6 kyu Persistent Bugger..py

"""
https://www.codewars.com/kata/55bf01e5a717a0d57e0000ec/train/python

Write a function, persistence, that takes in a positive parameter num and returns its multiplicative persistence, which is the number of times you must multiply the digits in num until you reach a single digit.

For example:

 persistence(39) # returns 3, because 3*9=27, 2*7=14, 1*4=4
                 # and 4 has only one digit
                  
 persistence(999) # returns 4, because 9*9*9=729, 7*2*9=126,
                  # 1*2*6=12, and finally 1*2=2

 persistence(4) # returns 0, because 4 is already a one-digit number
"""


def persistence(n):
    total = 1
    cnt = 0
    
    while n > 9:
        for i in list(str(n)):
            total *= int(i)
        n = total
        total = 1
        cnt += 1
    
    return cnt