Multiply list in list string items with list in list integers

I have 2 lists: list1 and list2

list1 is a list in list and contains various strings
list2 is also a list in list and consists of different integer values

the dimensions of both lists is the same

how do I multiply each string in list1 with each integer in list2 in order to achieve something like that:

list1 = [['ABC','DEF'],['GHI','JKL'],['MNO','PQR']]
list2 = [[2, 2], [3, 3], [4, 4]]


result = [['ABC', 'ABC', 'DEF', 'DEF'], ['GHI', 'GHI', 'GHI', 'JKL', 'JKL', 'JKL'], ...]

Thank you very much

I already tried the zip() function and worked with the * operator but nothing worked

Expecting that sublists in list2 can have varying values. I coded this:

list1 = [['ABC','DEF'],['GHI','JKL'],['MNO','PQR']]
list2 = [[2, 2], [3, 3], [4, 4], [1, 1]]

combined = []
for sublist1, sublist2 in zip(list1, list2):
	sublist = []
	for elem1, elem2 in zip(sublist1, sublist2):
		sublist.extend([elem1]* elem2)

	combined.append(sublist)

print(combined)

Result:

[['ABC', 'ABC', 'DEF', 'DEF'], ['GHI', 'GHI', 'GHI', 'JKL', 'JKL', 'JKL'], ['MNO', 'MNO', 'MNO', 'MNO', 'PQR', 'PQR', 'PQR', 'PQR']]

Here's a simple 1-liner:

[
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 for l1,l2 in zip(list1, list2)]

@Zero's code can be simplified to this list comprehension oneliner:

list1 = [['ABC','DEF'],['GHI','JKL'],['MNO','PQR']]
list2 = [[2, 2], [3, 3], [4, 4], [1, 1]]
[[i for i, j in zip(a, b) for _ in range(j)] for a, b in zip(list1, list2)]

Performance:

In [153]: %timeit [[i for i, j in zip(a, b) for _ in range(j)] for a, b in zip(list1, list2)]
3.86 µs ± 234 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

In [154]: def func(list1, list2):
     ...:     combined = []
     ...:     for sublist1, sublist2 in zip(list1, list2):
     ...:         sublist = []
     ...:         for elem1, elem2 in zip(sublist1, sublist2):
     ...:             sublist.extend([elem1]* elem2)
     ...:
     ...:         combined.append(sublist)
     ...:     return combined

In [155]: %timeit func(list1, list2)
2.76 µs ± 135 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

I am surprised to see that my method is slower.

I came up with another method, but it is even slower somehow:

from itertools import chain

[list(chain(*([i]*j for i, j in zip(a, b)))) for a, b in zip(list1, list2)]

In [158]: %timeit [list(chain(*([i]*j for i, j in zip(a, b)))) for a, b in zip(list1, list2)]
4.76 µs ± 78.9 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

So I came up with the smartest method I can think of, but it is somehow still slow:

from functools import reduce
from operator import iconcat

def repeat_elements(a, b):
    return reduce(iconcat,([i]*j for i, j in zip(a, b)),[])

list(map(repeat_elements, list1, list2))

In [169]: %timeit list(map(repeat_elements, list1, list2))
3.86 µs ± 90.1 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

Somehow my smart methods are all less performant than the idiotic straight-forward approach. This is humbling. But I liked the tiny exercise.

list1 = [['ABC','DEF'],['GHI','JKL'],['MNO','PQR']]
list2 = [[2, 2], [3, 3], [4, 4]]
res = []
for l1, l2 in zip(list1, list2):
    res.append([])
    for t1, t2 in zip(l1, l2):
        res[-1].extend([t1]*t2)
print(res)