CODEWITHSUNDEEP
pythonfunctionrecursion
naveen ch
naveen ch

Reputation: 15

Passing value to a function in recursion in Python

I have been trying to pass value from my recursive helper function to the parent, but it is passing None. I am learning Python and recursion and trying to understand how the values are passed between functions. 

def superDigitSum(n,k):
    num2str = [int(i) for i in list(str(n))*k] #creates the list[1,4,8,1,4,8,1,4,8]
    result=0
    return superDigitSumHelper(num2str, result)

def superDigitSumHelper(num2str, helper_result):
    if not num2str :
        return helper_result
    else:
        helper_result+=num2str[-1]
        superDigitSumHelper(num2str[:-1],helper_result)

print(superDigitSum(148,3))

The code creates the list [1,4,8,1,4,8,1,4,8] and sums the values. I expect the result to be 39 (3 * (1+4+8)).

Upvotes: 0

Views: 749

Answers (4)

Mulan
Mulan

Reputation: 135227

If you're allowed to do things like str(n) or list*k, or even * at all, it kind of defeats the purpose of the exercise

def super_digit_sum (n, k):
  return sum (map (int, list (str (n)))) * k

print (super_digit_sum (148, 3)) # 39

But I'm sure the point of this is to use primitive features and operations as a means of learning how to write simple recursive programs...

We start with a simple function that sums the digits of a single input number

def sum_digits (n):
  if n < 10:
    return n
  else:
    return n % 10 + sum_digits (n // 10)

print (sum_digits (148))
# 13

Then we make a simple function that can multiply two numbers

def mult (n, m):
  if n is 0:
    return 0
  else:
    return m + mult (n - 1, m)

print (mult (3, 7))
# 21

Then combining the two, we write your main function

def super_digit_sum (n, k):
  return mult (sum_digits (n), k)

print (super_digit_sum (148, 3))
# 39

If you're allowed to use *, you can skip mult altogether and simply write

def super_digit_sum (n, k):
  return sum_digits (n) * k

print (super_digit_sum (148, 3))
# 39

This is a better program because you arrive at the answer in a straightforward manner. Compare that to the other approach of taking a number, converting it into a string, chopping it up into single characters, converting those characters back to digits, then summing the digits, and finally multiplying the result by a constant.

Both approaches arrive at the same result, but only one results in a program worth reading, imo. I talk about this roundabout-style of programming in another python Q&A: How to count neighbors in a list? I think you will find it helpful.

Upvotes: 0

Prune
Prune

Reputation: 77847

This clause returns None to the next call up the stack. You need to return the last value:

else:
    helper_result+=num2str[-1]
    superDigitSumHelper(num2str[:-1],helper_result)

Instead,

else:
    helper_result+=num2str[-1]
    return superDigitSumHelper(num2str[:-1],helper_result)

Upvotes: 1

Fallenreaper
Fallenreaper

Reputation: 10704

Your function is slightly wrong.

You need to add a return the following to your superDigitSumHelper function:

return superDigitSumHelper(num2str[:-1], helper_result)

in your else conditional.

Essentially, it is returning none because after the function ends, it isnt returning anything so the default value is none.

Upvotes: 1

Sam
Sam

Reputation: 1552

try

else:
    helper_result+=num2str[-1]
    return superDigitSumHelper(num2str[:-1],helper_result)

Without returning the results of the recursive call you are just returning None

Upvotes: 1

Related Questions