Python recursive function to return the mean of the digits in a number

Question

I'm trying to create a recursive function that returns the average of the digits in a number. For example the average of the number 123 is 2. I know how to write a function that sums the digits but without using for I am no able to return the number to divide the function. The print value cannot be change so i can't pass a number of digits or something like that.

def media(x):
  if x<1:
    return 0
  else:
    return x%10+media(x//10)

print(media(91234))

Answer 1

Use a for -loop

There is really no reason not to use a for -loop for this.

def media(x):
  if x == 0:
    return 0
  else:
    digits = []
    while x > 0:
      digits.append(x % 10)
      x = x // 10
    return sum(digits) / len(digits)

If you insist on recursion

You can use additional accumulator parameters to store the sum of digits and the number of digits:

def media(x, sumSoFar=0, nDigitsSoFar=0):
  if x == 0 and sumSoFar==0:
    return 0
  elif x == 0:
    return sumSoFar / nDigitsSoFar
  else:
    return media(x // 10, sumSoFar + (x % 10), nDigitsSoFar + 1)

Please note that python is one of the worst languages for recursion, and in particular it doesn't optimize tail-calls.

Cheating: letting python handle the conversion to string

The builtin function str will convert an integer to a string of digits. You can use this in conjunction with ord to get the value of a digit:

def media(x):
  s = str(x)
  return sum((ord(c) - ord('0')) for c in s) / len(s)

Answer 2

I assume that it is the mean and not the median as it seems like you've written. To find the mean of the digits, you can convert the number into a type str then extract each digit by iterating over the string with a for... in string . Next, convert each digit back and find the average of those numbers.

EDIT: It's not recursive, but it doesn't seem meaningful to use recursion for this kind of simple calculation.