Python: get all months in range?

Question

I want to get all months between now and August 2010, as a list formatted like this:

[\'2010-08-01\', \'2010-09-01\', .... , \'2016-02-01\']

<

Answer 1

It's seems like there's a very simple and clean way to do this by generating a list of dates and subsetting to take only the first day of each month, as shown in the example below.

import datetime
import pandas as pd

start_date = datetime.date(2010,8,1)
end_date = datetime.date(2016,2,1)

date_range = pd.date_range(start_date, end_date)
date_range = date_range[date_range.day==1]

print(date_range)

Answer 2

I had a look at the dateutil documentation. Turns out it provides an even more convenient way than using dateutil.relativedelta: recurrence rules (examples)

For the task at hand, it's as easy as

from dateutil.rrule import *
from datetime import date

months = map(
    date.isoformat,
    rrule(MONTHLY, dtstart=date(2010, 8, 1), until=date.today())
)

The fine print

Note that we're cheating a little bit, here. The elements dateutil.rrule.rrule produces are of type datetime.datetime, even if we pass dtstart and until of type datetime.date, as we do above. I let map feed them to date's isoformat function, which just turns out to convert them to strings as if it were just dates without any time-of-day information.

Therefore, the seemingly equivalent list comprehension

[day.isoformat()
    for day in rrule(MONTHLY, dtstart=date(2010, 8, 1), until=date.today())]

would return a list like

['2010-08-01T00:00:00',
 '2010-09-01T00:00:00',
 '2010-10-01T00:00:00',
 '2010-11-01T00:00:00',
 ⋮
 '2015-12-01T00:00:00',
 '2016-01-01T00:00:00',
 '2016-02-01T00:00:00']

Thus, if we want to use a list comprehension instead of map, we have to do something like

[dt.date().isoformat()
    for dt in rrule(MONTHLY, dtstart=date(2010, 8, 1), until=date.today())]

Answer 3

fresh pythonic one-liner from me

from dateutil.relativedelta import relativedelta
import datetime

[(start_date + relativedelta(months=+m)).isoformat() for m in range(0,relativedelta(start_date,end_date).months+1)] 

Answer 4

I got another way using datetime, timedelta and calender:

from calendar import monthrange
from datetime import datetime, timedelta

def monthdelta(d1, d2):
    delta = 0
    while True:
        mdays = monthrange(d1.year, d1.month)[1]
        d1 += timedelta(days=mdays)
        if d1 <= d2:
            delta += 1
        else:
            break
    return delta

start_date = datetime(2016, 1, 1)
end_date = datetime(2016, 12, 1)

num_months = [i-12 if i>12 else i for i in range(start_date.month, monthdelta(start_date, end_date)+start_date.month+1)]
monthly_daterange = [datetime(start_date.year,i, start_date.day, start_date.hour) for i in num_months]

Answer 5

dateutil.relativedelta is handy here.

I've left the formatting out as an exercise.

from dateutil.relativedelta import relativedelta
import datetime

result = []

today = datetime.date.today()
current = datetime.date(2010, 8, 1)    

while current <= today:
    result.append(current)
    current += relativedelta(months=1)

Answer 6

You could reduce the number of if statements to two lines instead of four lines because having a second if statement that does the same thing with the previous if statement is a bit redundant.

if (y == 2010 and m < 8) or (y == 2016 and m > 2):
    continue