This function returns the datetime object by substracting n months (time won't change) #python

get_starting_date_minus_n_months.py

def get_starting_date_minus_n_months(curr_date, num_months):
    """
    Given the curr_date (datetime) object and num_months, returns the starting date
    minus "num_months" months. Note that time won't change

    for example: curr_date is "2019-10-22 23:59:59" and num_months is 3
    it returns "2019-07-01 23:59:59"
    """
    if num_months < 0:
        raise ValueError(
            "as the function name suggests it only supports for positive values of num_months"
        )
    number_of_years_to_travel_back = num_months // 12
    calculated_year = curr_date.year - number_of_years_to_travel_back
    calculated_month = (curr_date.month - num_months) % 12
    if calculated_month == 0:
        calculated_month = 12
        if (
            num_months > 0 # pylint: disable=R1716
            and number_of_years_to_travel_back <= 0
        ):  # special case : say if the curr_date month is 3 and num_months = 3, in that case we need to go a year back
            calculated_year -= 1
    elif calculated_month > curr_date.month:
        calculated_year -= 1
    return timezone.make_aware(
        datetime.datetime(
            calculated_year,
            calculated_month,
            1,
            curr_date.hour,
            curr_date.minute,
            curr_date.second,
            curr_date.microsecond if curr_date.microsecond else 0,
        )
    )