Iterating n python iterators at once

Question

I'm trying to iterate a bunch of data ranges at once, to get the combinations of all their values.

The number of ranges can differ, but I have them collected in a list.

Is there a way to iterate them using list comprehension or a similar clean, pythonic way?

This is what I mean by iterating together:

[print(i, j) for i in r1 for j in r2]

So that's a simple example with two known ranges, but what I need is more like

[print(i, j, ...) for i in r1 for j in r2 for k in r3...]

Note: i don't just need a list of number combinations, the iterators are my own iterator class which works similarly to range() but allows me to also get current state without calling next(), which would alter the state.

My iterator class sets its value back to the start on StopIteration, so it can be looped through more than once.

Here you can see the class:

@dataclass
class Range:
    start: float
    end: float
    step: float = field(default=1)
    includeEnd: bool = field(default=True)
    
    def __post_init__(self):
        self.value = self.start
        
    def __next__(self):
        v = self.value
        end = v > self.end if self.includeEnd else v >= self.end
        if not end:
            self.value += self.step
            return v
        else:
            self.value = self.start
            raise StopIteration

    def __iter__(self):
        return self

Answer 1

But how would you get the product of n iterators using itertools.product(), when you have a list of n iterators?

itertools.product(*the_list). Nothing special about product() there. The leading * is general Python syntax for treating a list (more generally, an iterable) as a sequence of individual arguments.

>>> from itertools import product
>>> args = [range(2), range(3), (i**2 for i in [5, 9])]
>>> args
[range(0, 2), range(0, 3), <generator object <genexpr> at 0x000001E2E7A710B0>]
>>> for x in product(*args):
...     print(x)
    
(0, 0, 25)
(0, 0, 81)
(0, 1, 25)
(0, 1, 81)
(0, 2, 25)
(0, 2, 81)
(1, 0, 25)
(1, 0, 81)
(1, 1, 25)
(1, 1, 81)
(1, 2, 25)
(1, 2, 81)