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Kevin Faust
15,353 Pointsmessy code
since we cant use a string for the key, is this the only way to do the challenge?:
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(integer, string):
if string == 'years':
return starter + datetime.timedelta(days = integer * 365)
elif string == 'days':
return starter + datetime.timedelta(days = integer)
elif string == 'weeks':
return starter + datetime.timedelta(weeks = integer)
elif string == 'seconds':
return starter + datetime.timedelta(seconds = integer)
elif string == 'microseconds':
return starter + datetime.timedelta(microseconds = integer)
elif string == 'milliseconds':
return starter + datetime.timedelta(milliseconds = integer)
elif string == 'minutes':
return starter + datetime.timedelta(minutes = integer)
elif string == 'hours':
return starter + datetime.timedelta(hours = integer)
1 Answer

Jason Anello
Courses Plus Student 94,596 PointsHi Kevin,
Your code could be about half as short since the challenge states that only "minutes", "hours", "days", or "years" will be passed in.
Beyond that, you could create a dictionary out of the string
, and integer
and then unpack the dictionary when you pass it into timedelta. This sets up the proper keyword argument while avoiding the problem of not being able to use a string as the key.
Since "years" is not a valid keyword argument, you first have to check for that and do some adjusting.
def time_machine(integer, string):
if string == "years":
# if years were passed in then adjust the variables to represent days instead
string = "days"
integer *= 365
return starter + datetime.timedelta(**{string : integer}) # create a dictionary with string and integer and then unpack it
Kevin Faust
15,353 PointsKevin Faust
15,353 Pointsthanks jason