Question

这个问题类似to this question about subtracting dates with Python，但不完全相同。我不是在处理字符串，我必须弄清楚两个纪元时间戳之间的区别，并以人类可读的格式产生差异。

例如：

32 Seconds
17 Minutes
22.3 Hours
1.25 Days
3.5 Weeks
2 Months
4.25 Years

或者，我想表达这样的差异：

4 years, 6 months, 3 weeks, 4 days, 6 hours 21 minutes and 15 seconds

我认为我不能使用strptime，因为我正在使用两个纪元日期的差异。我可以写一些东西来做这件事，但我很确定已经写了一些我可以使用的东西。

什么模块适合？我只是遗漏了time中的内容吗？我的Python之旅刚刚开始，如果这确实是重复的，那是因为我没有找到要搜索的内容。

附录

为了准确，我真的非常关心当年的日历。

Answer 1

您可以使用精彩的dateutil模块及其relativedelta类：

import datetime
import dateutil.relativedelta

dt1 = datetime.datetime.fromtimestamp(123456789) # 1973-11-29 22:33:09
dt2 = datetime.datetime.fromtimestamp(234567890) # 1977-06-07 23:44:50
rd = dateutil.relativedelta.relativedelta (dt2, dt1)

print "%d years, %d months, %d days, %d hours, %d minutes and %d seconds" % (rd.years, rd.months, rd.days, rd.hours, rd.minutes, rd.seconds)
# 3 years, 6 months, 9 days, 1 hours, 11 minutes and 41 seconds

它不计算数周，但这不应该太难添加。

Answer 2

与@ Schnouki解决方案相比，单线列表理解略有改进。在多个实体（如小时）的情况下也显示复数

导入relativedelta

>>> from dateutil.relativedelta import relativedelta

lambda函数

>>> attrs = ['years', 'months', 'days', 'hours', 'minutes', 'seconds']
>>> human_readable = lambda delta: ['%d %s' % (getattr(delta, attr), getattr(delta, attr) > 1 and attr or attr[:-1]) 
...     for attr in attrs if getattr(delta, attr)]

使用示例：

>>> human_readable(relativedelta(minutes=125))
['2 hours', '5 minutes']
>>> human_readable(relativedelta(hours=(24 * 365) + 1))
['365 days', '1 hour']

Answer 3

我今天早些时候遇到了同样的问题，我在标准库中找不到任何可以使用的东西，所以这就是我写的：

humanize_time.py

    #!/usr/bin/env python

    INTERVALS = [1, 60, 3600, 86400, 604800, 2419200, 29030400]
    NAMES = [('second', 'seconds'),
             ('minute', 'minutes'),
             ('hour', 'hours'),
             ('day', 'days'),
             ('week', 'weeks'),
             ('month', 'months'),
             ('year', 'years')]

    def humanize_time(amount, units):
    """
    Divide `amount` in time periods.
    Useful for making time intervals more human readable.

    >>> humanize_time(173, 'hours')
    [(1, 'week'), (5, 'hours')]
    >>> humanize_time(17313, 'seconds')
    [(4, 'hours'), (48, 'minutes'), (33, 'seconds')]
    >>> humanize_time(90, 'weeks')
    [(1, 'year'), (10, 'months'), (2, 'weeks')]
    >>> humanize_time(42, 'months')
    [(3, 'years'), (6, 'months')]
    >>> humanize_time(500, 'days')
    [(1, 'year'), (5, 'months'), (3, 'weeks'), (3, 'days')]
    """
       result = []

       unit = map(lambda a: a[1], NAMES).index(units)
       # Convert to seconds
       amount = amount * INTERVALS[unit]

       for i in range(len(NAMES)-1, -1, -1):
          a = amount / INTERVALS[i]
          if a > 0:
             result.append( (a, NAMES[i][1 % a]) )
             amount -= a * INTERVALS[i]

       return result

    if __name__ == "__main__":
        import doctest
        doctest.testmod()

您可以使用dateutil.relativedelta()计算准确的时间增量并使用此脚本将其人性化。

Answer 4

def humanize_time(amount, units = 'seconds'):    

    def process_time(amount, units):

        INTERVALS = [   1, 60, 
                        60*60, 
                        60*60*24, 
                        60*60*24*7, 
                        60*60*24*7*4, 
                        60*60*24*7*4*12, 
                        60*60*24*7*4*12*100,
                        60*60*24*7*4*12*100*10]
        NAMES = [('second', 'seconds'),
                 ('minute', 'minutes'),
                 ('hour', 'hours'),
                 ('day', 'days'),
                 ('week', 'weeks'),
                 ('month', 'months'),
                 ('year', 'years'),
                 ('century', 'centuries'),
                 ('millennium', 'millennia')]

        result = []

        unit = map(lambda a: a[1], NAMES).index(units)
        # Convert to seconds
        amount = amount * INTERVALS[unit]

        for i in range(len(NAMES)-1, -1, -1):
            a = amount // INTERVALS[i]
            if a > 0: 
                result.append( (a, NAMES[i][1 % a]) )
                amount -= a * INTERVALS[i]

        return result

    rd = process_time(int(amount), units)
    cont = 0
    for u in rd:
        if u[0] > 0:
            cont += 1

    buf = ''
    i = 0
    for u in rd:
        if u[0] > 0:
            buf += "%d %s" % (u[0], u[1])
            cont -= 1

        if i < (len(rd)-1):
            if cont > 1:
                buf += ", "
            else:
                buf += " and "

        i += 1

    return buf

使用示例：

>>> print humanize_time(234567890 - 123456789)
3 years, 9 months, 3 weeks, 5 days, 11 minutes and 41 seconds
>>> humanize_time(9, 'weeks')
2 months and 1 week

优势（您不需要第三方！）。

从“Liudmil Mitev”算法改进。（谢谢！）

Answer 5

老问题，但我个人最喜欢这种方法：

import datetime
import math

def human_time(*args, **kwargs):
    secs  = float(datetime.timedelta(*args, **kwargs).total_seconds())
    units = [("day", 86400), ("hour", 3600), ("minute", 60), ("second", 1)]
    parts = []
    for unit, mul in units:
        if secs / mul >= 1 or mul == 1:
            if mul > 1:
                n = int(math.floor(secs / mul))
                secs -= n * mul
            else:
                n = secs if secs != int(secs) else int(secs)
            parts.append("%s %s%s" % (n, unit, "" if n == 1 else "s"))
    return ", ".join(parts)

human_time(seconds=3721)
# -> "1 hour, 2 minutes, 1 second"

如果您想将秒部分与“和”分开：

"%s and %s" % tuple(human_time(seconds=3721).rsplit(", ", 1))
# -> "1 hour, 2 minutes and 1 second"

Answer 6

查看人性化软件包

https://github.com/jmoiron/humanize

import datetime

humanize.naturaltime(datetime.datetime.now() - datetime.timedelta(seconds=1))

'a second ago'

humanize.naturaltime(datetime.datetime.now() - datetime.timedelta(seconds=3600))

'an hour ago'

Answer 7

一个很老的问题，但我发现这个解决方案在 Python3 中似乎很简单：

print(datetime.timedelta(seconds=3600))
# output: 1:00:00
print(datetime.timedelta(hours=360.1245))
# output: 15 days, 0:07:28.200000

Answer 8

这里是较短的，间隔为几秒，一天之内（t <86400）。如果您使用的是unix时间戳（自大纪元以来的秒数，UTC），则很有用。

t = 45678
print('%d hours, %d minutes, %d seconds' % (t//3600, t%3600//60, t%60))

可能会进一步扩展（t // 86400，...）。

在使用Python减去两个UNIX时间戳时，如何产生人类可读的差异？

附录

8 个答案: