Question

This turned out not to be a trivial task for me and I couldn’t find any receipt so maybe you can point me to one or you have a ready, proper and well-tuned solution for that? Proper meaning works also for iterators that do not know own length (without __len__) and works for exhaustible iterators (e.g. chained iterators); well-tuned meaning fast.

Note: in place solution is not possible due to necessity to cache iterators outputs to re-iterate them (Glenn Maynard pointed that out).

Example usage:

>>> list(izip_cycle(range(2), range(5), range(3)))
[(0, 0, 0), (1, 1, 1), (0, 2, 2), (1, 3, 0), (0, 4, 1)]
>>> from itertools import islice, cycle, chain
>>> list(islice(izip_cycle(cycle(range(1)), chain(range(1), range(2))), 6))
[(0, 0), (0, 0), (0, 1), (0, 0), (0, 0), (0, 1)]

Answer 1

Here is something inspired by itertools.tee and itertools.cycle. It works for any kind of iterable:

class izip_cycle(object):
    def __init__(self, *iterables ):
        self.remains = len(iterables)
        self.items = izip(*[self._gen(it) for it in iterables])

    def __iter__(self):
        return self.items

    def _gen(self, src):
        q = []
        for item in src:
            yield item
            q.append(item)

        # done with this src
        self.remains -=1
        # if there are any other sources then cycle this one
        # the last souce remaining stops here and thus stops the izip
        if self.remains:
            while True:
                for item in q:
                    yield item