Complicated Recursive function

Question

I need to write a function nested_sum(L) that will sum all the ints inside a list no matter if they are inside another list. This with calling recrusively to another function mult2(n).

Example:

>>> nestedSum(mult2( [1,['a',3,'b',2],[4,['h',8,[10]]], -5]))
24

I tried to code this:

def mult2(n):
    if type(n) == int and n%2 ==0:
        return n

def nested_sum(L):
    total = 0 
    for i in L:
        if isinstance(i, list):  
            total += nested_sum(i)
        else:
            total += i
    return total

And unfortanetly I can't change the code of mult2(n) function. I can only change the nested_sum(L) function.

Can someone please give me a clue what to do? Thank you.

You still aren't saying what the problem is. What is the actual problem. What is the error? What behavior do you want to see and what do you actually see? — Daenyth, May 06 '16 at 13:04
@Daenyth I'm trying to code the function nested_sum(L) that it will cal recrusively to the unchangeable function mult2(n). it will sum all the ints inside a list no matter if they are inside another list. — Netta, May 06 '16 at 13:04
I don't understand. What do you want to achieve, what kind of use of mult2 you must implement, and why — kmaork, May 06 '16 at 13:05
this can be of interest: [flatten-an-irregular-list-of-lists-in-python](http://stackoverflow.com/questions/2158395/flatten-an-irregular-list-of-lists-in-python) — Copperfield, May 06 '16 at 13:28

score 2 · Answer 1 · answered May 06 '16 at 13:14

It would be better to separate out the recursion into a function that only does that like in the following:

sum(x for x in iflatten(L) if cond)

where iflatten is an iterator returning version of the flatten function and cond is in your case type(x) == int and x % 2 == 0.

def iflatten(L):
    for i in L:
        if isinstance(i, list):  
            for r in iflatten(i):
                yield r
        else:
            yield i

The code tests out like this:

L = [1,['a',3,'b',2],[4,['h',8,[10]]], -5]

def iflatten(L):
    for i in L:
        if isinstance(i, list):  
            for r in iflatten(i):
                yield r
        else:
            yield i

sum(x for x in iflatten(L) if type(x) == int and x % 2 == 0)

results in 24.

As you can't change

def mult2(n):
    if type(n) == int and n%2 ==0:
        return n

We can change the condition from type(x) == int and x % 2 == 0 to mult2(x) is not None as all functions have an implicit return None at their end.

Lets test that too:

L = [1,['a',3,'b',2],[4,['h',8,[10]]], -5]

def iflatten(L):
    for i in L:
        if isinstance(i, list):  
            for r in iflatten(i):
                yield r
        else:
            yield i

def mult2(n):
    if type(n) == int and n%2 ==0:
        return n

sum(x for x in iflatten(L) if mult2(x) is not None)

Also results in 24

mhawke · Accepted Answer · 2016-05-06T13:36:18.033

2

I'm not sure that the example invocation correct:

nestedSum(mult2( [1,['a',3,'b',2],[4,['h',8,[10]]], -5]))

Calling mult2() on a list will always return None. Passing None to nestedSum() will always cause a TypeError.

mult2() is designed to filter out non-integers and odd integers. I think that you are supposed to pass mult2 as a filter function to nestedSum():

def mult2(n):
    if type(n) == int and n%2 ==0:
        return n

def nested_sum(L, predicate):
    total = 0
    for item in L:
        if isinstance(item, list):  
            total += nested_sum(item, predicate)
        elif predicate(item):
            total += item
    return total

Now to invoke the function, pass mult2() as the predicate function to nested_sum():

>>> nested_sum([1,['a',3,'b',2],[4,['h',8,[10]]], -5], mult2)
24

edited May 06 '16 at 13:36

answered May 06 '16 at 13:20

mhawke

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Thank you so much!!! Can you just tell me what is the meaning of "if L"? why did you use that? – Netta May 06 '16 at 13:24
1

@Netta: `if L:`? Was supposed to guard against `None` being passed in ( as done in the original invocation), but it can be removed. I've updated the answer to remove it. – mhawke May 06 '16 at 13:36

score 0 · Answer 3 · answered May 06 '16 at 13:24

If you can do it without using mult2 but still must use recursion, this would do the job:

def sumArr(arr):
    return sum([numVal(item) for item in arr])

def numVal(item):
    if instanceof(item, int): return item
    elif instance(item, list): return sumArr(item)
    else: return 0

sumArr([1,['a',3,'b',2],[4,['h',8,[10]]], -5]) # >>> 23

Complicated Recursive function

3 Answers3