How to calculate number of days between two given dates

Question

If I have two dates  ex   8 18 2008  and  9 26 2008    what is the best way to get the number of days between these two dates

User · Answer

everyone has answered excellently using the date, let me try to answer it using pandas

dt = pd.to_datetime('2008/08/18', format='%Y/%m/%d')
dt1 = pd.to_datetime('2008/09/26', format='%Y/%m/%d')

(dt1-dt).days

This will give the answer. In case one of the input is dataframe column. simply use dt.days in place of days

(dt1-dt).dt.days

User · Answer

Here are three ways to go with this problem    from datetime import datetime  Now   datetime now   StartDate   datetime strptime str Now year    -01-01     Y- m- d   NumberOfDays    Now - StartDate   print NumberOfDays days                        Starts at 0 print datetime now   timetuple   tm yday       Starts at 1 print Now strftime   j                         Starts at 1

User · Answer

If you have two date objects  you can just subtract them  which computes a timedelta object   from datetime import date  d0   date 2008  8  18  d1   date 2008  9  26  delta   d1 - d0 print delta days    The relevant section of the docs  https   docs python org library datetime html   See this answer for another example

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For calculating dates and times  there are several options but I will write the simple way  from datetime import timedelta  datetime  date import dateutil relativedelta    current time date and time   datetime now   date only   date today   time only   datetime now   time      calculate date and time result   date and time - timedelta hours 26  minutes 25  seconds 10     calculate dates  years  -    result   date only - dateutil relativedelta relativedelta years 10     months result   date only - dateutil relativedelta relativedelta months 10     days result   date only - dateutil relativedelta relativedelta days 10     calculate time  result   date and time - timedelta hours 26  minutes 25  seconds 10  result time    Hope it helps

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from datetime import date def d s      month  day  year    map int  s split         return date year  month  day  def days start  end     return  d end  - d start   days print days  8 18 2008    9 26 2008     This assumes  of course  that you ve already verified that your dates are in the format r  d   d   d

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You want the datetime module     gt  gt  gt  from datetime import datetime  timedelta   gt  gt  gt  datetime 2008 08 18  - datetime 2008 09 26   datetime timedelta 4     Another example    gt  gt  gt  import datetime   gt  gt  gt  today   datetime date today     gt  gt  gt  print today  2008-09-01   gt  gt  gt  last year   datetime date 2007  9  1    gt  gt  gt  print today - last year  366 days  0 00 00    As pointed out here

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from datetime import datetime start date   datetime strptime  8 18 2008     m  d  Y   end date   datetime strptime  9 26 2008     m  d  Y   print abs  end date-start date  days

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Using the power of datetime   from datetime import datetime date format     m  d  Y  a   datetime strptime  8 18 2008   date format  b   datetime strptime  9 26 2008   date format  delta   b - a print delta days   that s it

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It also can be easily done with arrow   import arrow  a   arrow get  2017-05-09   b   arrow get  2017-05-11    delta    b-a  print delta days   For reference  http   arrow readthedocs io en latest

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without using Lib just pure code    Calculate the Days between Two Date  daysOfMonths     31  28  31  30  31  30  31  31  30  31  30  31   def isLeapYear year          Pseudo code for this algorithm is found at       http   en wikipedia org wiki Leap year Algorithm        if  year is not divisible by 4  then  it is a common Year       else if  year is not divisable by 100  then  ut us a leap year       else if  year is not disible by 400  then  it is a common year       else it is aleap year      return  year   4    0 and year   100    0  or year   400    0  def Count Days year1  month1  day1       if month1   2          if isLeapYear year1               if day1  lt  daysOfMonths month1-1  1                  return year1  month1  day1 1             else                  if month1   12                      return year1 1 1 1                 else                      return year1  month1  1   1         else               if day1  lt  daysOfMonths month1-1                   return year1  month1  day1 1             else                  if month1   12                      return year1 1 1 1                 else                      return year1  month1  1   1     else          if day1  lt  daysOfMonths month1-1                return year1  month1  day1 1         else              if month1   12                  return year1 1 1 1             else                      return year1  month1  1   1   def daysBetweenDates y1  m1  d1  y2  m2  d2 end day        if y1  gt  y2          m1 m2   m2 m1         y1 y2   y2 y1         d1 d2   d2 d1     days 0     while not m1  m2 and y1  y2 and d1  d2            y1 m1 d1   Count Days y1 m1 d1          days  1     if end day          days  1     return days     Test Case  def test        test cases      2012 1 1 2012 2 28 False   58                        2012 1 1 2012 3 1 False   60                       2011 6 30 2012 6 30 False   366                       2011 1 1 2012 8 8 False   585                        1994 5 15 2019 8 31 False   9239                       1999 3 24 2018 2 4 False   6892                       1999 6 24 2018 8 4 False  6981                       1995 5 24 2018 12 15 False  8606                       1994 8 24 2019 12 15 True  9245                       2019 12 15 1994 8 24 True  9245                       2019 5 15 1994 10 24 True  8970                       1994 11 24 2019 8 15 True  9031        for  args  answer  in test cases          result   daysBetweenDates  args          if result    answer              print  Test with data    args   failed          else              print  Test case passed    test

User · Answer

There is also a datetime toordinal   method that was not mentioned yet   import datetime print datetime date 2008 9 26  toordinal   - datetime date 2008 8 18  toordinal       39   https   docs python org 3 library datetime html datetime date toordinal     date toordinal        Return the proleptic Gregorian ordinal of the date  where January 1 of year 1 has ordinal 1  For any date object d    date fromordinal d toordinal       d    Seems well suited for calculating days difference  though not as readable as timedelta days

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Days until Christmas    gt  gt  gt  import datetime  gt  gt  gt  today   datetime date today    gt  gt  gt  someday   datetime date 2008  12  25   gt  gt  gt  diff   someday - today  gt  gt  gt  diff days 86   More arithmetic here

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Without using datetime object in python    A date has day  d   month  m  and year  y   class Date      def   init   self  d  m  y               self d   d             self m   m             self y   y    To store number of days in all months from    January to Dec   monthDays    31  28  31  30  31  30                                              31  31  30  31  30  31      This function counts number of leap years    before the given date  def countLeapYears d        years   d y        Check if the current year needs to be considered        for the count of leap years or not      if  d m  lt   2                years-  1        An year is a leap year if it is a multiple of 4         multiple of 400 and not a multiple of 100       return int years   4 - years   100   years   400       This function returns number of days between two    given dates  def getDifference dt1  dt2           COUNT TOTAL NUMBER OF DAYS BEFORE FIRST DATE  dt1          initialize count using years and day      n1   dt1 y   365   dt1 d        Add days for months in given date      for i in range 0  dt1 m - 1                n1    monthDays i         Since every leap year is of 366 days         Add a day for every leap year      n1    countLeapYears dt1         SIMILARLY  COUNT TOTAL NUMBER OF DAYS BEFORE  dt2        n2   dt2 y   365   dt2 d     for i in range 0  dt2 m - 1                n2    monthDays i      n2    countLeapYears dt2         return difference between two counts      return  n2 - n1      Driver program  dt1   Date 31  12  2018   dt2   Date 1  1  2019    print getDifference dt1  dt2    quot days quot

[python] How to calculate number of days between two given dates?

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