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

我接着MarredCheese的回答,加上了年、月、毫秒和微秒

除秒外,所有数字都被格式化为整数,因此秒的分数可以自定义。

@kfmfe04要求几分之一秒,所以我发布了这个解决方案

大体上有一些例子。

from string import Formatter
from datetime import timedelta

def strfdelta(tdelta, fmt='{D:02}d {H:02}h {M:02}m {S:02.0f}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:02.0f}s' --> '05d 08h 04m 02s' (default)
        '{W}w {D}d {H}:{M:02}:{S:02.0f}'     --> '4w 5d 8:04:02'
        '{D:2}d {H:2}:{M:02}:{S:02.0f}'      --> ' 5d  8:04:02'
        '{H}h {S:.0f}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 = tdelta.total_seconds()
    elif inputtype in ['s', 'seconds']:
        remainder = float(tdelta)
    elif inputtype in ['m', 'minutes']:
        remainder = float(tdelta)*60
    elif inputtype in ['h', 'hours']:
        remainder = float(tdelta)*3600
    elif inputtype in ['d', 'days']:
        remainder = float(tdelta)*86400
    elif inputtype in ['w', 'weeks']:
        remainder = float(tdelta)*604800

    f = Formatter()
    desired_fields = [field_tuple[1] for field_tuple in f.parse(fmt)]
    possible_fields = ('Y','m','W', 'D', 'H', 'M', 'S', 'mS', 'µS')
    constants = {'Y':86400*365.24,'m': 86400*30.44 ,'W': 604800, 'D': 86400, 'H': 3600, 'M': 60, 'S': 1, 'mS': 1/pow(10,3) , 'µS':1/pow(10,6)}
    values = {}
    for field in possible_fields:
        if field in desired_fields and field in constants:
            Quotient, remainder = divmod(remainder, constants[field])
            values[field] = int(Quotient) if field != 'S' else Quotient + remainder
    return f.format(fmt, **values)

if __name__ == "__main__":
    td = timedelta(days=717, hours=3, minutes=5, seconds=8, microseconds=3549)
    print(strfdelta(td,'{Y} years {m} months {W} weeks {D} days {H:02}:{M:02}:{S:02}'))  
    print(strfdelta(td,'{m} months {W} weeks {D} days {H:02}:{M:02}:{S:02.4f}'))  
    td = timedelta( seconds=8, microseconds=8549)
    print(strfdelta(td,'{S} seconds {mS} milliseconds {µS} microseconds'))  
    print(strfdelta(td,'{S:.0f} seconds {mS} milliseconds {µS} microseconds'))  
    print(strfdelta(pow(10,7),inputtype='s'))

输出:

1 years 11 months 2 weeks 3 days 01:09:56.00354900211096
23 months 2 weeks 3 days 00:12:20.0035
8.008549 seconds 8 milliseconds 549 microseconds
8 seconds 8 milliseconds 549 microseconds
115d 17h 46m 40s

其他回答

import datetime
hours = datetime.timedelta(hours=16, minutes=30)
print((datetime.datetime(1,1,1) + hours).strftime('%H:%M'))
# Format seconds to days, hours, minutes and seconds string
def ptime(seconds):
if(seconds >= 86400):
    d = seconds // 86400 # // floor division
    return (f"{round(d)}d") + ptime(seconds - d * 86400)
else:
    if(seconds >= 3600):
        h = seconds // 3600 
        return (f"{round(h)}h") + ptime(seconds - h * 3600)
    else:
        if(seconds >= 60):
            m = seconds // 60
            return(f"{round(m)}m" + ptime(seconds - m * 60))
        else:
            if (seconds > 0):
                return(f"{round(seconds)}s")
            else:
                return("")

我知道这是一个老问题,但我使用datetime.utcfromtimestamp()来解决这个问题。它接受秒数并返回一个datetime,该datetime可以像任何其他datetime一样格式化。

duration = datetime.utcfromtimestamp(end - begin)
print duration.strftime('%H:%M')

只要您停留在时间部分的合法范围内,这就应该工作,即它不会返回1234:35,因为小时<= 23。

我使用人类友好的python库来做这件事,它工作得非常好。

import humanfriendly
from datetime import timedelta
delta = timedelta(seconds = 321)
humanfriendly.format_timespan(delta)

'5 minutes and 21 seconds'

网址:https://pypi.org/project/humanfriendly/

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'