如何判断两个 datetime
对象之间的时差(以分钟为单位)?
原文由 Hobhouse 发布,翻译遵循 CC BY-SA 4.0 许可协议
如何判断两个 datetime
对象之间的时差(以分钟为单位)?
原文由 Hobhouse 发布,翻译遵循 CC BY-SA 4.0 许可协议
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(以秒为单位)
[!] 请参阅本文底部关于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
持续时间(以微秒为单位)
[!] 请参阅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:
>>> print(now - then)
编辑 2019 由于这个答案获得了关注,我将添加一个功能,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:注意内置的 .seconds 和 .microseconds
datetime.seconds
和 datetime.microseconds
分别上限为 [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 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 start
之后的 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
原文由 Nicolai Lissau 发布,翻译遵循 CC BY-SA 4.0 许可协议
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从第一个时间减去后面的时间
difference = later_time - first_time
创建一个只保留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。