# 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?

``````>>> 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, Ill 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 .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.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
``````

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

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`.

``````>>> 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
``````