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

Write a function named time_machine that takes an integer and a string of "minutes", "hours", "days", or "years". This d

Is that OK to write datetime.timedelta(string = integer)?

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(integer, string):
    if string == "years":   
        duration = datetime.timedelta(days = integer* 365)
    else:
        duration = datetime.timedelta(string = integer)
    return starter + duration

2 Answers

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,082 Points

The left-side of a keyword argument is not interpreted which causes the error string is not a valid keyword argument. If it were interpreted, then days, in the previous statement, would also need defining.

There are two ways to use a variable as a keyword argument:

  • use an if/elif/else statement to have a unique duration assignment for each timedelta argument type
  • create a dict with the key string and value integer, then use ** to expand this dict in the argument list

Post back if you need more help. Good luck!!!

Anupam Kumar
Anupam Kumar
3,795 Points

What could be wrong with this Code

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(strin,integer):
    return starter+datetime.timedelta(strin=integer)
time_machine("minutes",5)

thanks in advance

[MOD: added ```python formatting -cf]

Chris Freeman
Chris Freeman
Treehouse Moderator 68,082 Points

There are two issues:

  • the code does not account for the “years” option that is not a valid argument to timedelta. You will need to check for this case and convert to “days”
  • the left-side of a keyword argument assignment is always interpreted as the string **name* of a keyword argument*. It is never evaluated directly as a variable.

You can use the form

# create dict with value of strin used as key
kwargs = {strin: integer}
datetime.timedelta(**kwargs)

Post back if you need more help. Good luck!!!