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Python Dates and Times in Python Let's Build a Timed Quiz App Harder Time Machine

Sena Sari
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.a{fill-rule:evenodd;}techdegree
Sena Sari
Python Web Development Techdegree Student 9,855 Points

Don't understand why my solution is not correct

Hi! I know I can solve this challenge with if-else blocks but I wanted to solve it with unpacker method. I'm not sure where I was wrong. Any help would be appreciated :)

Write a function named time_machine that takes an integer and a string of "minutes", "hours", "days", or "years". This describes a timedelta. Return a datetime that is the timedelta's duration from the starter datetime.

time_machine.py
import datetime

starter = datetime.datetime(2015, 10, 21, 16, 29)

# Remember, you can't set "years" on a timedelta!
# Consider a year to be 365 days.

## Example
# time_machine(5, "minutes") => datetime(2015, 10, 21, 16, 34)

def time_machine(duration, time_type):
    if time_type == 'years':
        return starter + datetime.timedelta(days = 365)
    else:
        return starter + datetime.timedelta(**{str(time_type) : duration})

Kenneth Love

nakalkucing
nakalkucing
12,964 Points

Just so you know, Sena Sari, Kenneth is no longer with Treehouse.

1 Answer

Wade Williams
Wade Williams
24,476 Points

What you're looking for is keyword unpacking which you can do with a dictionary.

def time_machine(int1, str1):
    time_dict = {str1: int1}

    if str1 == "years":
        time_dict = {"days": int1*365}

    return starter + datetime.timedelta(**time_dict)

You can also just pass in a dictionary directly into the function as well, which would be really clean if we didn't have to worry about turning "years" into "days".

def time_machine(int1, str1):

    if str1 == "years":
        str1 = "days"
        int1 *= 365

    return starter + datetime.timedelta(**{str1: int1})