Summation of values in nested lists

Question

I have encountered a problem where a return gives None back, though the variable had a value just a single line of code before.

mylist = [[7]]

def sumcalc(the_list,n):
    s = n
    for x,y in enumerate(the_list):
        if type(y) == list:
            sumcalc(y,s)
        elif type(y) == int:
            s += y
            if x < len(the_list)-1:
                sumcalc(the_list[x+1], s)
            else:
                print (s)
                return s
    
print(sumcalc(mylist,0))

The print command in the second to last line gives me 7, as expected. But the return is None. Any help on this would be appreciated :)

You have to replace `sumcalc(y,s)` with `return sumcalc(y,s)` — Abdul Niyas P M, Jan 23 '21 at 17:14
Does this answer your question? [Why does my recursive function return None?](https://stackoverflow.com/questions/17778372/why-does-my-recursive-function-return-none) — wjandrea, Jan 23 '21 at 17:49

score 1 · Answer 1 · edited Jan 23 '21 at 18:20

If you are writing a recursive function. You must put return key word before recursive call.

mylist = [[7]]

def sumcalc(the_list,n):
    s = n
    for x,y in enumerate(the_list):
        if type(y) == list:
            return sumcalc(y,s)
        elif type(y) == int:
            s += y
            if x < len(the_list)-1:
                sumcalc(the_list[x+1], s)
            else:
                print (s)
                return s
    
print(sumcalc(mylist,0))

score 0 · Answer 2 · answered Jan 23 '21 at 17:23

sumcalc() doesn't do anything on its own; you need to use the return value. Beyond this, the index checking is unecessary, as the return could simply be placed after the loop. Here is a simpler way of writing the nested sum code.

def nested_sum(x):
    if type(x)==int:
        return x
    if type(x)==list:
        return sum(
            nested_sum(item)
            for item in x
        )

Summation of values in nested lists

2 Answers2