我个人使用humanize库:

>>> import datetime
>>> humanize.naturalday(datetime.datetime.now())
'today'
>>> humanize.naturalday(datetime.datetime.now() - datetime.timedelta(days=1))
'yesterday'
>>> humanize.naturalday(datetime.date(2007, 6, 5))
'Jun 05'
>>> humanize.naturaldate(datetime.date(2007, 6, 5))
'Jun 05 2007'
>>> humanize.naturaltime(datetime.datetime.now() - datetime.timedelta(seconds=1))
'a second ago'
>>> humanize.naturaltime(datetime.datetime.now() - datetime.timedelta(seconds=3600))
'an hour ago'

当然，它并没有给您所寻找的确切答案(实际上是str(timeA - timeB)，但我发现，一旦超过几个小时，显示很快就变得不可读了。Humanize支持更大的值，这些值是人类可读的，而且本地化得很好。

它的灵感来自于Django的贡献。humanize模块，显然，既然你在使用Django，你应该使用它。

如果你已经有一个timedelta obj，那么只需将obj转换为字符串。删除字符串的最后3个字符并打印。这将截断秒部分，并以小时:分钟的格式打印其余部分。

t = str(timedeltaobj) 

print t[:-3]

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

请检查这个函数-它将timedelta对象转换为字符串'HH:MM:SS'

def format_timedelta(td):
    hours, remainder = divmod(td.total_seconds(), 3600)
    minutes, seconds = divmod(remainder, 60)
    hours, minutes, seconds = int(hours), int(minutes), int(seconds)
    if hours < 10:
        hours = '0%s' % int(hours)
    if minutes < 10:
        minutes = '0%s' % minutes
    if seconds < 10:
        seconds = '0%s' % seconds
    return '%s:%s:%s' % (hours, minutes, seconds)

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'

针对这个问题的一个直接的模板过滤器。内置函数int()从不四舍五入。f - string(即f'')需要python 3.6。

@app_template_filter()
def diffTime(end, start):
    diff = (end - start).total_seconds()
    d = int(diff / 86400)
    h = int((diff - (d * 86400)) / 3600)
    m = int((diff - (d * 86400 + h * 3600)) / 60)
    s = int((diff - (d * 86400 + h * 3600 + m *60)))
    if d > 0:
        fdiff = f'{d}d {h}h {m}m {s}s'
    elif h > 0:
        fdiff = f'{h}h {m}m {s}s'
    elif m > 0:
        fdiff = f'{m}m {s}s'
    else:
        fdiff = f'{s}s'
    return fdiff