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 .microsecondsdatetime.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