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On Sat, 14 Sep 2019 at 03:26, Oscar Benjamin <oscar.j.benjamin at gmail.com> wrote: > > I've been staring at this for a little while: > > from itertools import product > > class Naturals: > def __iter__(self): > i = 1 > while True: > yield i > i += 1 > > N = Naturals() > print(iter(N)) > print(product(N)) # <--- hangs > > When I run the above the call to product hangs but I can't see why. There is already an issue for this: https://bugs.python.org/issue10109 It seems product always consumes all of its iterators entirely before yielding anything (in fact before even iter is called). That seemed surprising to me when looking at the trivial case of a product of one iterable since in *that* case it is clearly unnecessary. I wrote my own versions that get around this. This is a simple version: def iproduct_simple(*iterables): '''Returns the Cartesian product of iterables but doesn't hang for infinite iterators''' if len(iterables) == 0: yield () return first, others = iterables[0], iterables[1:] for f in first: for o in iprod(*others): yield (f,) + o This gives output in normal order for finite inputs and doesn't hang in the case of infinite inputs. However if any of the inputs is infinite iproduct_simple won't actually yield all elements. So I made a more complicated version that yields all possibilities (analogous to enumerating rational numbers): def iproduct(*iterables): '''Returns Cartesian product of iterables. Yields every element eventually''' if len(iterables) == 0: yield () return elif len(iterables) == 1: for e in iterables[0]: yield (e,) elif len(iterables) == 2: for e12 in iproduct2(*iterables): yield e12 else: first, others = iterables[0], iterables[1:] for ef, eo in iproduct2(first, iprod(*others)): yield (ef,) + eo def iproduct2(iterable1, iterable2): '''Cartesian product of two possibly infinite iterables''' it1 = iter(iterable1) it2 = iter(iterable2) elems1 = [] elems2 = [] sentinel = object() def append(it, elems): e = next(it, sentinel) if e is not sentinel: elems.append(e) n = 0 append(it1, elems1) append(it2, elems2) while n <= len(elems1) + len(elems2): for m in range(n-len(elems1)+1, len(elems2)): yield (elems1[n-m], elems2[m]) n += 1 append(it1, elems1) append(it2, elems2) -- Oscar

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