How to Calculate the number of days in the year(s) between 2 dates in python

Question

For example:

date 1 : 1 january 2000  
date 2 : 17 november 2006  

I want to know how many days there are between date 1 and date 2 in the year 2000, 2001, ..., 2006 so I need something that returns something like this (doesn't matter if it's in a list or something):
2000: 365, 2001: 365, ..., 2006: 320

I've looked for something like this on the internet but that only turned up ways to calculate the number of days/months/years between 2 dates

Answer 1

hm, try something like this:

import datetime, calendar
date1 = datetime.date(year1, month1, day1) # month and day are 1-base
date2 = datetime.date(year2, month2, day2)
days_in_first_year = (datetime.date(year1,12,31)-date1).days
days_in_last_year = (date2 - datetime.date(year2, 1, 1)).days
if year1 != year2:
    n_days_list = [days_in_first_year]
    for year in range(year1+1, year2): n_days_list.append(365 + (1*calendar.isleap(year)))
    n_days_list.append(days_in_last_year)
else: n_days_list = [days_in_first_year + days_in_last_year]

haven't tested this, might be some off-by-one errors; make sure it does what you expect.

edit: correct the boundaries of the range() call, correctly handle year1 == year2

Answer 2

>>> start_date = datetime.datetime(2006, 7, 3)
>>> end_date = datetime.datetime(2012, 12, 21)
>>> years = range(start_date.year, end_date.year + 1)
>>> start, end = start_date, end_date + datetime.timedelta(1)
>>> for year in years:
...     year_start = datetime.datetime(year, 1, 1, 0, 0)
...     year_end = datetime.datetime(year + 1, 1, 1, 0, 0)
...     print(year, min(end, year_end) - max(start, year_start))
... 
2006 182 days, 0:00:00
2007 365 days, 0:00:00
2008 366 days, 0:00:00
2009 365 days, 0:00:00
2010 365 days, 0:00:00
2011 365 days, 0:00:00
2012 356 days, 0:00:00

UPDATE: You should probably add a datetime.timedelta(1) to the end date, because otherwise you'd be off with one day at the end. Fixed. But that depends on whether you want to include it or exclude it.

Answer 3

from datetime import date

DATE_END = date(2006, 11, 17)

def get_days(date_start):
    return (DATE_END - date_start).days

starting_dates = [
    date(2000, 1, 1),
    date(2001, 1, 1),
    date(2002, 1, 1),
]

print map(get_days, starting_dates)

Answer 4

Use this pseudocode to see if a year is a leap-year or not

if year modulo 400 is 0
       then is_leap_year
else if year modulo 100 is 0
       then not_leap_year
else if year modulo 4 is 0
       then is_leap_year
else
       not_leap_year

to create a list of all leap-years and the years that's not.

Answer 5

two parts: build the date ranges as tuples with a start and end date, build a dictionary whose key is the year and the values are the days. I don't need to account for leap year in the calculation because that is automatic in the range calculation

date_ranges=[]
def buildDateRanges(start_year,start_month,start_day, end_year,end_month,end_day):
     start_date=datetime.datetime(start_year,start_month,start_day)
     end_date=datetime.datetime(end_year,end_month,end_day)
     start_year=start_date.year
     end_year=end_date.year
     date_ranges=[]
     if start_year != end_year:
          for year in range(start_year,end_year+1):
              if year==start_year:
            
  date_ranges.append((start_date,datetime.datetime(start_date.year,12,31)))
             elif year==end_year:
                 date_ranges.append((datetime.datetime(start_year+1,1,1),end_date))
     else:
          date_ranges.append((start_date,end_date))
      return date_ranges

 date_ranges=buildDateRanges(2006, 7, 3,2012,12,21)

def years_days(days):
    years = days // 365
    days = days % 365
    return years, days

 results={}
 for i in range(len(date_ranges)):
     start_date=date_ranges[i][0]
     end_date=date_ranges[i][1]
     days=(end_date-start_date).days
     years,days=years_days(days)
     year=start_date.year
     print(years,days)
     while True:
         if year==end_date.year:
              results[year]=days    
         else:
              results[year]=365
         year+=1
         if year>end_date.year:
             break    
 print(results)

output: {2006: 181, 2007: 365.0, 2008: 365.0, 2009: 365.0, 2010: 365.0, 2011: 365.0, 2012: 356.0}

Answer 6

You can have something simply by doing this:

>>> from datetime import datetime
>>> d1 = datetime.strptime("30 Nov 00", "%d %b %y")
>>> (datetime.now() - d1).days
3907

Answer 7

1. Convert both days into seconds since the epoch (ie. 1 Jan 1970 Midnight)
2. Subtract.
3. The do the division to convert seconds into days for the difference.