persistent_bugger..py

"""
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) => 3  # Because 3*9 = 27, 2*7 = 14, 1*4=4
                      # and 4 has only one digit.

persistence(999) => 4 # Because 9*9*9 = 729, 7*2*9 = 126,
                      # 1*2*6 = 12, and finally 1*2 = 2.

persistence(4) => 0   # Because 4 is already a one-digit number.
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
"""

from operator import mul
from functools import reduce


def persistence(n):
    count = 0
    while True:
        if len(str(n)) == 1:
            break
        n = reduce(mul, [int(c) for c in str(n)], 1)
        count += 1
    return count