253

必须有一种更简单的方法来做到这一点。我有想要经常刷新的对象，所以我想记录它们的创建时间，检查当前时间戳，并在必要时刷新。

datetime.datetime 已被证明是困难的，我不想深入 ctime 库。这种事情有什么更容易的吗？

4

7 回答 7

659

如果要计算两个已知日期之间的差异，请total_seconds像这样使用：

import datetime as dt

a = dt.datetime(2013,12,30,23,59,59)
b = dt.datetime(2013,12,31,23,59,59)

(b-a).total_seconds()

86400.0

#note that seconds doesn't give you what you want:
(b-a).seconds

0

于 2013-06-19T12:35:26.703 回答
42 
import time  
current = time.time()

...job...
end = time.time()
diff = end - current

那对你有用吗？

于 2010-12-06T01:34:18.917 回答
16 
>>> from datetime import datetime

>>>  a = datetime.now()

# wait a bit 
>>> b = datetime.now()

>>> d = b - a # yields a timedelta object
>>> d.seconds
7

（7 将是您在上面等待的时间）

我发现 datetime.datetime 相当有用，所以如果您遇到复杂或尴尬的情况，请告诉我们。

编辑：感谢@WoLpH 指出，人们不一定总是希望如此频繁地刷新，以至于日期时间会很接近。通过考虑增量中的天数，您可以处理更长的时间戳差异：

>>> a = datetime(2010, 12, 5)
>>> b = datetime(2010, 12, 7)
>>> d = b - a
>>> d.seconds
0
>>> d.days
2
>>> d.seconds + d.days * 86400
172800
于 2010-12-06T01:38:12.967 回答
12

我们有 Python 2.7 的函数 total_seconds() 请参见下面的 python 2.6 代码

import datetime
import time  

def diffdates(d1, d2):
    #Date format: %Y-%m-%d %H:%M:%S
    return (time.mktime(time.strptime(d2,"%Y-%m-%d %H:%M:%S")) -
               time.mktime(time.strptime(d1, "%Y-%m-%d %H:%M:%S")))

d1 = datetime.now()
d2 = datetime.now() + timedelta(days=1)
diff = diffdates(d1, d2)
于 2013-09-11T13:45:20.970 回答
2

这是为我工作的那个。

from datetime import datetime

date_format = "%H:%M:%S"

# You could also pass datetime.time object in this part and convert it to string.
time_start = str('09:00:00') 
time_end = str('18:00:00')

# Then get the difference here.    
diff = datetime.strptime(time_end, date_format) - datetime.strptime(time_start, date_format)

# Get the time in hours i.e. 9.60, 8.5
result = diff.seconds / 3600;

希望这可以帮助！

于 2019-07-07T08:18:08.120 回答
0

另一种方法是使用时间戳值：

end_time.timestamp() - start_time.timestamp()
于 2021-01-15T15:54:22.167 回答
0

通过阅读源代码，我得出一个结论：无法通过以下方式获得时差.seconds：

@property
def seconds(self):
    """seconds"""
    return self._seconds

# in the `__new__`, you can find the `seconds` is modulo by the total number of seconds in a day
def __new__(cls, days=0, seconds=0, microseconds=0,
            milliseconds=0, minutes=0, hours=0, weeks=0):
    seconds += minutes*60 + hours*3600
    # ...
    if isinstance(microseconds, float):
        microseconds = round(microseconds + usdouble)
        seconds, microseconds = divmod(microseconds, 1000000)
        # ! 
        days, seconds = divmod(seconds, 24*3600)
        d += days
        s += seconds
    else:
        microseconds = int(microseconds)
        seconds, microseconds = divmod(microseconds, 1000000)
        # ! 
        days, seconds = divmod(seconds, 24*3600)
        d += days
        s += seconds
        microseconds = round(microseconds + usdouble)
    # ...

total_seconds 可以得到两个时间之间的准确差异

def total_seconds(self):
    """Total seconds in the duration."""
    return ((self.days * 86400 + self.seconds) * 10**6 +
        self.microseconds) / 10**6

综上所述：

from datetime import datetime
dt1 = datetime.now()
dt2 = datetime.now()

print((dt2 - dt1).total_seconds())
于 2021-07-06T09:47:52.373 回答