如何在Python中找到两个日期时间对象之间的时间差？

Question

如何判断两个 datetime 对象之间的时间差（以分钟为单位）？

Answer 1

我用这样的东西：

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 2

这将给出以秒为单位的差异（然后除以 60 即可得到分钟）：

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())

输出：

10.009222

在我看来，这是最简单的方法，您无需担心精度或溢出。

例如，使用elapsedTime.seconds会丢失很多精度（它返回一个整数）。此外，正如 这个答案 所指出的，elapsedTime.microseconds 的上限为 10^6。例如，对于 10 秒的 sleep()，elapsedTime.microseconds 给出 8325（这是错误的 ，应约为 10,000,000）。

Answer 3

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 4

基于@Attaque伟大的答案，我提出了一个更短的简化版本的日期时间差计算器：

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]

我已经更改它以避免声明重复函数，删除了漂亮的打印默认间隔，并添加了对毫秒、周和 ISO 月的支持（请记住，月份只是近似值，基于每个月等于 365/12 的假设）。

其产生：

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 5

这是我使用 mktime 的方法。

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 6

以其他方式获取日期之间的差异；

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)

所以得到最小的输出。

谢谢

Answer 7

我利用时间差进行持续集成测试来检查和改进我的功能。如果有人需要的话，这是简单的代码

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()

使用：

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')

并且我在日志输出中有类似的内容

2019-12-20 17:19:23.635297 tl -----------> 0.00007
2019-12-20 17:19:28.147656 tl -----------> 4.51234   Tests finished

Answer 8

您可能会发现这个快速片段在不太长的时间间隔内很有用：

    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(':'))

在 Python v.3.8.6 上测试

Answer 9

这是一个易于概括或转化为函数的答案，并且合理紧凑且易于遵循。

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)

它以 Duration=270 天、10 小时、32 分钟和 42.13 秒的形式给出答案

Answer 10

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

开始时间 -->日期时间对象的开始时间
结束时间--> datetime 对象的结束时间

我们不能直接减去 datetime.time 对象
因此我们需要添加一个随机日期（我们使用组合）
或者您可以使用“今天”而不是 (1,1,1)

希望这会有所帮助

Answer 11

这可能会帮助某人，用这种方法计算天数来查找是否过期。还有 dt.seconds 和 dt.microseconds 也可用

from datetime import datetime
# updated_at = "2022-10-20T07:18:56.950563"
def is_expired(updated_at):
    expires_in = 7 #days
    datetime_format = '%Y-%m-%dT%H:%M:%S.%f'
    time_difference = datetime.now() - datetime.strptime(updated_at, datetime_format)

    return True if time_difference.days > expires_in else False

Answer 12

>>> 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

从第一次中减去后来的时间 difference = later_time - first_time 创建一个仅保存差异的日期时间对象。
在上面的示例中，它是 0 分 8 秒 562000 微秒。

Answer 13

使用日期时间示例

>>> 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

以年为单位的持续时间

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

以天为单位的持续时间

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

以小时为单位的持续时间

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

以分钟为单位的持续时间

>>> minutes = divmod(duration_in_s, 60)[0]        # Seconds in a minute = 60

以秒为单位的持续时间

[!] 请参阅本文底部有关使用以秒为单位的持续时间的警告

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

以微秒为单位的持续时间

[!] 请参阅关于在本文底部使用以微秒为单位的持续时间的警告

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

两个日期之间的总持续时间

>>> 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]))

或简单地：

>>> print(now - then)

---

编辑 2019
的使用

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

由于这个答案已经受到关注，我将添加一个函数，这可能会简化某些警告：关于内置 .seconds 和 .microseconds 的警告
datetime.seconds 和 datetime.microseconds 的上限分别为 [0,86400) 和 [0,10^6)。

如果 timedelta 大于最大返回值，则应谨慎使用它们。

示例：

end 是 start 之后 1h 和 200μs：

>>> 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 是 start 之后 1d 和 1h：

>>> 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 14

Python 2.7 中的新功能是 timedelta 实例方法 .total_seconds()。从 Python 文档来看，这相当于 (td.microseconds + (td.seconds + td.days * 24 * 3600) * 10**6) / 10**6。

参考：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 15

只需将其中一个减去另一个即可。您将获得一个具有差异的 timedelta 对象。

>>> 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

Answer 16

如果 a、b 是日期时间对象，那么要在 Python 3 中查找它们之间的时间差：

from datetime import timedelta

time_difference = a - b
time_difference_in_minutes = time_difference / timedelta(minutes=1)

在早期的 Python 版本上：

time_difference_in_minutes = time_difference.total_seconds() / 60

If a, b 是简单的日期时间对象，例如由 datetime.now() 返回，那么如果对象表示具有不同 UTC 偏移量的本地时间（例如，在 DST 转换前后或过去），结果可能是错误的/未来的日期。更多详细信息：查找日期时间之间是否已过去 24 小时 - Python。

要获得可靠的结果，请使用 UTC 时间或时区感知的日期时间对象。

Answer 17

使用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 18

要查找天数：timedelta 有一个“days”属性。你可以简单地查询一下。

>>>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 19

这就是我获取两个 datetime.datetime 对象之间经过的小时数的方法：

before = datetime.datetime.now()
after  = datetime.datetime.now()
hours  = math.floor(((after - before).seconds) / 3600)

Answer 20

只是认为提及有关时间增量的格式化可能会有用。 strptime() 根据格式解析表示时间的字符串。

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)

这将输出：
0:05:00.518000

Answer 21

要获取小时、分钟和秒，您可以这样做

>>> 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))