Creating a function for my code with datetime and timedelta - python-3.x

Question

I have finished my code and it works when I run it. However, I need to turn this into a function, that if I call the function and pass in any list of number, i can get the same results. This is my code:
from datetime import datetime, timedelta
dateStr = 'user-input'
dateObj = datetime.strptime(dateStr, '%Y%m%d')
timeStep = timedelta(days=1)
dateObj2 = dateObj + timeStep
days15 = [dateObj + timeStep*i for i in range(15)]
print(days15)
------------------ output:
datetime.datetime(2017, 1, 1, 0, 0),..
I need to be able to pass in
date_str = "20170817"
results = days_15(date_str)
print(results)
And then get the same results. Any hints? or any help - Thank you

Answer 1

You just need to add a def statement before your code to define the function, and rather than printing the result, return it:
from datetime import datetime, timedelta
def days_15(dateStr):
dateObj = datetime.strptime(dateStr, '%Y%m%d')
timeStep = timedelta(days=1)
return [dateObj + timeStep*i for i in range(15)]
my_date_str = "20170817"
results = days_15(my_date_str)
print(results)
Output:
[
datetime.datetime(2017, 8, 17, 0, 0), datetime.datetime(2017, 8, 18, 0, 0),
datetime.datetime(2017, 8, 19, 0, 0), datetime.datetime(2017, 8, 20, 0, 0),
datetime.datetime(2017, 8, 21, 0, 0), datetime.datetime(2017, 8, 22, 0, 0),
datetime.datetime(2017, 8, 23, 0, 0), datetime.datetime(2017, 8, 24, 0, 0),
datetime.datetime(2017, 8, 25, 0, 0), datetime.datetime(2017, 8, 26, 0, 0),
datetime.datetime(2017, 8, 27, 0, 0), datetime.datetime(2017, 8, 28, 0, 0),
datetime.datetime(2017, 8, 29, 0, 0), datetime.datetime(2017, 8, 30, 0, 0),
datetime.datetime(2017, 8, 31, 0, 0)
]