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Python Dates and Times in Python (2014) Let's Build a Timed Quiz App Harder Time Machine

Posted by markmneimneh
markmneimneh
14,132 Points

can someone please elaborate on what this challenge is asking for?

This challenge is ambiguous at best:

https://teamtreehouse.com/library/dates-and-times-in-python/lets-build-a-timed-quiz-app/harder-time-machine

I tried this, which returns date time:

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(an_int, a_string): a_begin_string = "datetime(" an_end_string = ")" a_separator = ", "

a_return=""
if a_string.lower() == 'minutes':
    new_time = starter +  datetime.timedelta(minutes=an_int)
    a_return = str(new_time.year) + a_separator
    a_return += str(new_time.month) + a_separator
    a_return += str(new_time.day) + a_separator
    a_return += str(new_time.hour) + a_separator
    a_return += str(new_time.minute)
    return datetime.datetime.strptime(a_return,'%Y, %m, %d, %H, %M')
elif a_string.lower() == 'hours':
    new_time = starter +  datetime.timedelta(hours=an_int)
    a_return = str(new_time.year) + a_separator
    a_return += str(new_time.month) + a_separator
    a_return += str(new_time.day) + a_separator
    a_return += str(new_time.hour) + a_separator
    a_return += str(new_time.minute)
    return datetime.datetime.strptime(a_return,'%Y, %m, %d, %H, %M')
elif a_string.lower() == 'years':
    new_time = starter + datetime.timedelta(days=an_int*365)
    a_return = str(new_time.year) + a_separator
    a_return += str(new_time.month) + a_separator
    a_return += str(new_time.day) + a_separator
    a_return += str(new_time.hour) + a_separator
    a_return += str(new_time.minute)
    return datetime.datetime.strptime(a_return,'%Y, %m, %d, %H, %M')

I got a bummer

I then tried this, which returns exact string as requested in the challenge ... e.g.

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

import datetime

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

def time_machine(an_int, a_string): a_begin_string = "datetime(" an_end_string = ")" a_separator = ", "

a_return=""
if a_string.lower() == 'minutes':
    new_time = starter +  datetime.timedelta(minutes=an_int)
    a_return = a_begin_string +  str(new_time.year) + a_separator
    a_return += str(new_time.month) + a_separator
    a_return += str(new_time.day) + a_separator
    a_return += str(new_time.hour) + a_separator
    a_return += str(new_time.minute) + an_end_string
    return a_return
elif a_string.lower() == 'hours':
    new_time = starter +  datetime.timedelta(hours=an_int)
    a_return = a_begin_string +  str(new_time.year) + a_separator
    a_return += str(new_time.month) + a_separator
    a_return += str(new_time.day) + a_separator
    a_return += str(new_time.hour) + a_separator
    a_return += str(new_time.minute) + an_end_string
    return a_return
elif a_string.lower() == 'years':
    new_time = starter + datetime.timedelta(days=an_int*365)
    a_return = a_begin_string + str(new_time.year) + a_separator
    a_return += str(new_time.month) + a_separator
    a_return += str(new_time.day) + a_separator
    a_return += str(new_time.hour) + a_separator
    a_return += str(new_time.minute) + an_end_string
    return a_return:

I still got a bummer

What exactly is being asked?

Thank you for your help / clarifications

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(an_int, a_string):
    a_begin_string = "datetime("
    an_end_string = ")"
    a_separator = ", "

    a_return=""
    if a_string.lower() == 'minutes':
        new_time = starter +  datetime.timedelta(minutes=an_int)
        a_return = str(new_time.year) + a_separator
        a_return += str(new_time.month) + a_separator
        a_return += str(new_time.day) + a_separator
        a_return += str(new_time.hour) + a_separator
        a_return += str(new_time.minute)
        return datetime.datetime.strptime(a_return,'%Y, %m, %d, %H, %M')
    elif a_string.lower() == 'hours':
        new_time = starter +  datetime.timedelta(hours=an_int)
        a_return = str(new_time.year) + a_separator
        a_return += str(new_time.month) + a_separator
        a_return += str(new_time.day) + a_separator
        a_return += str(new_time.hour) + a_separator
        a_return += str(new_time.minute)
        return datetime.datetime.strptime(a_return,'%Y, %m, %d, %H, %M')
    elif a_string.lower() == 'years':
        new_time = starter + datetime.timedelta(days=an_int*365)
        a_return = str(new_time.year) + a_separator
        a_return += str(new_time.month) + a_separator
        a_return += str(new_time.day) + a_separator
        a_return += str(new_time.hour) + a_separator
        a_return += str(new_time.minute)
        return datetime.datetime.strptime(a_return,'%Y, %m, %d, %H, %M')

1 Answer

Steven Parker
Steven Parker
229,744 Points
Steven Parker
Steven Parker
229,744 Points

:point_right: You didn't implement all of the required cases.

While your solution is perhaps a good deal more complicated than it needs to be, the concept is essentially sound. But the challenge asks that your function handle "minutes", "hours", "days", or "years"; and your implementation handles only 3 of these four cases (all but "days"). If you add additional code to handle that case, you should pass.

I would, however, suggest that you take the opportunity to re-consider your approach and see if you can find a way to perform the required task that would require only one or two lines for each case.

markmneimneh
markmneimneh
14,132 Points

Hello

thanks for pointing out the missing days. It passed after adding another elif for days.

The complexity came about because of the bummer thing ... I did not realize that I was missing days and thought I am expected to return a string like ... datetime(2015, 10, 21, 16, 34);

in any case, this is the simple code

import datetime

starter = datetime.datetime(2015, 10, 21, 16, 29) def time_machine(an_int, a_string): if a_string.lower() == 'minutes': return starter + datetime.timedelta(minutes=an_int) elif a_string.lower() == 'hours': return starter + datetime.timedelta(hours=an_int) elif a_string.lower() == 'days': return starter + datetime.timedelta(days=an_int) elif a_string.lower() == 'years': return starter + datetime.timedelta(days=an_int * 365)

Thank you very much for the catch