如何在两个datetime对象之间以分钟为单位区分时间?
当前回答
以datetime为例
>>> 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)
[!参见本文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(微秒)
[!参见本文底部关于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]))
或者仅仅是:
>>> print(now - then)
编辑2019 由于这个答案已经获得了支持,我将添加一个函数,这可能会简化一些用法
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:关于内置的.seconds和.microseconds的警告 datetime。Seconds和datetime。微秒的上限分别为[0,86400)和[0,10^6)。
如果timedelta大于返回值的最大值,则应该谨慎使用。
例子:
开始后1h, 200μs结束:
>>> 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在开始后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
其他回答
我用的是这样的:
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
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
如果a, b是datetime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是天真的datetime对象,例如datetime.now()返回的,那么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。更多细节:查找datetimes之间是否已经过了24小时- Python。
要获得可靠的结果,请使用UTC时间或时区感知的datetime对象。
用其他方法得到日期之间的差异;
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)
得到最小值的输出。
谢谢
这将给出以秒为单位的差值(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
在我看来,这是最简单的方法,您不需要担心精度或溢出问题。
例如,使用elapsedTime。秒你失去了很多精度(它返回一个整数)。同时,elapsedTime。正如这个答案所指出的,微秒的上限是10^6。因此,例如,对于10秒的sleep(), elapsedTime。微秒给出8325(这是错误的,应该是10,000,000左右)。
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