[python] 파이썬에서 두 datetime 객체의 시간 차이를 어떻게 알 수 있습니까?

날짜 시간 예제 사용

>>> 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

년의 기간

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

일수

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

시간 내

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

분 단위

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

초 단위의 시간

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

마이크로 초 단위의 지속 시간

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

두 날짜 사이의 총 기간

>>> 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]))

또는 간단히 :

>>> print(now - then)

2019 수정
이 답변에 큰 관심을 끌기 때문에 일부 기능을 단순화 할 수있는 기능을 추가하겠습니다.

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

답변

>>> import datetime
>>> d1 = datetime.datetime.now()
>>> d2 = datetime.datetime.now() # after a 5-second or so pause
>>> d2 - d1
datetime.timedelta(0, 5, 203000)

from datetime import timedelta

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

time_difference_in_minutes = time_difference.total_seconds() / 60

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)

before = datetime.datetime.now()
after  = datetime.datetime.now()
hours  = math.floor(((after - before).seconds) / 3600)