Digital sum, Python

Digital sum, Python

I need to write a code that counts the sum of the digits of a number, these is the exact text of the problem:The digital sum of a number n is the sum of its digits. Write a recursive function digitalSum(n) that takes a positive integer n and returns its digital sum. For example, digitalSum(2019) should return 12 because 2+0+1+9=12. These is the code I wrote :

def digitalSum(n):     L=[]      if n < 10:        return n     else:        S=str(n)        for i in S:           L.append(int(i))        return sum(L)  

These code works fine, but it's not a recursive function, and I'm not allowed to change any int to str. May you help me?

Answer by unwind for Digital sum, Python

Some hints:

1. You can define inner functions in Python
2. You can use the modulus operator (look up its syntax and usage) to good effect, here
3. There's no need to build up an explicit list representation with a proper recursive solution

EDIT The above is a bit "bad" as a general answer, what if someone else has this problem in a non-homework context? Then Stack Overflow fails ...

So, here's how I would implement it, and you need to decide whether or not you should continue reading. :)

def digitalSum(n):    def process(n, sum):      if n < 10:        return sum + n      return process(n / 10, sum + n % 10)    return process(n, 0)  

This might be a bit too much, but even in a learning situation having access to one answer can be instructive.

My solution is more a verbose than some, but it's also more friendly towards a tail call optimizing compiler, which I think is a feature.

Answer by Lev Levitsky for Digital sum, Python

It's homework, so I'm not writing much code. Recursion can be used in the following way:

• get the first (or last) digit
• format the rest as a shorter number
• add the digit and the digital sum of the shorter number (recursion!)

Answer by Grampa for Digital sum, Python

Try this:

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

Edit: The logic behind this algorithm is that for every call of the recursive function, we chop off the number's last digit and add it to the sum. First we obtain the last digit with n % 10 and then we call the function again, passing the number with the last digit truncated: n // 10. We only stop when we reach a one-digit number. After we stop, the sum of the digits is computed in reverse order, as the recursive calls return.

Example for the number 12345 :

5 + digitalSum( 1234 )  5 + 4 + digitalSum( 123 )  5 + 4 + 3 + digitalSum( 12 )  5 + 4 + 3 + 2 + 1 <- done recursing  5 + 4 + 3 + 3  5 + 4 + 6  5 + 10  15  

Answer by John La Rooy for Digital sum, Python

def digitalSum(n):     if n < 10:        return n     else:        return ???  

The 1st part is from your existing code. The ??? is the part you need to work out. It could take one digit off n and add it to the digitalSum of the remaining digits.

You don't really need the else, but I left it there so the code structure looks the same

Answer by Mike Vella for Digital sum, Python

This is more of an algorithms question.

Here is your answer:

def digit_sum(a):        if a== 0:          return 0        return a % 10 + digit_sum(a/10)  

Let me know if you don't understand why it works and I'll provide an explanation.

Answer by rxdazn for Digital sum, Python

def digital_sum(number):      if number < 10:           return number      else:          return number % 10 + digital_sum(number / 10)   

Answer by Gerry for Digital sum, Python

def drs_f(p):

 drs = sum([int (q) for q in str(p)])   while drs >= 10:       drs = sum([int(q) for q in str(drs)])  return drs  

Answer by pvkcse for Digital sum, Python

Still you can do it in O(log10 n)...cancel out all the digits that adds to 9 then if no numbers left,9 is the answer else sum up all the left out digits...

def rec_sum_Reduce(n) : ans = 0 for i in map(int,str(n)) : ans = 1+(ans+i-1)%9 return ans

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