# Choose at random from combinations

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### Question :

Choose at random from combinations

I can make a list of all combinations using `list(itertools.combinations(range(n), m))` but this will typically be very large.

Given `n` and `m`, how can I choose a combination uniformly at random without first constructing a massive list??

``````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 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)
``````

``````def random_combination(iterable,r):