How To Sample From Cartesian Product Without Repetition

I have a list of sets, and I wish to sample n different samples each containing an item from each set. What I do not want is to have it in order, so, for example, I will get all t

Solution 1:

All the above solutions waste a lot of resources for filtering repeated results when it comes to the end of the iteration. That's why I have thought of a method that has (almost) linear speed from start until the very end.

The idea is: Give (only in your head) each result of the standard order cartesian product an index. That would be for example for AxBxC with 2000x1x2 = 4000 elements:

0: (A[0], B[0], C[0])1: (A[1], B[0], C[0])
...
1999: (A[1999], B[0], C[0])2000: (A[0], B[0], C[1])
...
3999: (A[1999], B[0], C[1])
done.

So there are still some questions open:

• How do I get a list of possible indices? Answer: Just multiply 2000*1*2=4000 and every number below that will be a valid index.
• How do I generate random indices sequentially without repetition? There are two answers: If you want samples with a known sample size n, just use random.sample(xrange(numer_of_indices), n). But if you don't know the sample size yet (more general case), you have to generate indices on the fly to not waste memory. In that case, you can just generate index = random.randint(0, k - 1) with k = numer_of_indices to get the first index and k = number_of_indices - n for the nth result. Just check my code below (be aware, that I use a one sided linked list there to store the done indices. It makes insert operations O(1) operations and we need a lot of insertions here).
• How do I generate the output from the index? Answer: Well, say our index is i. Then i % 2000 will be the index of A for the result. Now i // 2000 can be treated recursively as the index for the cartesian product of the remaining factors.

So this is the code I came up with:

defrandom_order_cartesian_product(*factors):
    amount = functools.reduce(lambda prod, factor: prod * len(factor), factors, 1)
    index_linked_list = [None, None]
    for max_index inreversed(range(amount)):
        index = random.randint(0, max_index)
        index_link = index_linked_list
        while index_link[1] isnotNoneand index_link[1][0] <= index:
            index += 1
            index_link = index_link[1]
        index_link[1] = [index, index_link[1]]
        items = []
        for factor in factors:
            items.append(factor[index % len(factor)])
            index //= len(factor)
        yield items

Solution 2:

The following generator function generates non-repetitive samples. It will only work performantly if the number of samples generated is much smaller than the number of possible samples. It also requires the elements of the sets to be hashable:

defsamples(list_of_sets):
    list_of_lists = list(map(list, list_of_sets))  # choice only works on sequences
    seen = set()  # keep track of seen sampleswhileTrue:
        x = tuple(map(random.choice, list_of_lists))  # tuple is hashableif x notin seen:
            seen.add(x)
            yield x

>>> lst = [{'b', 'a'}, {'c', 'd'}, {'f', 'e'}, {'g', 'h'}]
>>> gen = samples(lst)
>>> next(gen)
('b', 'c', 'f', 'g')
>>> next(gen)
('a', 'c', 'e', 'g')
>>> next(gen)
('b', 'd', 'f', 'h')
>>> next(gen)
('a', 'c', 'f', 'g')

Solution 3:

You can use sample from the random lib:

import random
[[random.sample(x,1)[0] for x in list_of_sets] for _ in range(n)]

for example:

list_of_sets = [{1,2,3}, {4,5,6}, {1,4,7}]
n = 3

A possible output will be:

[[2, 4, 7], [1, 4, 7], [1, 6, 1]]

EDIT:

If we want to avoid repetitions we can use a while loop and collect the results to a set. In addition you can check that n is valid and return the Cartesian product for invalid n values:

chosen = set()
if0 < n < reduce(lambda a,b: a*b,[len(x) for x in list_of_sets]):
    whilelen(chosen) < n:
        chosen.add(tuple([random.sample(x,1)[0] for x in list_of_sets]))
else:
    chosen = itertools.product(*list_of_sets)

Solution 4:

Matmarbon's answer is valid, this is a complete version with an example and some modifies for easy understanding and easy use:

import functools
import random

defrandom_order_cartesian_product(factors):
    amount = functools.reduce(lambda prod, factor: prod * len(factor), factors, 1)
    print(amount)
    print(len(factors[0]))
    index_linked_list = [None, None]
    for max_index inreversed(range(amount)):
        index = random.randint(0, max_index)
        index_link = index_linked_list
        while index_link[1] isnotNoneand index_link[1][0] <= index:
            index += 1
            index_link = index_link[1]
        index_link[1] = [index, index_link[1]]
        items = []
        for factor in factors:
            items.append(factor[index % len(factor)])
            index //= len(factor)
        yield items


factors=[
    [1,2,3],
    [4,5,6],
    [7,8,9]
]

n = 5all = random_order_cartesian_product(factors)

count = 0for comb inall:
  print(comb)
  count += 1if count == n:
    break

Solution 5:

As I want no repetition, and sometimes it is not possible the code is not that short. But as @andreyF said, random.sample does the work. Perhaps there is also a better way that avoids resampling with repetition until enough non repetitive ones exist, this is the best I have so far.

import operator
import random
defget_cart_product(list_of_sets, n=None):
    max_products_num = reduce(operator.mul, [len(cluster) for cluster in list_of_sets], 1)
    if n isnotNoneand n < max_products_num:
        refs = set()
        whilelen(refs) < n:
            refs.add(tuple(random.sample(cluster, 1)[0] for cluster in list_of_sets))
        return refs
        return (prod for i, prod inzip(range(n), itertools.product(*list_of_sets)))
    return itertools.product(*list_of_sets)

Note that the code assumes a list of frozen sets, a conversion of random.sample(cluster, 1)[0] should be done otherwise.