itertools.product in dataframe

Inputs:

arr1  = ["A","B"]
arr2   = [[1,2],[3,4,5]]

Expected output:

Current output:

Current Code (using itertools):

import pandas as pd
from  itertools import product

def custom_product(arr1, arr2):
    expand_short_list = [[a1]*len(a2) for a1, a2 in zip(arr1,arr2)]
    return [[a1,a2] for a1, a2 in zip(sum(expand_short_list,[]),sum(arr2,[]))]

arr1  = ["A","B"]
arr2   = [[1,2],[3,4,5]]

df2 = pd.DataFrame(data = product(arr1,arr2),columns=["short_list", "long_list"])

Alternative code using nested list comprehensions to get the desired output:

import pandas as pd

def custom_product(arr1, arr2):
    expand_short_list = [[a1]*len(a2) for a1, a2 in zip(arr1,arr2)]
    return [[a1,a2] for a1, a2 in zip(sum(expand_short_list,[]),sum(arr2,[]))]

arr1  = ["A","B"]
arr2   = [[1,2],[3,4,5]]

df1 = pd.DataFrame(data = custom_product(arr1, arr2),columns=["short_list", "long_list"])

Question:

I'm wondering how could I achieve the desired output using itertools?

IIUC use DataFrame contructor with DataFrame.explode:

arr1  = ["A","B"]
arr2   = [[1,2],[3,4,5]]

df = (pd.DataFrame({'short_list':arr1, 'long_list':arr2})
        .explode('long_list')
        .reset_index(drop=True))
print (df)
  short_list long_list
0          A         1
1          A         2
2          B         3
3          B         4
4          B         5

Another idea is use flattening zipped arrays to list of tuples and pass to DataFrame constructor:

df = pd.DataFrame([(a, x) for a, b in zip(arr1, arr2) for x in b],
                  columns=['short_list','long_list'])
print (df)
  short_list  long_list
0          A          1
1          A          2
2          B          3
3          B          4
4          B          5