根据另一个列表列表从列表列表中选择元素

Question

I have two lists of lists that will always have the same lengths.

l1 = [[0, 1, 2, 2, 0, 1, 2], [0, 1, 0, 0, 0], [0, 1, 2, 0, 0]]
l2 = [[0,(1,'a','b'), (1,'a','b'), (1,'a','b'), 0, (3, 'x','y'), 0], [0, (2,'c','d'), 0, 0, 0], [0, (3, 'e','f'), 0, 0, 0]]

What I want is to return the non 0 elements of l2 only if the quantity of identical elements much the sequence of 1-2s from l1 (or just 1s). In this case for example, I would like it to return just

out = [[(1,'a','b'), (1,'a','b'), (1,'a','b')], [(2,'c','d')], []]

because there are three identical tuples and a sequence of 1-2-2, while for (3, 'e','f'), there is only on but there is a sequence of 1-2 (so it should have been (3,'e','f'),(3, 'e','f') instead in l2.

What I have been trying to do is use Counter to count the occurrences of non '0' elements, but when counting 1 and 2s it does not differentiate between sequences, so I cannot do a comparison

for x, y in zip(l1,l2):
    print(Counter(x),Counter(y))

returns

Counter({2: 3, 0: 2, 1: 2}) Counter({0: 3, (1, 'a', 'b'): 3, (3, 'x', 'y'): 1})
Counter({0: 4, 1: 1}) Counter({0: 3, (2, 'c', 'd'): 2})
Counter({0: 5}) Counter({0: 4, (3, 'e', 'f'): 1})

where there is no way to no what the count of 1 and 2 correspond to.

Any help on how to go about doing this?

Some input-expected output examples:

l1 = [[0,1,2,0], [0,1]]
l2 = [[0, (1,'a','b'), (1,'a','b'), 0],[(2,'a','b'),(2,'a','b')]]
out = [[(1,'a','b'), (1,'a','b')],[]]

l1 = [[2,2,2], [1,2]]
l2 = [[(1,'a','b'), (1,'a','b'), (1,'a','b')],[(2,'a','b'),(2,'a','b')]]
out = [[(1,'a','b'), (1,'a','b'), (1,'a','b')],[(2,'a','b'),(2,'a','b')]]

l1 = [[0,1], [0,0]]
l2 = [[0, (3, 'x','y')],[0,0]]
out = [[(3, 'x','y')],[]]

Answer 1

Doesn't this just gives the desired output.

op =[]
def ml(l1, l2):
    for x,y in zip(l1,l2):
        sl =[] 
        for a,b in zip(x,y):
            if a:
                sl.append(b)
        op.append(sl)
    return op

print(ml(l1, l2))

Please clarify if output is not as desired.