如何分辨两个datetime对象之间的时差（以分钟为单位）？
问问题
764437 次
20 回答
474
>>> 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
从第一次减去后面的时间difference = later_time - first_time会创建一个仅包含差异的日期时间对象。在上面的示例中，它是 0 分钟、8 秒和 562000 微秒。
于 2009-08-28T09:08:13.897 回答
209
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
持续时间（年）
>>> years = divmod(duration_in_s, 31536000)[0] # Seconds in a year=365*24*60*60 = 31536000.
持续时间（天）
>>> days = duration.days # Build-in datetime function
>>> days = divmod(duration_in_s, 86400)[0] # Seconds in a day = 86400
持续时间（小时）
>>> hours = divmod(duration_in_s, 3600)[0] # Seconds in an hour = 3600
持续时间（分钟）
>>> minutes = divmod(duration_in_s, 60)[0] # Seconds in a minute = 60
持续时间（秒）
[!] 请参阅这篇文章底部关于使用持续时间（以秒为单位）的警告
>>> seconds = duration.seconds # Build-in datetime function
>>> seconds = duration_in_s
持续时间（以微秒为单位）
[!] 请参阅这篇文章底部有关使用持续时间（以微秒为单位）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]))
或者简单地说：
>>> print(now - then)
编辑 2019 由于这个答案已经获得关注，我将添加一个函数，这可能会简化某些人的使用
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
警告：关于内置 .seconds 和 .microseconds 的警告
datetime.seconds，datetime.microseconds分别限制为 [0,86400) 和 [0,10^6)。
如果 timedelta 大于最大返回值，则应谨慎使用它们。
例子：
end在 1h 和 200μs 之后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是 1d 和 1h 之后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
于 2017-11-09T16:39:40.400 回答
167
Python 2.7 的新功能是timedelta实例方法.total_seconds()。在 Python 文档中，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
于 2012-09-29T16:35:40.887 回答
39
只需从另一个中减去一个。你得到一个timedelta不同的对象。
>>> 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
于 2009-08-28T09:06:47.893 回答
27
如果a,b是日期时间对象，那么在 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 版本中：
time_difference_in_minutes = time_difference.total_seconds() / 60
如果a,b是原始日期时间对象，例如由datetime.now()then 返回，则如果对象表示具有不同 UTC 偏移量的本地时间，例如 DST 转换前后或过去/未来日期，则结果可能是错误的。更多详细信息：查找日期时间之间是否经过了 24 小时 - Python。
要获得可靠的结果，请使用 UTC 时间或可识别时区的日期时间对象。
于 2014-10-27T11:49:16.807 回答
19
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)
于 2014-03-31T18:05:09.040 回答
12
这就是我如何获得两个 datetime.datetime 对象之间经过的小时数：
before = datetime.datetime.now()
after = datetime.datetime.now()
hours = math.floor(((after - before).seconds) / 3600)
于 2013-05-26T01:11:21.103 回答
12
只求天数：timedelta 有一个“天”属性。您可以简单地查询它。
>>>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
于 2015-09-11T15:49:20.163 回答
12
只是认为在 timedelta 方面提及格式可能很有用。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
于 2016-05-27T21:46:44.390 回答
4
我使用这样的东西：
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
于 2018-03-07T22:15:01.120 回答
4
要获得hour,minute和second，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))
于 2020-11-03T10:07:52.803 回答
3
这将以秒为单位给出差异（然后除以 60 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
这是我认为最简单的方法，您无需担心精度或溢出。
例如，使用elapsedTime.seconds会损失很多精度（它返回一个整数）。此外，elapsedTime.microseconds正如这个答案所指出的，上限为 10^6。因此，例如，对于 10 秒sleep()，elapsedTime.microseconds给出8325 （这是错误的，应该在附近10,000,000）。
于 2021-04-03T13:32:23.007 回答
2
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)
于 2018-10-30T13:27:27.990 回答
1
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
于 2014-10-27T11:28:24.773 回答
1
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)
所以在 Min 中获得输出。
谢谢
于 2018-02-09T04:02:18.760 回答
1
我在持续集成测试中使用了时间差异来检查和改进我的功能。如果有人需要，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')
我在日志输出中有类似的东西
2019-12-20 17:19:23.635297 tl -----------> 0.00007
2019-12-20 17:19:28.147656 tl -----------> 4.51234 Tests finished
于 2019-12-20T13:17:25.970 回答
1
基于@Attaque 很好的答案，我提出了一个更短的日期时间差计算器的简化版本：
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]
我对其进行了更改以避免声明重复功能，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，月只是近似值，基于每个月等于 的假设365/12）。
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
于 2020-03-10T05:51:27.977 回答
1
您可能会发现这个快速代码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
于 2020-11-13T20:57:42.960 回答
0
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)
它以表格形式给出答案Duration=270 days, 10 hrs, 32 mins and 42.13 secs
于 2021-08-28T03:41:41.780 回答
-1
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 --> datetime 对象的开始时间
end_time--> datetime 对象的结束时间
我们不能直接减去 datetime.time 对象，
因此我们需要向它添加一个随机日期（我们使用组合）
，或者您可以使用“今天”而不是 (1,1,1)
希望这可以帮助
于 2021-04-05T13:37:22.830 回答