问题：将时间量转换为天，小时和分钟
我有一个timedelta。我想要从中得到的天，小时和分钟-作为元组或字典…我不大惊小怪。
多年来，我肯定已经用十几种语言完成了十二次，但是Python通常对所有问题都有一个简单的答案，所以我想我会在这里提出一些令人讨厌的简单（但冗长）的数学之前先问一下。
福兹先生提出了一个很好的观点。
我正在处理“列表”（有点像ebay列表），每个列表都有持续时间。我试图通过做找到剩余的时间when_added + duration - now
我是不是说DST不正确？如果不是，加/减一小时的最简单方法是什么？
I’ve got a timedelta. I want the days, hours and minutes from that – either as a tuple or a dictionary… I’m not fussed.
I must have done this a dozen times in a dozen languages over the years but Python usually has a simple answer to everything so I thought I’d ask here before busting out some nauseatingly simple (yet verbose) mathematics.
Mr Fooz raises a good point.
I’m dealing with “listings” (a bit like ebay listings) where each one has a duration. I’m trying to find the time left by doing when_added + duration - now
Am I right in saying that wouldn’t account for DST? If not, what’s the simplest way to add/subtract an hour?
回答 0
如果您有一个datetime.timedelta值td，那么td.days已经为您提供了所需的“天数”。timedelta值将天的分数保持为秒（而不是小时或分钟），因此您确实必须执行“令人作呕的简单数学”，例如：
def days_hours_minutes(td):
return td.days, td.seconds//3600, (td.seconds//60)%60
If you have a datetime.timedelta value td, td.days already gives you the “days” you want. timedelta values keep fraction-of-day as seconds (not directly hours or minutes) so you’ll indeed have to perform “nauseatingly simple mathematics”, e.g.:
def days_hours_minutes(td):
return td.days, td.seconds//3600, (td.seconds//60)%60
回答 1
这更加紧凑，您可以在两行中获得小时，分钟和秒。
days = td.days
hours, remainder = divmod(td.seconds, 3600)
minutes, seconds = divmod(remainder, 60)
# If you want to take into account fractions of a second
seconds += td.microseconds / 1e6
This is a bit more compact, you get the hours, minutes and seconds in two lines.
days = td.days
hours, remainder = divmod(td.seconds, 3600)
minutes, seconds = divmod(remainder, 60)
# If you want to take into account fractions of a second
seconds += td.microseconds / 1e6
回答 2
days, hours, minutes = td.days, td.seconds // 3600, td.seconds // 60 % 60
对于DST，我认为最好的方法是将两个datetime对象都转换为秒。这样，系统将为您计算DST。
>>> m13 = datetime(2010, 3, 13, 8, 0, 0) # 2010 March 13 8:00 AM
>>> m14 = datetime(2010, 3, 14, 8, 0, 0) # DST starts on this day, in my time zone
>>> mktime(m14.timetuple()) - mktime(m13.timetuple()) # difference in seconds
82800.0
>>> _/3600 # convert to hours
23.0
days, hours, minutes = td.days, td.seconds // 3600, td.seconds // 60 % 60
As for DST, I think the best thing is to convert both datetime objects to seconds. This way the system calculates DST for you.
>>> m13 = datetime(2010, 3, 13, 8, 0, 0) # 2010 March 13 8:00 AM
>>> m14 = datetime(2010, 3, 14, 8, 0, 0) # DST starts on this day, in my time zone
>>> mktime(m14.timetuple()) - mktime(m13.timetuple()) # difference in seconds
82800.0
>>> _/3600 # convert to hours
23.0
回答 3
我不明白
days, hours, minutes = td.days, td.seconds // 3600, td.seconds // 60 % 60
这个怎么样
days, hours, minutes = td.days, td.seconds // 3600, td.seconds % 3600 / 60.0
您会得到几分钟和一分钟的秒数浮动。
I don’t understand
days, hours, minutes = td.days, td.seconds // 3600, td.seconds // 60 % 60
how about this
days, hours, minutes = td.days, td.seconds // 3600, td.seconds % 3600 / 60.0
You get minutes and seconds of a minute as a float.
回答 4
我使用了以下内容：
delta = timedelta()
totalMinute, second = divmod(delta.seconds, 60)
hour, minute = divmod(totalMinute, 60)
print(f"{hour}h{minute:02}m{second:02}s")
I used the following:
delta = timedelta()
totalMinute, second = divmod(delta.seconds, 60)
hour, minute = divmod(totalMinute, 60)
print(f"{hour}h{minute:02}m{second:02}s")
回答 5
timedeltas具有daysand seconds属性..您可以轻松地自己转换它们。
timedeltas have a days and seconds attribute .. you can convert them yourself with ease.
