如何在python中找到两个日期时间对象之间的时差？

Question

如何判断两个datetime对象之间的时差（以分钟为单位）？

Answer 1

>>> 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.

Answer 2

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 .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

Answer 3

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

Answer 4

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)

You can convert dd.days , dd.seconds and dd.microseconds to minutes.

Answer 5

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 eg, 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.

Answer 6

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)

Answer 7

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)

Answer 8

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

Answer 9

To just find the number of days: timedelta has a 'days' attribute. You can simply query that.

>>>from datetime import datetime, timedelta
>>>d1 = datetime(2015, 9, 12, 13, 9, 45)
>>>d2 = datetime(2015, 8, 29, 21, 10, 12)
>>>d3 = d1- d2
>>>print d3
13 days, 15:59:33
>>>print d3.days
13

Answer 10

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

Answer 11

This will give the difference in seconds (then just divide by 60 to get minutes):

import time
import datetime

t_start = datetime.datetime.now()

time.sleep(10)

t_end = datetime.datetime.now()
elapsedTime = (t_end - t_start )

print(elapsedTime.total_seconds())

outputs:

10.009222

This is the simplest way in my opinion, and you don't need to worry about precision or overflow.

For instance, using elapsedTime.seconds you lose a lot of precision (it returns an integer). Also, elapsedTime.microseconds is capped at 10^6, as this answer pointed out. So, for example, for a 10 second sleep() , elapsedTime.microseconds gives 8325 (which is wrong , should be around 10,000,000 ).

Answer 12

这是为了找出当前时间和上午 9.30 之间的差异

t=datetime.now()-datetime.now().replace(hour=9,minute=30)

Answer 13

To get the hour , minute and second , you can do this

>>> import datetime
>>> first_time = datetime.datetime.now()
>>> later_time = datetime.datetime.now()
>>> difference = later_time - first_time
>>> m,s = divmod(difference.total_seconds(), 60)
>>> print("H:M:S is {}:{}:{}".format(m//60,m%60,s)

Answer 14

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

Answer 15

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

Answer 16

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) -> 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 -----------> %.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 -----------> 0.00007
2019-12-20 17:19:28.147656 tl -----------> 4.51234   Tests finished

Answer 17

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))      # => (0.0, 2678400.0015)
print(get_duration(d1, d2, 'm', True))      # => (1.0, 50397.12149999989)
print(get_duration(d1, d2, 'w', True))      # => (4.0, 259200.00149999978)
print(get_duration(d1, d2, 'd', True))      # => (31.0, 0.0014999997802078724)
print(get_duration(d1, d2, 'h', True))      # => (744.0, 0.0014999997802078724)
print(get_duration(d1, d2, 'min', True))    # => (44640.0, 0.0014999997802078724)
print(get_duration(d1, d2, 's', True))      # => (2678400.0, 0.0014999997802078724)
print(get_duration(d1, d2, 'mil', True))    # => (2678400001.0, 0.0004999997244524721)

print(get_duration(d1, d2, 'y', False))     # => 0.08493150689687975
print(get_duration(d1, d2, 'm', False))     # => 1.019176965856293
print(get_duration(d1, d2, 'w', False))     # => 4.428571431051587
print(get_duration(d1, d2, 'd', False))     # => 31.00000001736111
print(get_duration(d1, d2, 'h', False))     # => 744.0000004166666
print(get_duration(d1, d2, 'min', False))   # => 44640.000024999994
print(get_duration(d1, d2, 's', False))     # => 2678400.0015
print(get_duration(d1, d2, 'mil', False))   # => 2678400001.4999995

Answer 18

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

Answer 19

Here is an answer that is easy to generalise or turn into a function and which is reasonable compact and easy to follow.

ts_start=datetime(2020, 12, 1, 3, 9, 45)
ts_end=datetime.now()
ts_diff=ts_end-ts_start
secs=ts_diff.total_seconds()
days,secs=divmod(secs,secs_per_day:=60*60*24)
hrs,secs=divmod(secs,secs_per_hr:=60*60)
mins,secs=divmod(secs,secs_per_min:=60)
secs=round(secs, 2)
answer='Duration={} days, {} hrs, {} mins and {} secs'.format(int(days),int(hrs),int(mins),secs)
print(answer)

It gives an answer in the form Duration=270 days, 10 hrs, 32 mins and 42.13 secs

Answer 20

import datetime
date = datetime.date(1, 1, 1)
#combine a dummy date to the time
datetime1 = datetime.datetime.combine(date, start_time)
datetime2 = datetime.datetime.combine(date, stop_time)  
#compute the difference
time_elapsed = datetime1 - datetime2

start_time --> start time for datetime object
end_time--> end time for datetime object

we cannot directly subtract the datetime.time objects
hence we need to add a random date to it (we use combine)
or you can use the "today" instead of (1,1,1)

hope this helps