我在格式化日期时间时遇到了麻烦。timedelta对象。

这就是我想做的: 我有一个对象列表,对象类的成员之一是显示事件持续时间的timedelta对象。我想以小时:分钟的格式显示这个持续时间。

我尝试了各种方法来做这件事,但我有困难。我目前的方法是为返回小时和分钟的对象在类中添加方法。我可以通过除以time得到小时数。秒乘以3600,四舍五入。我在得到剩余的秒并将其转换为分钟时遇到了麻烦。

顺便说一下,我使用谷歌AppEngine和Django模板来表示。


当前回答

可能:

>>> import datetime
>>> dt0 = datetime.datetime(1,1,1)
>>> td = datetime.timedelta(minutes=34, hours=12, seconds=56)
>>> (dt0+td).strftime('%X')
'12:34:56'
>>> (dt0+td).strftime('%M:%S')
'34:56'
>>> (dt0+td).strftime('%H:%M')
'12:34'
>>>

其他回答

def td_format(td_object):
    seconds = int(td_object.total_seconds())
    periods = [
        ('year',        60*60*24*365),
        ('month',       60*60*24*30),
        ('day',         60*60*24),
        ('hour',        60*60),
        ('minute',      60),
        ('second',      1)
    ]

    strings=[]
    for period_name, period_seconds in periods:
        if seconds > period_seconds:
            period_value , seconds = divmod(seconds, period_seconds)
            has_s = 's' if period_value > 1 else ''
            strings.append("%s %s%s" % (period_value, period_name, has_s))

    return ", ".join(strings)

谢谢大家的帮助。我把你的很多想法放在一起,让我知道你的想法。

我像这样在类中添加了两个方法:

def hours(self):
    retval = ""
    if self.totalTime:
        hoursfloat = self.totalTime.seconds / 3600
        retval = round(hoursfloat)
    return retval

def minutes(self):
    retval = ""
    if self.totalTime:
        minutesfloat = self.totalTime.seconds / 60
        hoursAsMinutes = self.hours() * 60
        retval = round(minutesfloat - hoursAsMinutes)
    return retval

在我的django中,我使用了这个(sum是对象,它在一个字典中):

<td>{{ sum.0 }}</td>    
<td>{{ sum.1.hours|stringformat:"d" }}:{{ sum.1.minutes|stringformat:"#02.0d" }}</td>

下面是一个通用函数,用于将timedelta对象或常规数字(以秒或分钟等形式)转换为格式化良好的字符串。我采用了mpounsett对一个重复问题的出色回答,使其更加灵活,提高了可读性,并添加了文档。

你会发现这是迄今为止最灵活的答案,因为它允许你:

动态自定义字符串格式,而不是硬编码。 省略特定的时间间隔没有问题(参见下面的例子)。

功能:

from string import Formatter
from datetime import timedelta

def strfdelta(tdelta, fmt='{D:02}d {H:02}h {M:02}m {S:02}s', inputtype='timedelta'):
    """Convert a datetime.timedelta object or a regular number to a custom-
    formatted string, just like the stftime() method does for datetime.datetime
    objects.

    The fmt argument allows custom formatting to be specified.  Fields can 
    include seconds, minutes, hours, days, and weeks.  Each field is optional.

    Some examples:
        '{D:02}d {H:02}h {M:02}m {S:02}s' --> '05d 08h 04m 02s' (default)
        '{W}w {D}d {H}:{M:02}:{S:02}'     --> '4w 5d 8:04:02'
        '{D:2}d {H:2}:{M:02}:{S:02}'      --> ' 5d  8:04:02'
        '{H}h {S}s'                       --> '72h 800s'

    The inputtype argument allows tdelta to be a regular number instead of the  
    default, which is a datetime.timedelta object.  Valid inputtype strings: 
        's', 'seconds', 
        'm', 'minutes', 
        'h', 'hours', 
        'd', 'days', 
        'w', 'weeks'
    """

    # Convert tdelta to integer seconds.
    if inputtype == 'timedelta':
        remainder = int(tdelta.total_seconds())
    elif inputtype in ['s', 'seconds']:
        remainder = int(tdelta)
    elif inputtype in ['m', 'minutes']:
        remainder = int(tdelta)*60
    elif inputtype in ['h', 'hours']:
        remainder = int(tdelta)*3600
    elif inputtype in ['d', 'days']:
        remainder = int(tdelta)*86400
    elif inputtype in ['w', 'weeks']:
        remainder = int(tdelta)*604800

    f = Formatter()
    desired_fields = [field_tuple[1] for field_tuple in f.parse(fmt)]
    possible_fields = ('W', 'D', 'H', 'M', 'S')
    constants = {'W': 604800, 'D': 86400, 'H': 3600, 'M': 60, 'S': 1}
    values = {}
    for field in possible_fields:
        if field in desired_fields and field in constants:
            values[field], remainder = divmod(remainder, constants[field])
    return f.format(fmt, **values)

演示:

>>> td = timedelta(days=2, hours=3, minutes=5, seconds=8, microseconds=340)

>>> print strfdelta(td)
02d 03h 05m 08s

>>> print strfdelta(td, '{D}d {H}:{M:02}:{S:02}')
2d 3:05:08

>>> print strfdelta(td, '{D:2}d {H:2}:{M:02}:{S:02}')
 2d  3:05:08

>>> print strfdelta(td, '{H}h {S}s')
51h 308s

>>> print strfdelta(12304, inputtype='s')
00d 03h 25m 04s

>>> print strfdelta(620, '{H}:{M:02}', 'm')
10:20

>>> print strfdelta(49, '{D}d {H}h', 'h')
2d 1h
from django.utils.translation import ngettext

def localize_timedelta(delta):
    ret = []
    num_years = int(delta.days / 365)
    if num_years > 0:
        delta -= timedelta(days=num_years * 365)
        ret.append(ngettext('%d year', '%d years', num_years) % num_years)

    if delta.days > 0:
        ret.append(ngettext('%d day', '%d days', delta.days) % delta.days)

    num_hours = int(delta.seconds / 3600)
    if num_hours > 0:
        delta -= timedelta(hours=num_hours)
        ret.append(ngettext('%d hour', '%d hours', num_hours) % num_hours)

    num_minutes = int(delta.seconds / 60)
    if num_minutes > 0:
        ret.append(ngettext('%d minute', '%d minutes', num_minutes) % num_minutes)

    return ' '.join(ret)

这将产生:

>>> from datetime import timedelta
>>> localize_timedelta(timedelta(days=3660, minutes=500))
'10 years 10 days 8 hours 20 minutes'

他已经有一个timedelta对象,所以为什么不使用其内置方法total_seconds()将其转换为秒,然后使用divmod()获得小时和分钟?

hours, remainder = divmod(myTimeDelta.total_seconds(), 3600)
minutes, seconds = divmod(remainder, 60)

# Formatted only for hours and minutes as requested
print '%s:%s' % (hours, minutes)

无论时间增量是天还是年,这都是有效的。