How can I get the current week using Python?

Question

Using Python...

How can I get a list of the days in a specific week?

Something like...

{
\'1\' : [\'01/03/2010\',\'01/04/2010\',\'01/05/2010\         


        

Answer 1

How do you identify weeks? Here I'm identifying by a day in that week, using a function which gets the Sunday in that week (what you used in your example), and then returns it plus the next 6 days.

import datetime

one_day = datetime.timedelta(days=1)

def get_week(date):
  """Return the full week (Sunday first) of the week containing the given date.

  'date' may be a datetime or date instance (the same type is returned).
  """
  day_idx = (date.weekday() + 1) % 7  # turn sunday into 0, monday into 1, etc.
  sunday = date - datetime.timedelta(days=day_idx)
  date = sunday
  for n in xrange(7):
    yield date
    date += one_day

print list(get_week(datetime.datetime.now().date()))
# [datetime.date(2010, 1, 3), datetime.date(2010, 1, 4),
#  datetime.date(2010, 1, 5), datetime.date(2010, 1, 6),
#  datetime.date(2010, 1, 7), datetime.date(2010, 1, 8),
#  datetime.date(2010, 1, 9)]
print [d.isoformat() for d in get_week(datetime.datetime.now().date())]
# ['2010-01-03', '2010-01-04', '2010-01-05', '2010-01-06', '2010-01-07',
#  '2010-01-08', '2010-01-09']

Answer 2

current_week = datetime.datetime.now().isocalendar()[1]

Answer 3

Here's some code:

import datetime

now = datetime.datetime.now()

now_day_1 = now - datetime.timedelta(days=now.weekday())

dates = {}

for n_week in range(3):
    dates[n_week] = [(now_day_1 + datetime.timedelta(days=d+n_week*7)).strftime("%m/%d/%Y") for d in range(7)]

print dates

prints:

{
 0: ['01/04/2010', '01/05/2010', '01/06/2010', '01/07/2010', '01/08/2010', '01/09/2010', '01/10/2010'], 
 1: ['01/11/2010', '01/12/2010', '01/13/2010', '01/14/2010', '01/15/2010', '01/16/2010', '01/17/2010'], 
 2: ['01/18/2010', '01/19/2010', '01/20/2010', '01/21/2010', '01/22/2010', '01/23/2010', '01/24/2010']
}

Answer 4

You could use the datetime module. You can specify the format and everything. Here's the link: http://docs.python.org/library/datetime.html

Look into datetime.datetime( params ) and datetime.timedelta( params ). Hope it all goes well ;-)

Example:

import datetime

numweeks = 5
start_date = datetime.datetime(year=2010,month=1,day=4)    

weeks = {}

offset = datetime.timedelta(days=0)
for week in range(numweeks):
   this_week = []
   for day in range(7):
        date = start_date + offset
        date = date.strftime( some_format_string )
        this_week.append( date )
        offset += datetime.timedelta(days=1)
   weeks[week] = this_week 

Answer 5

If you're OK with the ISO standard:

>>> import collections
>>> dd = collections.defaultdict(list)
>>> jan1 = datetime.date(2010, 1, 1)
>>> oneday = datetime.timedelta(days=1)
>>> allyear = [jan1 + k*oneday for k in range(365 + 6)]
>>> for d in allyear:
...   y, w, wd = d.isocalendar()
...   if y == 2010: dd[w].append(d.strftime('%m/%d/%Y'))
... 

This produces slightly different results than the ones you're looking for (by ISO standard, weeks begin on Monday, not Sunday...), e.g.:

>>> dd[1]
['01/04/2010', '01/05/2010', '01/06/2010', '01/07/01/2010', '01/08/2010', '01/09/2010', '01/10/2010']

but you could tweak this by simulating an appropriate "off by one" error!-)

The calendar modules let you set any weekday as "first day of week", but offers no simple way to get all weeks (without duplications when a week is split between two months), so I think that working directly off datetime is probably a better idea.

Answer 6

There is a 3 lines method I developed after read that question:

from datetime import timedelta

def get_week_dates(base_date, start_day, end_day=None):
    """
    Return entire week of dates based on given date limited by start_day and end_day.
    If end_day is None, return only start_day.

    >>> from datetime import date
    >>> get_week_dates(date(2015,1,16), 3, 5)
    [datetime.date(2015, 1, 14), datetime.date(2015, 1, 15), datetime.date(2015, 1, 16)]

    >>> get_week_dates(date(2015,1,15), 2, 5)
    [datetime.date(2015, 1, 13), datetime.date(2015, 1, 14), datetime.date(2015, 1, 15), datetime.date(2015, 1, 16)]
    """
    monday = base_date - timedelta(days=base_date.isoweekday() - 1)
    week_dates = [monday + timedelta(days=i) for i in range(7)]
    return week_dates[start_day - 1:end_day or start_day]

Use get_week_dates(date.today(), 1, 7) to get current week dates.