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Start your free trialEm Noble
6,918 Pointscan someone please assist?
not sure how to do this any other way.
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(i,st):
st = str(st)
td = datetime.timedelta(st=i)
return starter + td
1 Answer
Mel Rumsey
Treehouse ModeratorHey Em Noble! So this one is tricky. It takes an int and a string. First, the st
won't need to be converted to a string because what is entered will already be a string. But the main issue lies with this line:
datetime.timedelta(st=i)
Consider that time_machine
is being called with the following:
time_machine(4, "minutes")
time_machine(7, "hours")
time_machine(1, "days")
time_machine(2, "years")
This function would be trying to create a timedelta with "minutes"=4
. This will not work because timedelta does not take a string. It takes days
, hours
, minutes
as attributes. For example to do the one with hours, we would want it to say datetime.timedelta(hours, 7)
rather than datetime.timedelta("hours", 7)
.
In order to do this we need some if
and elif
statements that check if the string is equal to "minutes", "hours", "days", or "years", set the timedelta or td
accordingly.
Remember that if "years" is entered in, it will need to get converted to years by multiplying by 365
.
Then you can return the starter + td.
Hopefully this helps!