doing it continuously first. And for a moment we do not care about final number, so let's sample uniformly X_i in the interval [0...1] so that their sum is equal to 1

X_1 + X_2 + ... X_n = 1

This is well-known distribution called Dirichlet Distribution, or gamma variate, or simplex sampling. See details and discussion at Generating N uniform random numbers that sum to M. One can use random.gammavariate(a,1) or for correct handling of corners gamma variate with parameter 1 is equivalent exponential distribution, with direct sampling code below

def simplex_sampling(n):
    r = []
    sum = 0.0
    for k in range(0,n):
        x = random.random()
        if x == 0.0:
            return (1.0, make_corner_sample(n, k))

        t = -math.log(x)
        r.append(t)
        sum += t

    return (sum, r)

def make_corner_sample(n, k):
    r = []
    for i in range(0, n):
       if i == k:
           r.append(1.0)
       else:
           r.append(0.0)

    return r

So from simplex_sampling you have vector and the sum to be used as normalization.

Thus, to use it for, say, N=5

N = 5

sum, r = simplex_sampling(N)

norm = float(N)/sum

# normalization together with matching back to integers
result = []
for k in range(N):
    # t is now float uniformly distributed in [0.0...N], with sum equal to N
    t = r[k] * norm 
    # not sure if you could have zeros,
    # and check for boundaries might be useful, but
    # conversion to integers is trivial anyway:
    # values in [0...1) shall be converted to 0,
    # values in [1...2) shall be converted to 1, etc
    result.append( int(t) )

Answer 4

here's a simple way to have, a random list of probabilities, where there sum is equal to one;

a = np.random.random(2)
a /=sum(a)
a