Dennis Sweeney <sweeney.dennis...@gmail.com> added the comment:
The trouble is that itertools.product accepts iterators, and there is no
guaranteed way of "restarting" an arbitrary iterator in Python. Consider:
>>> a = iter([1,2,3])
>>> b = iter([4,5,6])
>>> next(a)
1
>>> next(b)
4
>>> from itertools import product
>>> list(product(a, b))
[(2, 5), (2, 6), (3, 5), (3, 6)]
Since there's no way to get back to items you've already consumed, the current
approach is to consume all of the iterators to begin with and store their items
in arrays, then lazily produce tuples of the items at the right indices of
those arrays.
Perhaps one could consume lazily from the iterators, say, only filling up the
pools as they're needed and not storing the contents of the first iterator, but
this would mean sometimes producing a product iterator that was doomed to cause
a memory error eventually. If you really need this behavior you could do this
in Python:
def lazy_product(*iterables):
if not iterables:
yield ()
return
it0 = iterables[0]
for x in it0:
print(f"{x=}")
for rest in lazy_product(*iterables[1:]):
print(f"{rest=}")
yield (x,) + rest
The above could surely be optimized as maybe you're suggesting, but this would
be a backward-incompatible change for itertools.product.
----------
nosy: +Dennis Sweeney
_______________________________________
Python tracker <rep...@bugs.python.org>
<https://bugs.python.org/issue40230>
_______________________________________
_______________________________________________
Python-bugs-list mailing list
Unsubscribe:
https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
