如何在Python中找到两个日期时间对象之间的时间差？

Question

我怎么能计算两个datetime对象之间的时间差，单位是分钟呢？

Answer 1

在Python 2.7中，新增了一个叫做timedelta的实例方法.total_seconds()。根据Python的文档，这个方法的作用是计算时间差的总秒数。简单来说，它的计算方式是这样的：先把微秒数加上秒数，再加上天数转换成秒的部分（天数乘以24小时再乘以3600秒），最后把这个总和除以1000000，得到的就是总秒数。

参考链接：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 2

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）

[!] 请查看本文底部关于使用秒数的警告

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

持续时间（微秒）

[!] 请查看本文底部关于使用微秒的警告

>>> 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]))

或者简单来说：

>>> print(now - then)

---

编辑于2019年
由于这个回答受到了关注，我会添加一个函数，可能会让一些人使用起来更简单

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 的注意事项
datetime.seconds 和 datetime.microseconds 的范围分别是 [0,86400) 和 [0,10^6)。

如果时间间隔超过了这些最大值，使用时要小心。

例子：

end 是在 start 之后 1小时和200微秒：

>>> 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 是在 start 之后 1天和1小时：

>>> 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

>>> 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

把后面的时间减去前面的时间，difference = later_time - first_time，就会得到一个只表示时间差的对象。在上面的例子中，这个时间差是0分钟、8秒和562000微秒。