How do I find the time difference between two datetime objects in python

Question

How do I tell the time difference in minutes between two datetime objects

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To get the hour  minute and second  you can do this  gt  gt  gt  import datetime  gt  gt  gt  first time   datetime datetime now    gt  gt  gt  later time   datetime datetime now    gt  gt  gt  difference   later time - first time  gt  gt  gt  m s   divmod difference total seconds    60   gt  gt  gt  print  quot H M S is          quot  format m  60 m 60 s

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gt  gt  gt  import datetime  gt  gt  gt  first time   datetime datetime now    gt  gt  gt  later time   datetime datetime now    gt  gt  gt  difference   later time - first time  gt  gt  gt  seconds in day   24   60   60 datetime timedelta 0  8  562000   gt  gt  gt  divmod difference days   seconds in day   difference seconds  60   0  8         0 minutes  8 seconds   Subtracting the later time from the first time difference   later time - first time creates a datetime object that only holds the difference   In the example above it is 0 minutes  8 seconds and 562000 microseconds

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To just find the number of days  timedelta has a  days  attribute  You can simply query that    gt  gt  gt from datetime import datetime  timedelta  gt  gt  gt d1   datetime 2015  9  12  13  9  45   gt  gt  gt d2   datetime 2015  8  29  21  10  12   gt  gt  gt d3   d1- d2  gt  gt  gt print d3 13 days  15 59 33  gt  gt  gt print d3 days 13

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I have used time differences for continuous integration tests to check and improve my functions  Here is simple code if somebody need it  from datetime import datetime  class TimeLogger      time cursor   None      def pin time self           global time cursor         time cursor   datetime now        def log self  text None  - gt  float          global time cursor          if not time cursor              time cursor   datetime now            now   datetime now           t delta   now - time cursor          seconds   t delta total seconds            result   str now      tl ----------- gt    5f    seconds         if text              result            text         print result           self pin time            return seconds   time logger   TimeLogger     Using   from  tests time logger import time logger class Tests TestCase       def test workflow self       time logger pin time            my functions here          time logger log            other function s           time logger log text  Tests finished     and i have something like that in log output  2019-12-20 17 19 23 635297 tl ----------- gt  0 00007 2019-12-20 17 19 28 147656 tl ----------- gt  4 51234   Tests finished

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If a  b are datetime objects then to find the time difference between them in Python 3   from datetime import timedelta  time difference   a - b time difference in minutes   time difference   timedelta minutes 1    On earlier Python versions   time difference in minutes   time difference total seconds     60   If a  b are naive datetime objects such as returned by datetime now   then the result may be wrong if the objects represent local time with different UTC offsets e g   around DST transitions or for past future dates  More details  Find if 24 hrs have passed between datetimes - Python   To get reliable results  use UTC time or timezone-aware datetime objects

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Just thought it might be useful to mention formatting as well in regards to timedelta  strptime   parses a string representing a time according to a format   from datetime import datetime  datetimeFormat     Y  m  d  H  M  S  f      time1    2016 03 16 10 01 28 585  time2    2016 03 16 09 56 28 067    time dif   datetime strptime time1  datetimeFormat  - datetime strptime time2 datetimeFormat  print time dif    This will output      0 05 00 518000

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You may find this fast snippet useful in not so much long time intervals      from datetime import datetime as dttm     time ago   dttm 2017  3  1  1  1  1  1348      delta   dttm now   - time ago     days   delta days   can be converted into years which complicates a bit        hours  minutes  seconds   map int  delta   format       split      0  split      -1  split        tested on Python v 3 8 6

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This is how I get the number of hours that elapsed between two datetime datetime objects   before   datetime datetime now   after    datetime datetime now   hours    math floor   after - before  seconds    3600

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this is to find the difference between current time and 9 30 am  t datetime now  -datetime now   replace hour 9 minute 30

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I use somethign like this    from datetime import datetime  def check time difference t1  datetime  t2  datetime       t1 date   datetime          t1 year          t1 month          t1 day          t1 hour          t1 minute          t1 second       t2 date   datetime          t2 year          t2 month          t2 day          t2 hour          t2 minute          t2 second       t elapsed   t1 date - t2 date      return t elapsed    usage  f     Y- m- d  H  M  S 01 00  t1   datetime strptime  2018-03-07 22 56 57 01 00   f  t2   datetime strptime  2018-03-07 22 48 05 01 00   f  elapsed time   check time difference t1  t2   print elapsed time   return   0 08 52

User · Answer

Just subtract one from the other  You get a timedelta object with the difference    gt  gt  gt  import datetime  gt  gt  gt  d1   datetime datetime now    gt  gt  gt  d2   datetime datetime now     after a 5-second or so pause  gt  gt  gt  d2 - d1 datetime timedelta 0  5  203000    You can convert dd days  dd seconds and dd microseconds to minutes

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New at Python 2 7 is the timedelta instance method  total seconds    From the Python docs  this is equivalent to  td microseconds    td seconds   td days   24   3600    10  6    10  6   Reference  http   docs python org 2 library datetime html datetime timedelta total seconds   gt  gt  gt  import datetime  gt  gt  gt  time1   datetime datetime now    gt  gt  gt  time2   datetime datetime now     waited a few minutes before pressing enter  gt  gt  gt  elapsedTime   time2 - time1  gt  gt  gt  elapsedTime datetime timedelta 0  125  749430   gt  gt  gt  divmod elapsedTime total seconds    60   2 0  5 749430000000004    divmod returns quotient and remainder   2 minutes  5 74943 seconds

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This is my approach using mktime   from datetime import datetime  timedelta from time import mktime  yesterday   datetime now   - timedelta days 1  today   datetime now    difference in seconds   abs mktime yesterday timetuple    - mktime today timetuple     difference in minutes   difference in seconds   60

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Use divmod   now   int time time      epoch seconds then   now - 90000   some time in the past  d   divmod now-then 86400     days h   divmod d 1  3600     hours m   divmod h 1  60     minutes s   m 1     seconds  print   d days   d hours   d minutes   d seconds     d 0  h 0  m 0  s

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Using datetime example  gt  gt  gt  from datetime import datetime  gt  gt  gt  then   datetime 2012  3  5  23  8  15           Random date in the past  gt  gt  gt  now    datetime now                             Now  gt  gt  gt  duration   now - then                           For build-in functions  gt  gt  gt  duration in s   duration total seconds          Total number of seconds between dates  Duration in years  gt  gt  gt  years   divmod duration in s  31536000  0       Seconds in a year 365 24 60 60   31536000   Duration in days  gt  gt  gt  days    duration days                           Build-in datetime function  gt  gt  gt  days    divmod duration in s  86400  0          Seconds in a day   86400  Duration in hours  gt  gt  gt  hours   divmod duration in s  3600  0           Seconds in an hour   3600  Duration in minutes  gt  gt  gt  minutes   divmod duration in s  60  0           Seconds in a minute   60  Duration in seconds     See warning about using duration in seconds in the bottom of this post  gt  gt  gt  seconds   duration seconds                      Build-in datetime function  gt  gt  gt  seconds   duration in s  Duration in microseconds     See warning about using duration in microseconds in the bottom of this post  gt  gt  gt  microseconds   duration microseconds            Build-in datetime function  Total duration between the two dates  gt  gt  gt  days      divmod duration in s  86400           Get days  without  0     gt  gt  gt  hours     divmod days 1   3600                  Use remainder of days to calc hours  gt  gt  gt  minutes   divmod hours 1   60                   Use remainder of hours to calc minutes  gt  gt  gt  seconds   divmod minutes 1   1                  Use remainder of minutes to calc seconds  gt  gt  gt  print  quot Time between dates   d days   d hours   d minutes and  d seconds quot     days 0   hours 0   minutes 0   seconds 0     or simply   gt  gt  gt  print now - then    Edit 2019 Since this answer has gained traction  I ll add a function  which might simplify the usage for some from datetime import datetime  def getDuration then  now   datetime now    interval    quot default quot           Returns a duration as specified by variable interval       Functions  except totalDuration  returns  quotient  remainder       duration   now - then   For build-in functions     duration in s   duration total seconds             def years          return divmod duration in s  31536000    Seconds in a year 31536000       def days seconds   None         return divmod seconds if seconds    None else duration in s  86400    Seconds in a day   86400      def hours seconds   None         return divmod seconds if seconds    None else duration in s  3600    Seconds in an hour   3600      def minutes seconds   None         return divmod seconds if seconds    None else duration in s  60    Seconds in a minute   60      def seconds seconds   None         if seconds    None          return divmod seconds  1           return duration in s      def totalDuration            y   years           d   days y 1     Use remainder to calculate next variable         h   hours d 1           m   minutes h 1           s   seconds m 1            return  quot Time between dates     years     days     hours     minutes and    seconds quot  format int y 0    int d 0    int h 0    int m 0    int s 0         return            years   int years   0             days   int days   0             hours   int hours   0             minutes   int minutes   0             seconds   int seconds              default   totalDuration         interval     Example usage then   datetime 2012  3  5  23  8  15  now   datetime now    print getDuration then     E g  Time between dates  7 years  208 days  21 hours  19 minutes and 15 seconds print getDuration then  now   years           Prints duration in years print getDuration then  now   days                               days print getDuration then  now   hours                              hours print getDuration then  now   minutes                            minutes print getDuration then  now   seconds                            seconds  Warning  Caveat about built-in  seconds and  microseconds datetime seconds and datetime microseconds are capped to  0 86400  and  0 10 6  respectively  They should be used carefully if timedelta is bigger than the max returned value  Examples  end is 1h and 200  s after start   gt  gt  gt  start   datetime 2020 12 31 22 0 0 500   gt  gt  gt  end   datetime 2020 12 31 23 0 0 700   gt  gt  gt  delta   end - start  gt  gt  gt  delta microseconds RESULT  200 EXPECTED  3600000200  end is 1d and 1h after start   gt  gt  gt  start   datetime 2020 12 30 22 0 0   gt  gt  gt  end   datetime 2020 12 31 23 0 0   gt  gt  gt  delta   end - start  gt  gt  gt  delta seconds RESULT  3600 EXPECTED  90000

User · Answer

In Other ways to get difference between date   import dateutil parser import datetime last sent date        date string timeDifference   current date - dateutil parser parse last sent date  time difference in minutes    int timeDifference days    24   60    int  timeDifference seconds    60    So get output in Min   Thanks

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Based on  Attaque great answer  I propose a shorter simplified version of the datetime difference calculator   seconds mapping          y   31536000       m   2628002 88    this is approximate  365   12  use with caution      w   604800       d   86400       h   3600       min   60       s   1       mil   0 001     def get duration d1  d2  interval  with reminder False       if with reminder          return divmod  d2 - d1  total seconds    seconds mapping interval       else          return  d2 - d1  total seconds     seconds mapping interval    I ve changed it to avoid declaring repetetive functions  removed the pretty print default interval and added support for milliseconds  weeks and ISO months  bare in mind months are just approximate  based on assumption that each month is equal to 365 12    Which produces   d1   datetime 2011  3  1  1  1  1  1000  d2   datetime 2011  4  1  1  1  1  2500   print get duration d1  d2   y   True            gt   0 0  2678400 0015  print get duration d1  d2   m   True            gt   1 0  50397 12149999989  print get duration d1  d2   w   True            gt   4 0  259200 00149999978  print get duration d1  d2   d   True            gt   31 0  0 0014999997802078724  print get duration d1  d2   h   True            gt   744 0  0 0014999997802078724  print get duration d1  d2   min   True          gt   44640 0  0 0014999997802078724  print get duration d1  d2   s   True            gt   2678400 0  0 0014999997802078724  print get duration d1  d2   mil   True          gt   2678400001 0  0 0004999997244524721   print get duration d1  d2   y   False           gt  0 08493150689687975 print get duration d1  d2   m   False           gt  1 019176965856293 print get duration d1  d2   w   False           gt  4 428571431051587 print get duration d1  d2   d   False           gt  31 00000001736111 print get duration d1  d2   h   False           gt  744 0000004166666 print get duration d1  d2   min   False         gt  44640 000024999994 print get duration d1  d2   s   False           gt  2678400 0015 print get duration d1  d2   mil   False         gt  2678400001 4999995

[python] How do I find the time difference between two datetime objects in python?

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