Given a date range how can we break it up into N contiguous sub-intervals?

you should change date for datetime

from datetime import date, datetime, timedelta

begin = '20150101'
end = '20150228'

def get_yyyy_mm_dd(yyyymmdd):
  # given string 'yyyymmdd' return (yyyy, mm, dd)
  year = yyyymmdd[0:4]
  month = yyyymmdd[4:6]
  day = yyyymmdd[6:]
  return int(year), int(month), int(day)

y1, m1, d1 = get_yyyy_mm_dd(begin)
d1 = datetime(y1, m1, d1)
y2, m2, d2 = get_yyyy_mm_dd(end)
d2 = datetime(y2, m2, d2)

def remove_tack(dates_list):
  # given a list of dates in form YYYY-MM-DD return a list of strings in form 'YYYYMMDD'
  tackless = []
  for d in dates_list:
    s = str(d)
    tackless.append(s[0:4]+s[5:7]+s[8:])
  return tackless

def divide_date(date1, date2, intervals):
  dates = [date1]
  delta = (date2-date1).total_seconds()/4
  for i in range(0, intervals):
    dates.append(dates[i] + timedelta(0,delta))
  return remove_tack(dates)

listdates = divide_date(d1, d2, 4)
print listdates

result:

['20150101 00:00:00', '20150115 12:00:00', '20150130 00:00:00', '20150213 12:00:00', '20150228 00:00:00']

Using Datetimeindex and Periods from Pandas, together with dictionary comprehension:

import pandas as pd

begin = '20100101'
end = '20101231'

start = dt.datetime.strptime(begin, '%Y%m%d')
finish = dt.datetime.strptime(end, '%Y%m%d')

dates = pd.DatetimeIndex(start=start, end=finish, freq='D').tolist()
quarters = [d.to_period('Q') for d in dates]
df = pd.DataFrame([quarters, dates], index=['Quarter', 'Date']).T

quarterly_dates = {str(q): [ts.strftime('%Y%m%d') 
                            for ts in df[df.Quarter == q].Date.values.tolist()]
                           for q in quarters}

>>> quarterly_dates
{'2010Q1': ['20100101',
  '20100102',
  '20100103',
  '20100104',
  '20100105',
...
  '20101227',
  '20101228',
  '20101229',
  '20101230',
  '20101231']}

>>> quarterly_dates.keys()
['2010Q1', '2010Q2', '2010Q3', '2010Q4']

Could you use the datetime.date objects instead?

If you do:

import datetime
begin = datetime.date(2001, 1, 1)
end = datetime.date(2010, 12, 31)

intervals = 4

date_list = []

delta = (end - begin)/4
for i in range(1, intervals + 1):
    date_list.append((begin+i*delta).strftime('%Y%m%d'))

and date_list should have the end dates for each inteval.

I would actually follow a different approach and rely on timedelta and date addition to determine the non-overlapping ranges

Implementation

def date_range(start, end, intv):
    from datetime import datetime
    start = datetime.strptime(start,"%Y%m%d")
    end = datetime.strptime(end,"%Y%m%d")
    diff = (end  - start ) / intv
    for i in range(intv):
        yield (start + diff * i).strftime("%Y%m%d")
    yield end.strftime("%Y%m%d")

Execution

>>> begin = '20150101'
>>> end = '20150228'
>>> list(date_range(begin, end, 4))
['20150101', '20150115', '20150130', '20150213', '20150228']