问题： 编写一个程序，要求用户输入几秒钟，并按如下方式工作：
一分钟内有60秒。如果用户输入的秒数大于或等于60，则程序应显示该秒数的分钟数。
一小时内有3600秒。如果用户输入的秒数大于或等于3600，则程序应显示该秒数的小时数。
一天有86400秒。如果用户输入的秒数大于或等于86400，则程序应显示该秒数的天数。
到目前为止我所拥有的：
def time():
sec = int( input ('Enter the number of seconds:'.strip())
if sec <= 60:
minutes = sec // 60
print('The number of minutes is {0:.2f}'.format(minutes))
if sec (<= 3600):
hours = sec // 3600
print('The number of minutes is {0:.2f}'.format(hours))
if sec <= 86400:
days = sec // 86400
print('The number of minutes is {0:.2f}'.format(days))
return
答案 0 :(得分：43)
这会将 n 秒转换为 d 天， h 小时， m 分钟，以及 s 秒。
from datetime import datetime, timedelta
def GetTime():
sec = timedelta(seconds=int(input('Enter the number of seconds: ')))
d = datetime(1,1,1) + sec
print("DAYS:HOURS:MIN:SEC")
print("%d:%d:%d:%d" % (d.day-1, d.hour, d.minute, d.second))
答案 1 :(得分：43)
此花絮可用于显示不同粒度的经过时间。
我个人认为效率问题在这里几乎毫无意义，只要没有完成效率低下的事情。过早的优化是相当多的邪恶的根源。这足够快，它永远不会成为你的瓶颈。
intervals = (
('weeks', 604800), # 60 * 60 * 24 * 7
('days', 86400), # 60 * 60 * 24
('hours', 3600), # 60 * 60
('minutes', 60),
('seconds', 1),
)
def display_time(seconds, granularity=2):
result = []
for name, count in intervals:
value = seconds // count
if value:
seconds -= value * count
if value == 1:
name = name.rstrip('s')
result.append("{} {}".format(value, name))
return ', '.join(result[:granularity])
..这提供了不错的输出：
In [52]: display_time(1934815)
Out[52]: '3 weeks, 1 day'
In [53]: display_time(1934815, 4)
Out[53]: '3 weeks, 1 day, 9 hours, 26 minutes'
答案 2 :(得分：7)
要将秒（作为字符串）转换为日期时间，这也可能有所帮助。你得到天数和秒数。秒数可以进一步转换为分钟和小时。
from datetime import datetime, timedelta
sec = timedelta(seconds=(input('Enter the number of seconds: ')))
time = str(sec)
答案 3 :(得分：4)
我不完全确定你是否想要它，但我有一个类似的任务，如果它是零，则需要删除一个字段。例如，86401秒将显示“1天，1秒”而不是“1天，0小时，0分钟，1秒”。以下代码就是这样做的。
def secondsToText(secs):
days = secs//86400
hours = (secs - days*86400)//3600
minutes = (secs - days*86400 - hours*3600)//60
seconds = secs - days*86400 - hours*3600 - minutes*60
result = ("{} days, ".format(days) if days else "") + \
("{} hours, ".format(hours) if hours else "") + \
("{} minutes, ".format(minutes) if minutes else "") + \
("{} seconds, ".format(seconds) if seconds else "")
return result
编辑：一个稍微好一点的版本，用于处理单词的多元化。
def secondsToText(secs):
days = secs//86400
hours = (secs - days*86400)//3600
minutes = (secs - days*86400 - hours*3600)//60
seconds = secs - days*86400 - hours*3600 - minutes*60
result = ("{0} day{1}, ".format(days, "s" if days!=1 else "") if days else "") + \
("{0} hour{1}, ".format(hours, "s" if hours!=1 else "") if hours else "") + \
("{0} minute{1}, ".format(minutes, "s" if minutes!=1 else "") if minutes else "") + \
("{0} second{1}, ".format(seconds, "s" if seconds!=1 else "") if seconds else "")
return result
EDIT2：创建了gist以多种语言创建
答案 4 :(得分：4)
seconds_in_day = 86400
seconds_in_hour = 3600
seconds_in_minute = 60
seconds = int(input("Enter a number of seconds: "))
days = seconds // seconds_in_day
seconds = seconds - (days * seconds_in_day)
hours = seconds // seconds_in_hour
seconds = seconds - (hours * seconds_in_hour)
minutes = seconds // seconds_in_minute
seconds = seconds - (minutes * seconds_in_minute)
print("{0:.0f} days, {1:.0f} hours, {2:.0f} minutes, {3:.0f} seconds.".format(
days, hours, minutes, seconds))
答案 5 :(得分：2)
#1 min = 60
#1 hour = 60 * 60 = 3600
#1 day = 60 * 60 * 24 = 86400
x=input('enter a positive integer: ')
t=int(x)
day= t//86400
hour= (t-(day*86400))//3600
minit= (t - ((day*86400) + (hour*3600)))//60
seconds= t - ((day*86400) + (hour*3600) + (minit*60))
print( day, 'days' , hour,' hours', minit, 'minutes',seconds,' seconds')
答案 6 :(得分：2)
def convertSeconds(seconds):
h = seconds//(60*60)
m = (seconds-h*60*60)//60
s = seconds-(h*60*60)-(m*60)
return [h, m, s]
函数输入的秒数，返回的是小时，分钟和秒的列表，以秒表示。
答案 7 :(得分：1)
这些函数相当紧凑，只使用标准的Python 2.6及更高版本。
def ddhhmmss(seconds):
"""Convert seconds to a time string "[[[DD:]HH:]MM:]SS".
"""
dhms = ''
for scale in 86400, 3600, 60:
result, seconds = divmod(seconds, scale)
if dhms != '' or result > 0:
dhms += '{0:02d}:'.format(result)
dhms += '{0:02d}'.format(seconds)
return dhms
def seconds(dhms):
"""Convert a time string "[[[DD:]HH:]MM:]SS" to seconds.
"""
components = [int(i) for i in dhms.split(':')]
pad = 4 - len(components)
if pad < 0:
raise ValueError('Too many components to match [[[DD:]HH:]MM:]SS')
components = [0] * pad + components
return sum(i * j for i, j in zip((86400, 3600, 60, 1), components))
以下是与它们一起进行的测试。我使用pytest包作为测试异常的简单方法。
import ddhhmmss
import pytest
def test_ddhhmmss():
assert ddhhmmss.ddhhmmss(0) == '00'
assert ddhhmmss.ddhhmmss(2) == '02'
assert ddhhmmss.ddhhmmss(12 * 60) == '12:00'
assert ddhhmmss.ddhhmmss(3600) == '01:00:00'
assert ddhhmmss.ddhhmmss(10 * 86400) == '10:00:00:00'
assert ddhhmmss.ddhhmmss(86400 + 5 * 3600 + 30 * 60 + 1) == '01:05:30:01'
assert ddhhmmss.ddhhmmss(365 * 86400) == '365:00:00:00'
def test_seconds():
assert ddhhmmss.seconds('00') == 0
assert ddhhmmss.seconds('02') == 2
assert ddhhmmss.seconds('12:00') == 12 * 60
assert ddhhmmss.seconds('01:00:00') == 3600
assert ddhhmmss.seconds('1:0:0') == 3600
assert ddhhmmss.seconds('3600') == 3600
assert ddhhmmss.seconds('60:0') == 3600
assert ddhhmmss.seconds('10:00:00:00') == 10 * 86400
assert ddhhmmss.seconds('1:05:30:01') == 86400 + 5 * 3600 + 30 * 60 + 1
assert ddhhmmss.seconds('365:00:00:00') == 365 * 86400
def test_seconds_raises():
with pytest.raises(ValueError):
ddhhmmss.seconds('')
with pytest.raises(ValueError):
ddhhmmss.seconds('foo')
with pytest.raises(ValueError):
ddhhmmss.seconds('1:00:00:00:00')
答案 8 :(得分：1)
乍一看，我认为divmod会更快，因为它是单个语句和内置函数，但timeit似乎表明不是这样。考虑一下我想出的一个小例子，当我试图在一个循环中使用最快的方法时，该循环在gobject idle_add中连续运行，将秒计数器拆分为人类可读时间以更新进度条标签。
import timeit
def test1(x,y, dropy):
while x > 0:
y -= dropy
x -= 1
# the test
minutes = (y-x) / 60
seconds = (y-x) % 60.0
def test2(x,y, dropy):
while x > 0:
y -= dropy
x -= 1
# the test
minutes, seconds = divmod((y-x), 60)
x = 55 # litte number, also number of tests
y = 10000 # make y > x by factor of drop
dropy = 7 # y is reduced this much each iteration, for variation
print "division and modulus:", timeit.timeit( lambda: test1(x,y,dropy) )
print "divmod function:", timeit.timeit( lambda: test2(x,y,dropy) )
与使用简单的除法和模数相比，内置的divmod函数似乎要慢得多。
division and modulus: 12.5737669468
divmod function: 17.2861430645
答案 9 :(得分：1)
以相反的方式做，根据需要减去秒数，不要叫它时间;有一个包含该名称的包裹：
def sec_to_time():
sec = int( input ('Enter the number of seconds:'.strip()) )
days = sec / 86400
sec -= 86400*days
hrs = sec / 3600
sec -= 3600*hrs
mins = sec / 60
sec -= 60*mins
print days, ':', hrs, ':', mins, ':', sec
答案 10 :(得分：1)
time_in_sec = 65 #I took time as 65 sec for example
day = divmod(time_in_sec, 86_400)
hour = divmod(day[1], 3_600)
min = divmod(hour[1], 60)
sec = min[1]
print(f"Day {day[0]} |Hour {hour[0]} |Min {min[0]} |Sec {sec}")
说明：
divmod(x, y) 函数返回 (x//y, x%y)
在 day 变量中，x//y 是天数，x%y 是剩余秒数。
因此对于 hour 变量，day 变量的其余部分是输入。同样，您可以使用这个简单的程序最多计算一周。
答案 11 :(得分：0)
修补Mr.B's answer（抱歉，没有足够的代表评论），我们可以根据时间量返回变量粒度。例如，我们不会说＆＃34; 1周，5秒＆＃34;我们只是说＆＃34; 1周＆＃34;：
def display_time(seconds, granularity=2):
result = []
for name, count in intervals:
value = seconds // count
if value:
seconds -= value * count
if value == 1:
name = name.rstrip('s')
result.append("{} {}".format(value, name))
else:
# Add a blank if we're in the middle of other values
if len(result) > 0:
result.append(None)
return ', '.join([x for x in result[:granularity] if x is not None])
一些示例输入：
for diff in [5, 67, 3600, 3605, 3667, 24*60*60, 24*60*60+5, 24*60*60+57, 24*60*60+3600, 24*60*60+3667, 2*24*60*60, 2*24*60*60+5*60*60, 7*24*60*60, 7*24*60*60 + 24*60*60]:
print "For %d seconds: %s" % (diff, display_time(diff, 2))
...返回此输出：
For 5 seconds: 5 seconds
For 67 seconds: 1 minute, 7 seconds
For 3600 seconds: 1 hour
For 3605 seconds: 1 hour
For 3667 seconds: 1 hour, 1 minute
For 86400 seconds: 1 day
For 86405 seconds: 1 day
For 86457 seconds: 1 day
For 90000 seconds: 1 day, 1 hour
For 90067 seconds: 1 day, 1 hour
For 172800 seconds: 2 days
For 190800 seconds: 2 days, 5 hours
For 604800 seconds: 1 week
For 691200 seconds: 1 week, 1 day
答案 12 :(得分：0)
上面的宣布divmod变慢的“ timeit”答案在逻辑上存在严重缺陷。
Test1调用运算符。
Test2调用函数divmod， 并调用函数会产生开销。
一种更准确的测试方法是：
import timeit
def moddiv(a,b):
q= a/b
r= a%b
return q,r
a=10
b=3
md=0
dm=0
for i in range(1,10):
c=a*i
md+= timeit.timeit( lambda: moddiv(c,b))
dm+=timeit.timeit( lambda: divmod(c,b))
print("moddiv ", md)
print("divmod ", dm)
moddiv 5.806157339000492
divmod 4.322451676005585
divmod更快。
答案 13 :(得分：0)
尽管已经提到了divmod（），但我没有看到我认为是一个很好的示例。这是我的：
q=972021.0000 # For example
days = divmod(q, 86400)
# days[0] = whole days and
# days[1] = seconds remaining after those days
hours = divmod(days[1], 3600)
minutes = divmod(hours[1], 60)
print "%i days, %i hours, %i minutes, %i seconds" % (days[0], hours[0], minutes[0], minutes[1])
哪个输出：
11 days, 6 hours, 0 minutes, 21 seconds
答案 14 :(得分：0)
def seconds_to_dhms(time):
seconds_to_minute = 60
seconds_to_hour = 60 * seconds_to_minute
seconds_to_day = 24 * seconds_to_hour
days = time // seconds_to_day
time %= seconds_to_day
hours = time // seconds_to_hour
time %= seconds_to_hour
minutes = time // seconds_to_minute
time %= seconds_to_minute
seconds = time
print("%d days, %d hours, %d minutes, %d seconds" % (days, hours, minutes, seconds))
time = int(input("Enter the number of seconds: "))
seconds_to_dhms(time)
输出： 输入秒数：2434234232
结果： 28174天0小时10分钟32秒
答案 15 :(得分：0)
def normalize_seconds(seconds: int) -> tuple:
(days, remainder) = divmod(seconds, 86400)
(hours, remainder) = divmod(remainder, 3600)
(minutes, seconds) = divmod(remainder, 60)
return namedtuple("_", ("days", "hours", "minutes", "seconds"))(days, hours, minutes, seconds)
答案 16 :(得分：0)
也正在修补Ralph Bolton's answer。移至一类并将tulp（间隔）的tulp移动到字典。根据粒度添加可选的舍入函数（默认情况下启用）。准备使用gettext进行翻译（默认为禁用）。打算从模块中加载。这是针对python3（测试3.6-3.8）
import gettext
import locale
from itertools import chain
mylocale = locale.getdefaultlocale()
# see --> https://stackoverflow.com/a/10174657/11869956 thx
#localedir = os.path.join(os.path.dirname(__file__), 'locales')
# or python > 3.4:
try:
localedir = pathlib.Path(__file__).parent/'locales'
lang_translations = gettext.translation('utils', localedir,
languages=[mylocale[0]])
lang_translations.install()
_ = lang_translations.gettext
except Exception as exc:
print('Error: unexcept error while initializing translation:', file=sys.stderr)
print(f'Error: {exc}', file=sys.stderr)
print(f'Error: localedir={localedir}, languages={mylocale[0]}', file=sys.stderr)
print('Error: translation has been disabled.', file=sys.stderr)
_ = gettext.gettext
这是课程：
class FormatTimestamp:
"""Convert seconds to, optional rounded, time depending of granularity's degrees.
inspired by https://stackoverflow.com/a/24542445/11869956"""
def __init__(self):
# For now i haven't found a way to do it better
# TODO: optimize ?!? ;)
self.intervals = {
# 'years' : 31556952, # https://www.calculateme.com/time/years/to-seconds/
# https://www.calculateme.com/time/months/to-seconds/ -> 2629746 seconds
# But it's outputing some strange result :
# So 3 seconds less (2629743) : 4 weeks, 2 days, 10 hours, 29 minutes and 3 seconds
# than after 3 more seconds : 1 month ?!?
# Google give me 2628000 seconds
# So 3 seconds less (2627997): 4 weeks, 2 days, 9 hours, 59 minutes and 57 seconds
# Strange as well
# So for the moment latest is week ...
#'months' : 2419200, # 60 * 60 * 24 * 7 * 4
'weeks' : 604800, # 60 * 60 * 24 * 7
'days' : 86400, # 60 * 60 * 24
'hours' : 3600, # 60 * 60
'minutes' : 60,
'seconds' : 1
}
self.nextkey = {
'seconds' : 'minutes',
'minutes' : 'hours',
'hours' : 'days',
'days' : 'weeks',
'weeks' : 'weeks',
#'months' : 'months',
#'years' : 'years' # stop here
}
self.translate = {
'weeks' : _('weeks'),
'days' : _('days'),
'hours' : _('hours'),
'minutes' : _('minutes'),
'seconds' : _('seconds'),
## Single
'week' : _('week'),
'day' : _('day'),
'hour' : _('hour'),
'minute' : _('minute'),
'second' : _('second'),
' and' : _('and'),
',' : _(','), # This is for compatibility
'' : '\0' # same here BUT we CANNOT pass empty string to gettext
# or we get : warning: Empty msgid. It is reserved by GNU gettext:
# gettext("") returns the header entry with
# meta information, not the empty string.
# Thx to --> https://stackoverflow.com/a/30852705/11869956 - saved my day
}
def convert(self, seconds, granularity=2, rounded=True, translate=False):
"""Proceed the conversion"""
def _format(result):
"""Return the formatted result
TODO : numpy / google docstrings"""
start = 1
length = len(result)
none = 0
next_item = False
for item in reversed(result[:]):
if item['value']:
# if we have more than one item
if length - none > 1:
# This is the first 'real' item
if start == 1:
item['punctuation'] = ''
next_item = True
elif next_item:
# This is the second 'real' item
# Happened 'and' to key name
item['punctuation'] = ' and'
next_item = False
# If there is more than two 'real' item
# than happened ','
elif 2 < start:
item['punctuation'] = ','
else:
item['punctuation'] = ''
else:
item['punctuation'] = ''
start += 1
else:
none += 1
return [ { 'value' : mydict['value'],
'name' : mydict['name_strip'],
'punctuation' : mydict['punctuation'] } for mydict in result \
if mydict['value'] is not None ]
def _rstrip(value, name):
"""Rstrip 's' name depending of value"""
if value == 1:
name = name.rstrip('s')
return name
# Make sure granularity is an integer
if not isinstance(granularity, int):
raise ValueError(f'Granularity should be an integer: {granularity}')
# For seconds only don't need to compute
if seconds < 0:
return 'any time now.'
elif seconds < 60:
return 'less than a minute.'
result = []
for name, count in self.intervals.items():
value = seconds // count
if value:
seconds -= value * count
name_strip = _rstrip(value, name)
# save as dict: value, name_strip (eventually strip), name (for reference), value in seconds
# and count (for reference)
result.append({
'value' : value,
'name_strip' : name_strip,
'name' : name,
'seconds' : value * count,
'count' : count
})
else:
if len(result) > 0:
# We strip the name as second == 0
name_strip = name.rstrip('s')
# adding None to key 'value' but keep other value
# in case when need to add seconds when we will
# recompute every thing
result.append({
'value' : None,
'name_strip' : name_strip,
'name' : name,
'seconds' : 0,
'count' : count
})
# Get the length of the list
length = len(result)
# Don't need to compute everything / every time
if length < granularity or not rounded:
if translate:
return ' '.join('{0} {1}{2}'.format(item['value'], _(self.translate[item['name']]),
_(self.translate[item['punctuation']])) \
for item in _format(result))
else:
return ' '.join('{0} {1}{2}'.format(item['value'], item['name'], item['punctuation']) \
for item in _format(result))
start = length - 1
# Reverse list so the firsts elements
# could be not selected depending on granularity.
# And we can delete item after we had his seconds to next
# item in the current list (result)
for item in reversed(result[:]):
if granularity <= start <= length - 1:
# So we have to round
current_index = result.index(item)
next_index = current_index - 1
# skip item value == None
# if the seconds of current item is superior
# to the half seconds of the next item: round
if item['value'] and item['seconds'] > result[next_index]['count'] // 2:
# +1 to the next item (in seconds: depending on item count)
result[next_index]['seconds'] += result[next_index]['count']
# Remove item which is not selected
del result[current_index]
start -= 1
# Ok now recalculate everything
# Reverse as well
for item in reversed(result[:]):
# Check if seconds is superior or equal to the next item
# but not from 'result' list but from 'self.intervals' dict
# Make sure it's not None
if item['value']:
next_item_name = self.nextkey[item['name']]
# This mean we are at weeks
if item['name'] == next_item_name:
# Just recalcul
item['value'] = item['seconds'] // item['count']
item['name_strip'] = _rstrip(item['value'], item['name'])
# Stop to weeks to stay 'right'
elif item['seconds'] >= self.intervals[next_item_name]:
# First make sure we have the 'next item'
# found via --> https://stackoverflow.com/q/26447309/11869956
# maybe there is a faster way to do it ? - TODO
if any(search_item['name'] == next_item_name for search_item in result):
next_item_index = result.index(item) - 1
# Append to
result[next_item_index]['seconds'] += item['seconds']
# recalculate value
result[next_item_index]['value'] = result[next_item_index]['seconds'] // \
result[next_item_index]['count']
# strip or not
result[next_item_index]['name_strip'] = _rstrip(result[next_item_index]['value'],
result[next_item_index]['name'])
else:
# Creating
next_item_index = result.index(item) - 1
# get count
next_item_count = self.intervals[next_item_name]
# convert seconds
next_item_value = item['seconds'] // next_item_count
# strip 's' or not
next_item_name_strip = _rstrip(next_item_value, next_item_name)
# added to dict
next_item = {
'value' : next_item_value,
'name_strip' : next_item_name_strip,
'name' : next_item_name,
'seconds' : item['seconds'],
'count' : next_item_count
}
# insert to the list
result.insert(next_item_index, next_item)
# Remove current item
del result[result.index(item)]
else:
# for current item recalculate
# keys 'value' and 'name_strip'
item['value'] = item['seconds'] // item['count']
item['name_strip'] = _rstrip(item['value'], item['name'])
if translate:
return ' '.join('{0} {1}{2}'.format(item['value'],
_(self.translate[item['name']]),
_(self.translate[item['punctuation']])) \
for item in _format(result))
else:
return ' '.join('{0} {1}{2}'.format(item['value'], item['name'], item['punctuation']) \
for item in _format(result))
要使用它：
myformater = FormatTimestamp()
myconverter = myformater.convert(seconds)
粒度= 1-5，四舍五入=正确/错误，转换=正确/错误
一些测试来显示差异：
myformater = FormatTimestamp()
for firstrange in [131440, 563440, 604780, 2419180, 113478160]:
print(f'#### Seconds : {firstrange} ####')
print('\tFull - function: {0}'.format(display_time(firstrange, granularity=5)))
print('\tFull - class: {0}'.format(myformater.convert(firstrange, granularity=5)))
for secondrange in range(1, 6, 1):
print('\tGranularity this answer ({0}): {1}'.format(secondrange,
myformater.convert(firstrange,
granularity=secondrange, translate=False)))
print('\tGranularity Bolton\'s answer ({0}): {1}'.format(secondrange, display_time(firstrange,
granularity=secondrange)))
print()
秒：131440秒：563440Full - function: 1 day, 12 hours, 30 minutes, 40 seconds Full - class: 1 day, 12 hours, 30 minutes and 40 seconds Granularity this answer (1): 2 days Granularity Bolton's answer (1): 1 day Granularity this answer (2): 1 day and 13 hours Granularity Bolton's answer (2): 1 day, 12 hours Granularity this answer (3): 1 day, 12 hours and 31 minutes Granularity Bolton's answer (3): 1 day, 12 hours, 30 minutes Granularity this answer (4): 1 day, 12 hours, 30 minutes and 40 seconds Granularity Bolton's answer (4): 1 day, 12 hours, 30 minutes, 40 seconds Granularity this answer (5): 1 day, 12 hours, 30 minutes and 40 seconds Granularity Bolton's answer (5): 1 day, 12 hours, 30 minutes, 40 seconds
Full - function: 6 days, 12 hours, 30 minutes, 40 seconds
Full - class: 6 days, 12 hours, 30 minutes and 40 seconds
Granularity this answer (1): 1 week
Granularity Bolton's answer (1): 6 days
Granularity this answer (2): 6 days and 13 hours
Granularity Bolton's answer (2): 6 days, 12 hours
Granularity this answer (3): 6 days, 12 hours and 31 minutes
Granularity Bolton's answer (3): 6 days, 12 hours, 30 minutes
Granularity this answer (4): 6 days, 12 hours, 30 minutes and 40 seconds
Granularity Bolton's answer (4): 6 days, 12 hours, 30 minutes, 40 seconds
Granularity this answer (5): 6 days, 12 hours, 30 minutes and 40 seconds
Granularity Bolton's answer (5): 6 days, 12 hours, 30 minutes, 40 seconds
Full - function: 6 days, 23 hours, 59 minutes, 40 seconds
Full - class: 6 days, 23 hours, 59 minutes and 40 seconds
Granularity this answer (1): 1 week
Granularity Bolton's answer (1): 6 days
Granularity this answer (2): 1 week
Granularity Bolton's answer (2): 6 days, 23 hours
Granularity this answer (3): 1 week
Granularity Bolton's answer (3): 6 days, 23 hours, 59 minutes
Granularity this answer (4): 6 days, 23 hours, 59 minutes and 40 seconds
Granularity Bolton's answer (4): 6 days, 23 hours, 59 minutes, 40 seconds
Granularity this answer (5): 6 days, 23 hours, 59 minutes and 40 seconds
Granularity Bolton's answer (5): 6 days, 23 hours, 59 minutes, 40 seconds
Full - function: 3 weeks, 6 days, 23 hours, 59 minutes, 40 seconds
Full - class: 3 weeks, 6 days, 23 hours, 59 minutes and 40 seconds
Granularity this answer (1): 4 weeks
Granularity Bolton's answer (1): 3 weeks
Granularity this answer (2): 4 weeks
Granularity Bolton's answer (2): 3 weeks, 6 days
Granularity this answer (3): 4 weeks
Granularity Bolton's answer (3): 3 weeks, 6 days, 23 hours
Granularity this answer (4): 4 weeks
Granularity Bolton's answer (4): 3 weeks, 6 days, 23 hours, 59 minutes
Granularity this answer (5): 3 weeks, 6 days, 23 hours, 59 minutes and 40 seconds
Granularity Bolton's answer (5): 3 weeks, 6 days, 23 hours, 59 minutes, 40 seconds
Full - function: 187 weeks, 4 days, 9 hours, 42 minutes, 40 seconds
Full - class: 187 weeks, 4 days, 9 hours, 42 minutes and 40 seconds
Granularity this answer (1): 188 weeks
Granularity Bolton's answer (1): 187 weeks
Granularity this answer (2): 187 weeks and 4 days
Granularity Bolton's answer (2): 187 weeks, 4 days
Granularity this answer (3): 187 weeks, 4 days and 10 hours
Granularity Bolton's answer (3): 187 weeks, 4 days, 9 hours
Granularity this answer (4): 187 weeks, 4 days, 9 hours and 43 minutes
Granularity Bolton's answer (4): 187 weeks, 4 days, 9 hours, 42 minutes
Granularity this answer (5): 187 weeks, 4 days, 9 hours, 42 minutes and 40 seconds
Granularity Bolton's answer (5): 187 weeks, 4 days, 9 hours, 42 minutes, 40 seconds
我准备好法语翻译。但是翻译很快……只需几个单词。 希望这能对其他答案有帮助。