>>> 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

Subtracting the later time from the first time difference = later_time - first_time creates a datetime object that only holds the difference. In the example above it is 0 minutes, 8 seconds and 562000 microseconds.

www.un.org/Depts/DGACM/index_spanish.htm 采用日期实例

>>> 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

<Duration in years

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

<>日<>

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

<>0> 小时长度:>

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

<<>Durent>

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

<>二版>

[!] 见关于“这一员额底部使用期限”的警告。

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

缩略语

[!] 见关于在本员额底部使用微型秒钟的警告。

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

www.un.org/Depts/DGACM/index_spanish.htm 两个日期之间的总期限:

>>> 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:

>>> print(now - then)

Edit 2019 Since this answer has gained traction, I ll add a function, which might simplify the usage for some

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

Warning: Caveat about built-in .seconds and .microseconds
datetime.seconds and datetime.microseconds are capped to [0,86400) and [0,10^6) respectively.

如果时间选择大于最大回报值,就应当谨慎使用。

实例:

>>> 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

>>> 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

新的第2.7条是<条码> 准时 例法>。 这相当于<条码>(td.microseconds + 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

仅从另一边推倒。 页: 1

>>> 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

如果a,b为时标,则在3/403中找到时间间隔:

from datetime import timedelta

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

此前的Adhury版本:

time_difference_in_minutes = time_difference.total_seconds() / 60

如果a,b为日元件,例如通过<代码>datetime.now(<>/code>退回,则如果这些物体代表当地时间,且其大小为不同的自动交换机,例如,围绕ST转换或过去/未来日期,则结果可能是错误的。 更详细情况: 如果从日期到<<>之间已过24小时,则Find。

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())

10.009222

例如,使用<代码>过期失效时间.seconds,你失去许多精确度(重新计算)。 此外,elapsedtime.microseconds 封顶在10^6上,,该回答。 因此,例如,关于第二个<代码>sleep(>,elapsedtime.microseconds,8325<>> (wrong,应改为10,000,000。)