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。

我有一个函数:

def period(delta, pattern):
    d = {'d': delta.days}
    d['h'], rem = divmod(delta.seconds, 3600)
    d['m'], d['s'] = divmod(rem, 60)
    return pattern.format(**d)

例子:

>>> td = timedelta(seconds=123456789)
>>> period(td, "{d} days {h}:{m}:{s}")
'1428 days 21:33:9'
>>> period(td, "{h} hours, {m} minutes and {s} seconds, {d} days")
'21 hours, 33 minutes and 9 seconds, 1428 days'

提问者想要一个比典型的更好的格式:

  >>> import datetime
  >>> datetime.timedelta(seconds=41000)
  datetime.timedelta(0, 41000)
  >>> str(datetime.timedelta(seconds=41000))
  '11:23:20'
  >>> str(datetime.timedelta(seconds=4102.33))
  '1:08:22.330000'
  >>> str(datetime.timedelta(seconds=413302.33))
  '4 days, 18:48:22.330000'

所以，实际上有两种格式，一种是天数为0，它被省略了，另一种是文本“n天，h:m:s”。但是，秒可能有分数，打印输出中没有前导0，所以列很乱。

如果你喜欢，下面是我的日常工作:

def printNiceTimeDelta(stime, etime):
    delay = datetime.timedelta(seconds=(etime - stime))
    if (delay.days > 0):
        out = str(delay).replace(" days, ", ":")
    else:
        out = "0:" + str(delay)
    outAr = out.split(':')
    outAr = ["%02d" % (int(float(x))) for x in outAr]
    out   = ":".join(outAr)
    return out

返回dd:hh:mm:ss格式的输出:

00:00:00:15
00:00:00:19
02:01:31:40
02:01:32:22

我确实想过在上面加上年份，但这是留给读者的练习，因为超过1年的输出是安全的:

>>> str(datetime.timedelta(seconds=99999999))
'1157 days, 9:46:39'

def seconds_to_time_left_string(total_seconds):
    s = int(total_seconds)
    years = s // 31104000
    if years > 1:
        return '%d years' % years
    s = s - (years * 31104000)
    months = s // 2592000
    if years == 1:
        r = 'one year'
        if months > 0:
            r += ' and %d months' % months
        return r
    if months > 1:
        return '%d months' % months
    s = s - (months * 2592000)
    days = s // 86400
    if months == 1:
        r = 'one month'
        if days > 0:
            r += ' and %d days' % days
        return r
    if days > 1:
        return '%d days' % days
    s = s - (days * 86400)
    hours = s // 3600
    if days == 1:
        r = 'one day'
        if hours > 0:
            r += ' and %d hours' % hours
        return r 
    s = s - (hours * 3600)
    minutes = s // 60
    seconds = s - (minutes * 60)
    if hours >= 6:
        return '%d hours' % hours
    if hours >= 1:
        r = '%d hours' % hours
        if hours == 1:
            r = 'one hour'
        if minutes > 0:
            r += ' and %d minutes' % minutes
        return r
    if minutes == 1:
        r = 'one minute'
        if seconds > 0:
            r += ' and %d seconds' % seconds
        return r
    if minutes == 0:
        return '%d seconds' % seconds
    if seconds == 0:
        return '%d minutes' % minutes
    return '%d minutes and %d seconds' % (minutes, seconds)

for i in range(10):
    print pow(8, i), seconds_to_time_left_string(pow(8, i))


Output:
1 1 seconds
8 8 seconds
64 one minute and 4 seconds
512 8 minutes and 32 seconds
4096 one hour and 8 minutes
32768 9 hours
262144 3 days
2097152 24 days
16777216 6 months
134217728 4 years

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>

我想这样做，所以写了一个简单的函数。它对我来说非常有用，而且非常通用(支持年到微秒，以及任何粒度级别，例如，你可以选择“2天4小时48分钟”和“2天4小时”和“2天4.8小时”等。

def pretty_print_timedelta(t, max_components=None, max_decimal_places=2):
''' 
Print a pretty string for a timedelta. 
For example datetime.timedelta(days=2, seconds=17280) will be printed as '2 days, 4 hours, 48 minutes'. Setting max_components to e.g. 1 will change this to '2.2 days', where the 
number of decimal points can also be set. 
'''
time_scales = [timedelta(days=365), timedelta(days=1), timedelta(hours=1), timedelta(minutes=1), timedelta(seconds=1), timedelta(microseconds=1000), timedelta(microseconds=1)]
time_scale_names_dict = {timedelta(days=365): 'year',  
                         timedelta(days=1): 'day', 
                         timedelta(hours=1): 'hour', 
                         timedelta(minutes=1): 'minute', 
                         timedelta(seconds=1): 'second', 
                         timedelta(microseconds=1000): 'millisecond', 
                         timedelta(microseconds=1): 'microsecond'}
count = 0
txt = ''
first = True
for scale in time_scales:
    if t >= scale: 
        count += 1
        if count == max_components:
            n = t / scale
        else:
            n = int(t / scale)
            
        t -= n*scale
        
        n_txt = str(round(n, max_decimal_places))
        if n_txt[-2:]=='.0': n_txt = n_txt[:-2]
        txt += '{}{} {}{}'.format('' if first else ', ', n_txt, time_scale_names_dict[scale], 's' if n>1 else '', )
        if first:
            first = False
        
        
if len(txt) == 0: 
    txt = 'none'
return txt