Handling months in python datetimes

Question

I have a function which gets the start of the month before the datetime provided:

def get_start_of_previous_month(dt):
    '''
    Return the datetime corresponding to the start of the month
    before the provided datetime.
    '''
    target_month = (dt.month - 1)
    if target_month == 0:
        target_month = 12
    year_delta = (dt.month - 2) / 12
    target_year = dt.year + year_delta

    midnight = datetime.time.min
    target_date = datetime.date(target_year, target_month, 1)
    start_of_target_month = datetime.datetime.combine(target_date, midnight)
    return start_of_target_month

However, it seems very convoluted. Can anyone suggest a simpler way? I am using python 2.4.

Answer 1

Use a timedelta(days=1) offset of the beginning of this month:

import datetime

def get_start_of_previous_month(dt):
    '''
    Return the datetime corresponding to the start of the month
    before the provided datetime.
    '''
    previous = dt.date().replace(day=1) - datetime.timedelta(days=1)
    return datetime.datetime.combine(previous.replace(day=1), datetime.time.min)

.replace(day=1) returns a new date that is at the start of the current month, after which subtracting a day is going to guarantee that we end up in the month before. Then we pull the same trick again to get the first day of that month.

Demo (on Python 2.4 to be sure):

>>> get_start_of_previous_month(datetime.datetime.now())
datetime.datetime(2013, 2, 1, 0, 0)
>>> get_start_of_previous_month(datetime.datetime(2013, 1, 21, 12, 23))
datetime.datetime(2012, 12, 1, 0, 0)