How to consolidate the function of local_sum (list of mixed numbers and sublists) in one loop?

Given a list containing numbers, and sublists. The goal of this function – local_sum() is to get all local sum (or range sum) with these conditions: 1) if it’s a sublist, get its sum and remain a list, if consecutive numbers just add up them.

One example prob. will make it easier to follow:

(somehow I just got stuck in a nicer Pythonic way…)

from itertools import groupby

A =   [3, 4, 2, [10, 22, 32, 14], 9, 8, 6, [22, 11]]
expected = [9, 23,  [111]   
ans = [9, [78], 23, [33]]   <----   what I'm able to get

# is there a way to get this expected result in one shot, instead of doing another processing? 

# my current working code:
def local_sum(L):
    '''add up the numbers which are not in sublist,
       and sum the sublist as a one unit, if there're more than one -
       try to consolidate it into ONE.
    '''
    ans = []
    
    for k, g in groupby(L, key=lambda x: isinstance(x, list)):
        
        if k:                         # the group is a list
                                      # the whole list :: sum(list(g))X
            ans.append([sum(*g)])     # g is _grouper instance
            
        else:
            
            ans.append(sum(list(g)))           # each single number
    
    return ans

>Solution :

With only one example, it seems you want to keep a running total of everything in a sublist:

from itertools import groupby

A = [3, 4, 2, [10, 22, 32, 14], 9, 8, 6, [22, 11]]

def local_sum(L):
    ans = []
    sub = 0
    for k, g in groupby(A, key=lambda x: isinstance(x, list)):
        if k:
            sub += sum(*g)
        else:
            ans.append(sum(list(g)))
    ans.append([sub])
    return ans

print(local_sum(A))

Output:

[9, 23, [111]]