The fastest variant I know of this algorithm is the one found at https://stackoverflow.com/a/3796442/6899 :

    import itertools as it
    def erat3( ):
        D = { 9: 3, 25: 5 }
        yield 2
        yield 3
        yield 5
        MASK= 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0,
        MODULOS= frozenset( (1, 7, 11, 13, 17, 19, 23, 29) )

        for q in it.compress(
                it.islice(it.count(7), 0, None, 2),
                it.cycle(MASK)):
            p = D.pop(q, None)
            if p is None:
                D[q*q] = q
                yield q
            else:
                x = q + 2*p
                while x in D or (x%30) not in MODULOS:
                    x += 2*p
                D[x] = p