Cartesian product of large iterators (itertools)

Question

From a previous question I learned something interesting. If Python's itertools.product is fed a series of iterators, these iterators will be converted into tuples before the Cartesian product begins. Related questions look at the source code of itertools.product to conclude that, while no intermediate results are stored in memory, tuple versions of the original iterators are created before the product iteration begins.

Question: Is there a way to create an iterator to a Cartesian product when the (tuple converted) inputs are too large to hold in memory? Trivial example:

import itertools
A = itertools.permutations(xrange(100))
itertools.product(A)

A more practical use case would take in a series of (*iterables[, repeat]) like the original implementation of the function - the above is just an example. It doesn't look like you can use the current implementation of itertools.product, so I welcome in submission in pure python (though you can't beat the C backend of itertools!).

Answer 1

Here's an implementation which calls callables and iterates iterables, which are assumed restartable:

def product(*iterables, **kwargs):
    if len(iterables) == 0:
        yield ()
    else:
        iterables = iterables * kwargs.get('repeat', 1)
        it = iterables[0]
        for item in it() if callable(it) else iter(it):
            for items in product(*iterables[1:]):
                yield (item, ) + items

Testing:

import itertools
g = product(lambda: itertools.permutations(xrange(100)),
            lambda: itertools.permutations(xrange(100)))
print next(g)
print sum(1 for _ in g)