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

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


>>> import datetime>>> first_time = datetime.datetime.now()>>> later_time = datetime.datetime.now()>>> difference = later_time - first_timedatetime.timedelta(0, 8, 562000)>>> seconds_in_day = 24 * 60 * 60>>> 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.


Using datetime example

>>> from datetime import datetime>>> then = datetime(2012, 3, 5, 23, 8, 15)        # Random date in the past>>> now  = datetime.now()                         # Now>>> duration = now - then                         # For build-in functions>>> duration_in_s = duration.total_seconds()      # Total number of seconds between dates

Duration in years

>>> years = divmod(duration_in_s, 31536000)[0]    # Seconds in a year=365*24*60*60 = 31536000.

Duration in days

>>> days  = duration.days                         # Build-in datetime function>>> days  = divmod(duration_in_s, 86400)[0]       # Seconds in a day = 86400

Duration in hours

>>> hours = divmod(duration_in_s, 3600)[0]        # Seconds in an hour = 3600

Duration in minutes

>>> 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

>>> seconds = duration.seconds                    # Build-in datetime function>>> seconds = duration_in_s

Duration in microseconds

[!] See warning about using duration in microseconds in the bottom of this post

>>> microseconds = duration.microseconds          # Build-in datetime function

Total duration between the two dates

>>> days    = divmod(duration_in_s, 86400)        # Get days (without [0]!)>>> hours   = divmod(days[1], 3600)               # Use remainder of days to calc hours>>> minutes = divmod(hours[1], 60)                # Use remainder of hours to calc minutes>>> seconds = divmod(minutes[1], 1)               # Use remainder of minutes to calc seconds>>> print("Time between dates: %d days, %d hours, %d minutes and %d seconds" % (days[0], hours[0], minutes[0], seconds[0]))

or simply:

>>> print(now - then)

Edit 2019Since this answer has gained traction, I'll add a function, which might simplify the usage for some

from datetime import datetimedef getDuration(then, now = datetime.now(), interval = "default"):    # 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 "Time between dates: {} years, {} days, {} hours, {} minutes and {} seconds".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 usagethen = 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 secondsprint(getDuration(then, now, 'years'))      # Prints duration in yearsprint(getDuration(then, now, 'days'))       #                    daysprint(getDuration(then, now, 'hours'))      #                    hoursprint(getDuration(then, now, 'minutes'))    #                    minutesprint(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 200μs after start:

>>> start = datetime(2020,12,31,22,0,0,500)>>> end = datetime(2020,12,31,23,0,0,700)>>> delta = end - start>>> delta.microsecondsRESULT: 200EXPECTED: 3600000200

end is 1d and 1h after start:

>>> start = datetime(2020,12,30,22,0,0)>>> end = datetime(2020,12,31,23,0,0)>>> delta = end - start>>> delta.secondsRESULT: 3600EXPECTED: 90000


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

>>> import datetime>>> time1 = datetime.datetime.now()>>> time2 = datetime.datetime.now() # waited a few minutes before pressing enter>>> elapsedTime = time2 - time1>>> elapsedTimedatetime.timedelta(0, 125, 749430)>>> divmod(elapsedTime.total_seconds(), 60)(2.0, 5.749430000000004) # divmod returns quotient and remainder# 2 minutes, 5.74943 seconds


matomo