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

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 look开发者_StackOverflow中文版ed for something like this on the internet but that only turned up ways to calculate the number of days/months/years between 2 dates

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

>>> 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.

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)

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.

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

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}

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.