Choose at random from combinations

Solution 1:

In the itertools module there is a recipe for returning a random combination from an iterable. Below are two versions of the code, one for Python 2.x and one for Python 3.x - in both cases you are using a generator which means that you are not creating a large iterable in memory.

Assumes Python 2.x

def random_combination(iterable, r):
    "Random selection from itertools.combinations(iterable, r)"
    pool = tuple(iterable)
    n = len(pool)
    indices = sorted(random.sample(xrange(n), r))
    return tuple(pool[i] for i in indices)

In your case then it would be simple to do:

>>> import random
>>> def random_combination(iterable, r):
    "Random selection from itertools.combinations(iterable, r)"
    pool = tuple(iterable)
    n = len(pool)
    indices = sorted(random.sample(xrange(n), r))
    return tuple(pool[i] for i in indices)
>>> n = 10
>>> m = 3
>>> print(random_combination(range(n), m))
(3, 5, 9) # Returns a random tuple with length 3 from the iterable range(10)

In the case of Python 3.x

In the case of Python 3.x you replace the xrange call with range but the use-case is still the same.

def random_combination(iterable, r):
    "Random selection from itertools.combinations(iterable, r)"
    pool = tuple(iterable)
    n = len(pool)
    indices = sorted(random.sample(range(n), r))
    return tuple(pool[i] for i in indices)

Solution 2:

From http://docs.python.org/2/library/itertools.html#recipes

def random_combination(iterable, r):
    "Random selection from itertools.combinations(iterable, r)"
    pool = tuple(iterable)
    n = len(pool)
    indices = sorted(random.sample(xrange(n), r))
    return tuple(pool[i] for i in indices)

Solution 3:

A generator would be more memory efficient for iteration:

def random_combination(iterable,r):
    i = 0
    pool = tuple(iterable)
    n = len(pool)
    rng = range(n)
    while i < r:
        i += 1
        yield [pool[j] for j in random.sample(rng, r)]