Stepping dates with timedelta in months for Python

Question

I have a start date of datetime(2015, 01, 01) and I want my program to advance to the first of the following month with each step, it will be doing this for a couple of decades, but to troubleshoot it is just doing it for one year. The reason it is in a while loop is because I also want it to some scrapping based on the month and year. This is my code that I have written:

import datetime

dayS = 01; monthS = 01; yearS = 2015 #Start date
dayE = 01; monthE = 01; yearE = 2016 #End date
SDate = datetime.date(yearS, monthS, dayS)
EDate = datetime.date(yearE, monthE, dayE)
Date = []

while SDate <= EDate:
        Date.append(SDate.strftime('%d/%m/%Y'))
        if SDate.strftime('%m') == '02':
                step = datetime.timedelta(days=28)
        if SDate.strftime('%m') == '09' or '04' or '06' or '11':
                step = datetime.timedelta(days=30)
        if SDate.strftime('%m') == '01' or '03' or '05' or '07' or '08' or '10' or '12':
                step = datetime.timedelta(days=31)
        SDate += step

print Date

But the result I got is ['01/01/2015', '01/02/2015', '04/03/2015', '04/04/2015', '05/05/2015', '05/06/2015', '06/07/2015', '06/08/2015', '06/09/2015', '07/10/2015', '07/11/2015', '08/12/2015'], and as you can see for some months it doesn't add the correct amount of days.

Any help is much appreciated!

mgilson has correctly identified the problem with the above code, but I'd rework it a bit to do less of the work yourself. Given an arbitrary date, `(somedate.replace(day=1) + datetime.timedelta(days=31)).replace(day=1)` is always the first day of the next month. If you know you only start from the first of the month, you can skip the first `replace`. Or there are third-party libraries like `dateutil` that can help. — Peter DeGlopper, Oct 03 '14 at 19:46

score 4 · Accepted Answer · answered Oct 03 '14 at 19:43

What you have is the classic "or" problem.

"foo" == "1" or "2"

will always return True because it gets parsed as:

("foo" == "1") or "2"
         |         |
         False or "2"
               |
              "2"

Note that in this case, you might be better off using datetime.date.replace.

while SDate <= EDate:
    if SDate.month < 12:
        SDate = SDate.replace(month=SDate.month + 1)
    else:
        SDate = SDate.replace(month=1, year=SDate.year + 1)

Stepping dates with timedelta in months for Python

1 Answers1