[Solved] Python: Nested If Looping

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You must realize that 12 does not divide 52, and that there are not 4 weeks to every month. So to give an example that you can fine tune to get exactly what you want, I’ve defined a week to belong to the same month that its thursdays belong to. This dovetails nicely with the ISO 8601 definition of the first week of the year. If there’s a week left, then I add that week to december.

import datetime
from itertools import groupby

def get_week(date):
    return date.isocalendar()[1]

def group_by_month(weeks, year):
    """
    Group a list containing one item per week, starting with week 1, by month.

    If there are too few items to fill a year, stop after last item.
    If there are more items than weeks in the year, stop before new year.
    """
    day = datetime.timedelta(days=1)
    week = datetime.timedelta(days=7)

    # Find first Thursday (it's in week 1 by ISO 8601)
    date = datetime.date(year, 1, 1)
    while date.weekday() != 3:
        date += day

    # Create list of one day from each week
    thursdays = []
    while date.year == year:
        thursdays.append(date)
        date += week

    # Check if the last day is in the last week and if not, 
    # add the week of the last day
    last = tursdays[-1]
    if get_week(last.replace(day=31)) != get_week(last):
        # this will not be a Thursday, but what the hey
        thursdays.append(last.replace(day=31))

    # The thursdays are already sorted by month, so 
    # it's OK to use groupby without sorting first
    for k, g in groupby(zip(weeks, thursdays), key=lambda x: x[1].month):
        yield [x[0] for x in g]

list_1 = [500] * 52

print map(sum, group_by_month(list_1, 2012))

Result:

[2000, 2000, 2500, 2000, 2500, 2000, 2000, 2500, 2000, 2000, 2500, 2000]

You should also be aware of the fact that the year may contain 53 weeks, and if so you must supply a 53-item list instead of a 52-item list. If you don’t, the 53rd week is simply ignored.

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solved Python: Nested If Looping