Best way to find the months between two dates (in python)

Question

I have the need to be able to accurately find the months between two dates in python. I have a solution that works but its not very good (as in elegant) or fast.

dateRange = [datetime.strptime(dateRanges[0], "%Y-%m-%d"), datetime.strptime(dateRanges[1], "%Y-%m-%d")]
months = [] 

tmpTime = dateRange[0]
oneWeek = timedelta(weeks=1)
tmpTime = tmpTime.replace(day=1)
dateRange[0] = tmpTime
dateRange[1] = dateRange[1].replace(day=1)
lastMonth = tmpTime.month
months.append(tmpTime)
while tmpTime < dateRange[1]:
    if lastMonth != 12:
        while tmpTime.month <= lastMonth:
            tmpTime += oneWeek
        tmpTime = tmpTime.replace(day=1)
        months.append(tmpTime)
        lastMonth = tmpTime.month

    else:
        while tmpTime.month >= lastMonth:
            tmpTime += oneWeek
        tmpTime = tmpTime.replace(day=1)
        months.append(tmpTime)
        lastMonth = tmpTime.month

So just to explain, what I'm doing here is taking the two dates and converting them from iso format into python datetime objects. Then I loop through adding a week to the start datetime object and check if the numerical value of the month is greater (unless the month is December then it checks if the date is less), If the value is greater I append it to the list of months and keep looping through until I get to my end date.

It works perfectly it just doesn't seem like a good way of doing it...

Answer 1

Perhaps this works for you: http://bytes.com/topic/python/answers/803702-difference-between-two-dates

Answer 2

Try something like this. It presently includes the month if both dates happen to be in the same month.

from datetime import datetime,timedelta

def months_between(start,end):
    months = []
    cursor = start

    while cursor <= end:
        if cursor.month not in months:
            months.append(cursor.month)
        cursor += timedelta(weeks=1)

    return months

Output looks like:

>>> start = datetime.now() - timedelta(days=120)
>>> end = datetime.now()
>>> months_between(start,end)
[6, 7, 8, 9, 10]

Answer 3

If adding by a week, then it will approximately do work 4.35 times the work as needed. Why not just:

1. get start date in array of integer, set it to i: [2008, 3, 12], 
       and change it to [2008, 3, 1]
2. get end date in array: [2010, 10, 26]
3. add the date to your result by parsing i
       increment the month in i
       if month is >= 13, then set it to 1, and increment the year by 1
   until either the year in i is > year in end_date, 
           or (year in i == year in end_date and month in i > month in end_date)

just pseduo code for now, haven't tested, but i think the idea along the same line will work.

Answer 4

Assuming upperDate is always later than lowerDate and both are datetime.date objects:

if lowerDate.year == upperDate.year:
    monthsInBetween = range( lowerDate.month + 1, upperDate.month )
elif upperDate.year > lowerDate.year:
    monthsInBetween = range( lowerDate.month + 1, 12 )
    for year in range( lowerDate.year + 1, upperDate.year ):
        monthsInBetween.extend( range(1,13) )
    monthsInBetween.extend( range( 1, upperDate.month ) )

I haven't tested this thoroughly, but it looks like it should do the trick.

Answer 5

Define a "month" as ¹/₁₂ year, then do this:

def month_diff(d1, d2): 
    """Return the number of months between d1 and d2, 
    such that d2 + month_diff(d1, d2) == d1
    """
    diff = (12 * d1.year + d1.month) - (12 * d2.year + d2.month)
    return diff

You might try to define a month as "a period of either 29, 28, 30 or 31 days (depending on the year)". But you you do that, you have an additional problem to solve.

While it's usually clear that June 15^th + 1 month should be July 15^th, it's not usually not clear if January 30^th + 1 month is in February or March. In the latter case, you may be compelled to compute the date as February 30^th, then "correct" it to March 2^nd. But when you do that, you'll find that March 2^nd - 1 month is clearly February 2^nd. Ergo, reductio ad absurdum (this operation is not well defined).

Answer 6

Try this:

dateRange = [datetime.strptime(dateRanges[0], "%Y-%m-%d"),
             datetime.strptime(dateRanges[1], "%Y-%m-%d")]
delta_time = max(dateRange) - min(dateRange)
#Need to use min(dateRange).month to account for different length month
#Note that timedelta returns a number of days
delta_datetime = (datetime(1, min(dateRange).month, 1) + delta_time -
                           timedelta(days=1)) #min y/m/d are 1
months = ((delta_datetime.year - 1) * 12 + delta_datetime.month -
          min(dateRange).month)
print months

Shouldn't matter what order you input the dates, and it takes into account the difference in month lengths.

Answer 7

Get the ending month (relative to the year and month of the start month ex: 2011 January = 13 if your start date starts on 2010 Oct) and then generate the datetimes beginning the start month and that end month like so:

dt1, dt2 = dateRange
start_month=dt1.month
end_months=(dt2.year-dt1.year)*12 + dt2.month+1
dates=[datetime.datetime(year=yr, month=mn, day=1) for (yr, mn) in (
          ((m - 1) / 12 + dt1.year, (m - 1) % 12 + 1) for m in range(start_month, end_months)
      )]

if both dates are on the same year, it could also be simply written as:

dates=[datetime.datetime(year=dt1.year, month=mn, day=1) for mn in range(dt1.month, dt2.month + 1)]

Answer 8

Here is a method:

def months_between(start_dt, stop_dt):
    month_list = []
    total_months = 12*(stop_dt.year-start_dt.year)+(stop_dt.month-start_d.month)+1
    if total_months > 0:
        month_list=[ datetime.date(start_dt.year+int((start_dt+i-1)/12), 
                                   ((start_dt-1+i)%12)+1,
                                   1) for i in xrange(0,total_months) ]
    return month_list

This is first computing the total number of months between the two dates, inclusive. Then it creates a list using the first date as the base and performs modula arithmetic to create the date objects.

Answer 9

Start by defining some test cases, then you will see that the function is very simple and needs no loops

from datetime import datetime

def diff_month(d1, d2):
    return (d1.year - d2.year)*12 + d1.month - d2.month

assert diff_month(datetime(2010,10,1), datetime(2010,9,1)) == 1
assert diff_month(datetime(2010,10,1), datetime(2009,10,1)) == 12
assert diff_month(datetime(2010,10,1), datetime(2009,11,1)) == 11
assert diff_month(datetime(2010,10,1), datetime(2009,8,1)) == 14

You should add some testcases to your question, as there are lots of potential corner cases to cover - there is more than one way to define the number of months between two dates