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Create a function named is_over_13 with datetime woes.

Hello again!

So, here is my code for task one of this challenge. I've just come to accept that I'm not sure what is being asked of me. In this code you will see that:

a) it accepts 2 separate datetime (today as well as some other one) b) it creates a timedelta named how_many_days c) returns the truthiness of whether or not the day component of the timedelta is greater than or equal o 4745

I'm stumped.

birthdays.py
import datetime

birthdays = [
    datetime.datetime(2012, 4, 29),
    datetime.datetime(2006, 8, 9),
    datetime.datetime(1978, 5, 16),
    datetime.datetime(1981, 8, 15),
    datetime.datetime(2001, 7, 4),
    datetime.datetime(1999, 12, 30)
]

today = datetime.datetime.today()

def is_over_13(today, dt):
  how_many_days = today-dt
  return how_many_days.days >= 4745

In order to solve this challenge you need to remember the following:

  1. datetime objects have a year property
  2. the number of days in a year = 365

Knowing these two facts, we can tackle this challenge:

def is_over_13(datetime):
  return today.year * 365 - datetime.year * 365 >= 4745
Chris Freeman
Chris Freeman
Treehouse Moderator 68,063 Points

Nathan Krishnan, Your hint is in the right direction, but it only takes into account the year, but does not account for when in the year the date occurs.

1 Answer

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,063 Points

Adam Raitano, your code is very nearly correct. You don't need to pass today as an argument since it's defined as a global:

def is_over_13(dt):
    # compare total days
    delta = today - dt
    return delta.days >= 4745