Python：我需要在“1天前”，“两小时前”格式中显示文件修改时间。
有什么准备好的吗？它应该是英文。
答案 0 :(得分：107)
该代码最初发布在博客文章“Python Pretty Date function”（http://evaisse.com/post/93417709/python-pretty-date-function）
上此处转载此博客帐户已被暂停且页面不再可用。
def pretty_date(time=False):
"""
Get a datetime object or a int() Epoch timestamp and return a
pretty string like 'an hour ago', 'Yesterday', '3 months ago',
'just now', etc
"""
from datetime import datetime
now = datetime.now()
if type(time) is int:
diff = now - datetime.fromtimestamp(time)
elif isinstance(time,datetime):
diff = now - time
elif not time:
diff = now - now
second_diff = diff.seconds
day_diff = diff.days
if day_diff < 0:
return ''
if day_diff == 0:
if second_diff < 10:
return "just now"
if second_diff < 60:
return str(second_diff) + " seconds ago"
if second_diff < 120:
return "a minute ago"
if second_diff < 3600:
return str(second_diff / 60) + " minutes ago"
if second_diff < 7200:
return "an hour ago"
if second_diff < 86400:
return str(second_diff / 3600) + " hours ago"
if day_diff == 1:
return "Yesterday"
if day_diff < 7:
return str(day_diff) + " days ago"
if day_diff < 31:
return str(day_diff / 7) + " weeks ago"
if day_diff < 365:
return str(day_diff / 30) + " months ago"
return str(day_diff / 365) + " years ago"
答案 1 :(得分：29)
如果您正在使用Django，则版本1.4中的新功能是naturaltime模板过滤器。
要使用它，请先将'django.contrib.humanize'添加到settings.py中的INSTALLED_APPS设置，然后将{% load humanize %}添加到您正在使用过滤器的模板中。
然后，在您的模板中，如果您有一个日期时间变量my_date，则可以使用{{ my_date|naturaltime }}打印其与当前距离的距离，该4 minutes ago将呈现为类似{{1}}的内容。
Other new things in Django 1.4.
Documentation for naturaltime and other filters in the django.contrib.humanize set.
答案 2 :(得分：14)
在寻找处理未来日期的额外要求时，我发现了这一点： http://pypi.python.org/pypi/py-pretty/1
示例代码（来自网站）：
from datetime import datetime, timedelta
now = datetime.now()
hrago = now - timedelta(hours=1)
yesterday = now - timedelta(days=1)
tomorrow = now + timedelta(days=1)
dayafter = now + timedelta(days=2)
import pretty
print pretty.date(now) # 'now'
print pretty.date(hrago) # 'an hour ago'
print pretty.date(hrago, short=True) # '1h ago'
print pretty.date(hrago, asdays=True) # 'today'
print pretty.date(yesterday, short=True) # 'yest'
print pretty.date(tomorrow) # 'tomorrow'
答案 3 :(得分：5)
杰德·史密斯所说的答案很好，我用了一年左右，但我认为可以通过以下几种方式改进：
以下是我提出的建议：
def PrettyRelativeTime(time_diff_secs):
# Each tuple in the sequence gives the name of a unit, and the number of
# previous units which go into it.
weeks_per_month = 365.242 / 12 / 7
intervals = [('minute', 60), ('hour', 60), ('day', 24), ('week', 7),
('month', weeks_per_month), ('year', 12)]
unit, number = 'second', abs(time_diff_secs)
for new_unit, ratio in intervals:
new_number = float(number) / ratio
# If the new number is too small, don't go to the next unit.
if new_number < 2:
break
unit, number = new_unit, new_number
shown_num = int(number)
return '{} {}'.format(shown_num, unit + ('' if shown_num == 1 else 's'))
注意intervals中的每个元组如何易于解释和检查：'minute'是60秒; 'hour'为60分钟;唯一的软糖就是将weeks_per_month设置为其平均值;鉴于申请，这应该没问题。 （请注意，一目了然，最后三个常数乘以365.242，即每年的天数。）
我的功能的一个缺点是它不会在“## units”模式之外做任何事情：“昨天”，“刚才”，等等。然后，原始的海报并没有要求这些花哨的术语，所以我更喜欢我的功能，因为它的简洁性和数值常数的可读性。 ：）
答案 4 :(得分：5)
您也可以使用arrow包
执行此操作来自github page：
>>> import arrow >>> utc = arrow.utcnow() >>> utc = utc.replace(hours=-1) >>> local.humanize() 'an hour ago'
答案 5 :(得分：4)
ago包提供了此功能。在human对象上调用datetime以获得差异的可读描述。
from ago import human
from datetime import datetime
from datetime import timedelta
ts = datetime.now() - timedelta(days=1, hours=5)
print(human(ts))
# 1 day, 5 hours ago
print(human(ts, precision=1))
# 1 day ago
答案 6 :(得分：3)
>>> from datetime import datetime, timedelta
>>> import humanize # $ pip install humanize
>>> humanize.naturaltime(datetime.now() - timedelta(days=1))
'a day ago'
>>> humanize.naturaltime(datetime.now() - timedelta(hours=2))
'2 hours ago'
>>> _ = humanize.i18n.activate('ru_RU')
>>> print humanize.naturaltime(datetime.now() - timedelta(days=1))
день назад
>>> print humanize.naturaltime(datetime.now() - timedelta(hours=2))
2 часа назад
答案 7 :(得分：1)
我在http://sunilarora.org/17329071上为解决方案写了一篇详细的博文 我也在这里发布一个快速片段。
from datetime import datetime
from dateutil.relativedelta import relativedelta
def get_fancy_time(d, display_full_version = False):
"""Returns a user friendly date format
d: some datetime instace in the past
display_second_unit: True/False
"""
#some helpers lambda's
plural = lambda x: 's' if x > 1 else ''
singular = lambda x: x[:-1]
#convert pluran (years) --> to singular (year)
display_unit = lambda unit, name: '%s %s%s'%(unit, name, plural(unit)) if unit > 0 else ''
#time units we are interested in descending order of significance
tm_units = ['years', 'months', 'days', 'hours', 'minutes', 'seconds']
rdelta = relativedelta(datetime.utcnow(), d) #capture the date difference
for idx, tm_unit in enumerate(tm_units):
first_unit_val = getattr(rdelta, tm_unit)
if first_unit_val > 0:
primary_unit = display_unit(first_unit_val, singular(tm_unit))
if display_full_version and idx < len(tm_units)-1:
next_unit = tm_units[idx + 1]
second_unit_val = getattr(rdelta, next_unit)
if second_unit_val > 0:
secondary_unit = display_unit(second_unit_val, singular(next_unit))
return primary_unit + ', ' + secondary_unit
return primary_unit
return None
答案 8 :(得分：1)
将datetime对象与tzinfo一起使用：
def time_elapsed(etime):
# need to add tzinfo to datetime.utcnow
now = datetime.datetime.utcnow().replace(tzinfo=etime.tzinfo)
opened_for = (now - etime).total_seconds()
names = ["seconds","minutes","hours","days","weeks","months"]
modulos = [ 1,60,3600,3600*24,3600*24*7,3660*24*30]
values = []
for m in modulos[::-1]:
values.append(int(opened_for / m))
opened_for -= values[-1]*m
pretty = []
for i,nm in enumerate(names[::-1]):
if values[i]!=0:
pretty.append("%i %s" % (values[i],nm))
return " ".join(pretty)
答案 9 :(得分：0)
这是@sunil的帖子
的要点>>> from datetime import datetime
>>> from dateutil.relativedelta import relativedelta
>>> then = datetime(2003, 9, 17, 20, 54, 47, 282310)
>>> relativedelta(then, datetime.now())
relativedelta(years=-11, months=-3, days=-9, hours=-18, minutes=-17, seconds=-8, microseconds=+912664)
答案 10 :(得分：0)
您可以从以下链接下载并安装。它应该对你更有帮助。它一直提供用户友好的消息，从第二年到第二年。
经过充分测试。
https://github.com/nareshchaudhary37/timestamp_content
以下步骤安装到虚拟环境中
git clone https://github.com/nareshchaudhary37/timestamp_content
cd timestamp-content
python setup.py
答案 11 :(得分：0)
这是根据杰德·史密斯（Jed Smith）的实现方式更新的答案，该实现方式正确地处理了天真偏移时间和可感知偏移的日期时间。您还可以提供默认时区。 Python 3.5以上版本。
import datetime
def pretty_date(time=None, default_timezone=datetime.timezone.utc):
"""
Get a datetime object or a int() Epoch timestamp and return a
pretty string like 'an hour ago', 'Yesterday', '3 months ago',
'just now', etc
"""
# Assumes all timezone naive dates are UTC
if time.tzinfo is None or time.tzinfo.utcoffset(time) is None:
if default_timezone:
time = time.replace(tzinfo=default_timezone)
now = datetime.datetime.utcnow().replace(tzinfo=datetime.timezone.utc)
if type(time) is int:
diff = now - datetime.fromtimestamp(time)
elif isinstance(time, datetime.datetime):
diff = now - time
elif not time:
diff = now - now
second_diff = diff.seconds
day_diff = diff.days
if day_diff < 0:
return ''
if day_diff == 0:
if second_diff < 10:
return "just now"
if second_diff < 60:
return str(second_diff) + " seconds ago"
if second_diff < 120:
return "a minute ago"
if second_diff < 3600:
return str(second_diff / 60) + " minutes ago"
if second_diff < 7200:
return "an hour ago"
if second_diff < 86400:
return str(second_diff / 3600) + " hours ago"
if day_diff == 1:
return "Yesterday"
if day_diff < 7:
return str(day_diff) + " days ago"
if day_diff < 31:
return str(day_diff / 7) + " weeks ago"
if day_diff < 365:
return str(day_diff / 30) + " months ago"
return str(day_diff / 365) + " years ago"
答案 12 :(得分：0)
很长一段时间以来，我一直在将此代码从编程语言拖放到编程语言中，我不记得我最初从哪里得到的代码。在PHP，Java和TypeScript中，它为我提供了很好的服务，现在是时候使用Python。
它可以处理过去和将来的日期，以及边缘情况。
def unix_time() -> int:
return int(time.time())
def pretty_time(t: int, absolute=False) -> str:
if not type(t) is int:
return "N/A"
if t == 0:
return "Never"
now = unix_time()
if t == now:
return "Now"
periods = ["second", "minute", "hour", "day", "week", "month", "year", "decade"]
lengths = [60, 60, 24, 7, 4.35, 12, 10]
diff = now - t
if absolute:
suffix = ""
else:
if diff >= 0:
suffix = "ago"
else:
diff *= -1
suffix = "remaining"
i = 0
while diff >= lengths[i] and i < len(lengths) - 1:
diff /= lengths[i]
i += 1
diff = round(diff)
if diff > 1:
periods[i] += "s"
return "{0} {1} {2}".format(diff, periods[i], suffix)
答案 13 :(得分：0)
DAY_INCREMENTS = [
[365, "year"],
[30, "month"],
[7, "week"],
[1, "day"],
]
SECOND_INCREMENTS = [
[3600, "hour"],
[60, "minute"],
[1, "second"],
]
def time_ago(dt):
diff = datetime.now() - dt # use timezone.now() or equivalent if `dt` is timezone aware
if diff.days < 0:
return "in the future?!?"
for increment, label in DAY_INCREMENTS:
if diff.days >= increment:
increment_diff = int(diff.days / increment)
return str(increment_diff) + " " + label + plural(increment_diff) + " ago"
for increment, label in SECOND_INCREMENTS:
if diff.seconds >= increment:
increment_diff = int(diff.seconds / increment)
return str(increment_diff) + " " + label + plural(increment_diff) + " ago"
return "just now"
def plural(num):
if num != 1:
return "s"
return ""