Generate a random 3 element Numpy array of integers summing to 3

Question

I need to fill a numpy array of three elements with random integers such that the sum total of the array is three (e.g. [0,1,2]).

By my reckoning there are 10 possible arrays:

111, 012, 021, 102, 120, 201, 210, 300, 030, 003

My ideas is to randomly generate an integer between 1 and 10 using randint, and then use a look-up table to fill the array from the above list of combinations.

Does anyone know of a better approach?

Lots of discussion on this [here](http://stackoverflow.com/q/11380491/553404) for this problem at larger sizes. May be useful to read the points uniformity/bias. Though if the question here is about the best way to do it in this restricted case then IMO it's not a duplicate. — YXD, Nov 26 '13 at 09:50

score 2 · Answer 1 · answered Nov 26 '13 at 09:28

Here is how I did it:

>>> import numpy as np
>>> a=np.array([[1,1,1],[0,1,2],[0,2,1],[1,0,2],[1,2,0],[2,0,1],[2,1,0],[3,0,0],[0,3,0],[0,0,3]])
>>> a[np.random.randint(0,10)]
array([1, 2, 0])
>>> a[np.random.randint(0,10)]
array([0, 1, 2])
>>> a[np.random.randint(0,10)]
array([1, 0, 2])
>>> a[np.random.randint(0,10)]
array([3, 0, 0])

rroowwllaanndd · Answer 2 · 2013-11-26T12:51:34.320

Here’s a naive programmatic way to do this for arbitrary array sizes/sums:

def n_ints_summing_to_v(n, v):
  elements = (np.arange(n) == np.random.randint(0, n)) for i in range(v))
  return np.sum(elements, axis=0)

This will, of course, slow down proportionally to the desired sum, but would be ok for small values.

Alternatively, we can phrase this in terms of drawing samples from the Multinomial distribution, for which there is a function available in NumPy (see here), as follows:

def n_ints_summing_to_v(n, v):
  return np.random.multinomial(v, ones((n)) / float(n))

This is a lot quicker!

score 1 · Answer 3 · answered Sep 23 '15 at 14:46

This problem can be solved in the generic case, where the number of elements and their sum are both configurable. One advantage of the solution below is that it does not require generating a list of all the possibilities. The idea is to pick random numbers sequentially, each of which is less than the required sum. The required sum is reduced every time you pick a number:

import numpy

def gen(numel = 3, sum = 3):
    arr = numpy.zeros((numel,), dtype = numpy.int)
    for i in range(len(arr) - 1): # last element must be free to fill in the sum
        arr[i] = numpy.random.randint(0, sum + 1)
        sum -= arr[i]
        if sum == 0: break # Nothing left to do
    arr[-1] = sum # ensure that everything adds up
    return arr

print(gen())

This solution does not guarantee that the possibilities will all occur with the same frequency. Among the ten possibilities you list, four start with 0, three with 1, two with 2 and one with 3. This is clearly not the uniform distribution that numpy.random.randint() provides for the first digit.

Generate a random 3 element Numpy array of integers summing to 3

3 Answers3