python combinations from lists

Question

I have n generators in a list, each arbitrary length but none of them is empty.

lst = [gen_1, gen_2, ... gen_n]

I'd like to create the list A using the generators so that each element of A

contains zero or one element from each generator
all generators are exhausted when finsihing the creation of A

The content of A should be something like this:

gen_1[0]
.
.
gen_1[n]
gen_1[0], gen_2[0]
.
.
gen_1[n], gen_2[0]
.
.
gen_1[n], gen_2[m], ... gen_n[o]

In essence this is like creating the powerset using itertools.combinations (e.g. here), but we take zero to one element from each generator.

I imagine this would be solvable using recursion but can't wrap my head around it. Any help is appreciated.

Functions let you iterate multiple times (so you don't have to store the elements elsewhere). — Kelly Bundy
– Kelly Bundy, Commented Jun 8, 2022 at 5:38
I see. All generators in the questions are generator functions yielding an iterable. — jake77
– jake77, Commented Jun 8, 2022 at 5:43
@DanielHao it is a use case. Normally when iterating multiple iterables parallel e.g. with zip the iteration stops after the shortest iterable is exhausted. — jake77
– jake77, Commented Jun 20, 2022 at 16:08

Kelly Bundy · Accepted Answer · 2022-06-08 07:01:46Z

Starting with the empty combination and then one by one consider each further generator and the combinations it contributes:

def gen_1():
    yield from 'ab'
def gen_2():
    yield from 'XY'
def gen_3():
    yield from [1, 2]
lst = [gen_1, gen_2, gen_3]

A = [[]]
for gen in lst:
    A += [a + [g]
          for a in A
          for g in gen()]

import pprint
pprint.pprint(A)

You could also swap the order of the list comprehension's two for clauses. Would likely be more efficient. I've done it the above way partly in order to somewhat use that they're generator functions and support multiple iterations.

Output (Try it online!):

[[],
 ['a'],
 ['b'],
 ['X'],
 ['Y'],
 ['a', 'X'],
 ['a', 'Y'],
 ['b', 'X'],
 ['b', 'Y'],
 [1],
 [2],
 ['a', 1],
 ['a', 2],
 ['b', 1],
 ['b', 2],
 ['X', 1],
 ['X', 2],
 ['Y', 1],
 ['Y', 2],
 ['a', 'X', 1],
 ['a', 'X', 2],
 ['a', 'Y', 1],
 ['a', 'Y', 2],
 ['b', 'X', 1],
 ['b', 'X', 2],
 ['b', 'Y', 1],
 ['b', 'Y', 2]]

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