How do I find the time difference between two datetime objects in python?
How do I tell the time difference in minutes between two datetime objects?
>>> import datetime
>>> first_time = datetime.datetime.now()
>>> later_time = datetime.datetime.now()
>>> difference = later_time - first_time
datetime.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 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 = "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 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 .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.microseconds
RESULT: 200
EXPECTED: 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.seconds
RESULT: 3600
EXPECTED: 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
>>> elapsedTime
datetime.timedelta(0, 125, 749430)
>>> divmod(elapsedTime.total_seconds(), 60)
(2.0, 5.749430000000004) # divmod returns quotient and remainder
# 2 minutes, 5.74943 seconds
Just subtract one from the other. You get a timedelta object with the difference.
>>> import datetime
>>> d1 = datetime.datetime.now()
>>> d2 = datetime.datetime.now() # after a 5-second or so pause
>>> d2 - d1
datetime.timedelta(0, 5, 203000)
>>> dd = d2 - d1
>>> print (dd.days) # get days
>>> print (dd.seconds) # get seconds
>>> print (dd.microseconds) # get microseconds
>>> print (int(round(dd.total_seconds()/60, 0))) # get minutes