Python格式timedelta到字符串

>>> str(datetime.timedelta(hours=10.56))
10:33:36

>>> td = datetime.timedelta(hours=10.505) # any timedelta object
>>> ':'.join(str(td).split(':')[:2])
10:30

将timedelta对象传递给str()函数会调用与我们只需键入时相同的格式代码print td.由于您不需要秒数,我们可以用冒号(3个部分)拆分字符串,然后将它与仅前两个部分放回原处.

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)

他已经有了一个timedelta对象,为什么不使用它的内置方法total_seconds()将其转换为秒,然后使用divmod()来获取小时和分钟？

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)

无论时间增量是否有几天或几年,这都有效.

我个人使用这个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,它不会给你确切地你要找的人(这是确实的答案str(timeA - timeB),但我发现,一旦你超越了几个小时,显示变得不可读很快.humanize对那些更大价值的支持人类可读,也很好地本地化.

它的灵感来自Django的contrib.humanize模块,显然,因为你使用Django,你应该使用它.

这是一个通用函数,用于将timedelta对象或常规数字(以秒或分钟等形式)转换为格式良好的字符串.我在一个重复的问题上采用了mpounsett的精彩答案,使其更加灵活,提高了可读性并增加了文档.

你会发现这是迄今为止最灵活的答案,因为它允许你:

即时自定义字符串格式,而不是硬编码.

留出一定的时间间隔没有问题(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

我知道这是一个陈旧的问题,但我用datetime.utcfromtimestamp()它.它需要秒数并返回一个datetime可以像任何其他格式一样的格式datetime.

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

只要您保持时间部分的合法范围,这应该有效,即它不会返回1234:35,因为小时<= 23.

发问者想要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,真的有两种格式,一种是天数为0而它是遗漏的,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零,所以列很乱.

如果您喜欢,这是我的常规:

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

这将输出返回为dd:hh:mm:ss格式:

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

我确实考虑过多年才能做到这一点,但这是留给读者的练习,因为输出安全超过1年:

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

我的datetime.timedelta物品超过了一天.所以这是另一个问题.上面的所有讨论都假设不到一天.A timedelta实际上是天,秒和微秒的元组.上面的讨论应该td.seconds像乔一样使用,但是如果你有几天它不包括在秒值中.

我在2个日期时间和打印日期和时间之间得到了一段时间.

span = currentdt - previousdt
print '%d,%d\n' % (span.days,span.seconds/3600)

我会认真考虑奥卡姆的剃刀方法:

td = str(timedelta).split('.')[0]

这将返回不带微秒的字符串

如果要重新生成datetime.timedelta对象,只需执行以下操作:

h,m,s = re.split(':', td)
new_delta = datetime.timedelta(hours=int(h),minutes=int(m),seconds=int(s))

2年了,我喜欢这门语言!

按照Joe上面的示例值,我将使用模数算术运算符,因此:

td = datetime.timedelta(hours=10.56)
td_str = "%d:%d" % (td.seconds/3600, td.seconds%3600/60)

请注意,默认情况下,Python中的整数除法向下舍入; 如果您想更明确,请根据需要使用math.floor()或math.ceil().

我使用humanfriendlypython库来做到这一点，它工作得很好。

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

'5 minutes and 21 seconds'

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