product of different length list using itertools in Python

I have the following lists:

sectors = ["A", "B"]

rows = [['1', '2', '3'], ['1', '2', '3', '4']]

seats = [['ab', 'abcd', 'ab'], ['ab', 'abcd', 'ab', 'abcd']]

and I want to create products like A1a, A1b, A2a, A2b, A2c ...

this code

  combinations = []
  for i in range(len(rows)):
    c = list(zip_longest(repeat(sectors[i], len(rows[i])), rows[i], seats[i]))
    combinations += c

  for c in combinations:
    for x in product(*c):
       print("".join(x))

prints the desired results as A1a A1b A2a A2b A2c A2d A3a ...

Can this be solved in a better and more readable way, I am practicing itertools and it is a bit confusing for me.

Solution 1:

Not sure if that is what you want or if it is more elegant:

from itertools import chain, product

combinations = product(
    sectors,
    chain.from_iterable(chain.from_iterable(rows)),
    chain.from_iterable(chain.from_iterable(seats)),
)
joined_combinations = map(lambda t: "".join(t), combinations)
list(joined_combinations)
# returns
['A1a', 'A1b', 'A1a', 'A1b', 'A1c', 'A1d', 'A1a', ...]

Explanation: Applying two times chain.from_iterable you can "unpack" individual characters from the nested lists, then creating the product of the items of the unnested lists (which creates 3-tuples) and finally join the items of each 3-tuple together.
If you want to avoid duplicates you can put a set() around each argument in the product.

Solution 2:

Since you're creating triples from 3 nested lists, not Cartesian product, I think it may be more intuitive to zip the corresponding sublists and use a loop.

out = [sector + row + s for sector, rws, sts in zip(sectors, rows, seats) 
       for row, seat in zip(rws, sts) for s in seat] 

If we want to use itertools.product, first repeat elements in sectors to match the lengths of sublists of the other lists and instead of creating the product elements manually like in the inner-most loop above, we iterate over the product itself. Of course, this is a lot like your own approach.

repsectors = [repeat(sector, len(rows[i])) for i, sector in enumerate(sectors)]
out = [''.join(x) for tpl1 in zip(repsectors, rows, seats) 
       for tpl2 in zip(*tpl1) for x in itertools.product(*tpl2)]

Output:

['A1a', 'A1b', 'A2a', 'A2b', 'A2c', 'A2d', 'A3a', 'A3b', 'B1a', 'B1b', 'B2a', 'B2b', 'B2c', 'B2d', 'B3a', 'B3b', 'B4a', 'B4b', 'B4c', 'B4d']