Python函数将秒转换为分钟、小时和天
问题: 编写一个程序，要求用户输入秒数，工作如下：Python函数将秒转换为分钟、小时和天,python,Python,问题: 编写一个程序，要求用户输入秒数，工作如下： 一分钟有60秒。如果用户输入的秒数大于或等于60，程序应显示该秒数中的分钟数 一小时有3600秒。如果用户输入的秒数大于或等于3600，程序应显示该秒数中的小时数 一天有86400秒。如果用户输入的秒数大于或等于86400，程序应显示该秒数内的天数 到目前为止，我所拥有的： def time(): sec = int( input ('Enter the number of seconds:'.strip()) if sec
- 一分钟有60秒。如果用户输入的秒数大于或等于60，程序应显示该秒数中的分钟数
- 一小时有3600秒。如果用户输入的秒数大于或等于3600，程序应显示该秒数中的小时数
- 一天有86400秒。如果用户输入的秒数大于或等于86400，程序应显示该秒数内的天数
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
def time（）：
sec=int（输入（'输入秒数：'.strip（））
如果秒则按照需要减去秒的另一种方法进行，并且不要称之为时间；有一个包具有该名称：
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
这将n秒转换为d天、h小时、m分钟和s秒
首先，我认为DIVMOD会更快，因为它是一个单独的语句和一个内置函数，但是时间看起来似乎不一样。考虑一下这个小例子，当我试图找出最快的方法，它在一个循环中使用，它在GOBASIDIDLE中连续运行，增加一个秒计数器到一个人类读数。更新进度条标签所需的时间
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相比，内置的divmod函数似乎慢得令人难以置信
division and modulus: 12.5737669468
divmod function: 17.2861430645
此技巧对于显示不同粒度的运行时间非常有用
我个人认为，效率问题在这里实际上是毫无意义的，只要不去做一些非常低效的事情。过早的优化是相当多邪恶的根源。这足够快，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])
…这提供了良好的输出：
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'
这些函数相当紧凑，只使用标准的Python2.6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))
要将秒（作为字符串）转换为datetime，这也会有所帮助。您可以获得天数和秒数。秒可以进一步转换为分钟和小时
from datetime import datetime, timedelta
sec = timedelta(seconds=(int(input('Enter the number of seconds: '))))
time = str(sec)
我不完全确定您是否需要它，但我有一个类似的任务，需要删除一个字段，如果它为零。例如，86401秒将显示“1天，1秒”，而不是“1天，0小时，0分钟，1秒”。下面的代码就是这样做的
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，处理单词的复数化
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：创建了一个可以用多种语言实现这一功能的补丁（抱歉，没有足够的代表发表评论），我们可以根据时间量返回可变粒度。例如，我们不说“1周，5秒”，我们只说“1周”：
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
上面的“timeit”答案宣称divmod速度较慢，这有严重的逻辑缺陷
Test1调用操作符
Test2调用函数divmod，
调用函数会产生开销
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速度更快
def convertSeconds(seconds):
h = seconds//(60*60)
m = (seconds-h*60*60)//60
s = seconds-(h*60*60)-(m*60)
return [h, m, s]
函数输入是秒数，返回的是秒数所代表的小时、分钟和秒的列表。虽然提到了divmod（），但我没有看到我认为是一个很好的示例。下面是我的示例：
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])
哪些产出：
11 days, 6 hours, 0 minutes, 21 seconds
输出：
输入秒数：2434234232
结果：
28174天，0小时，10分钟，32秒，同时进行修补。移动到类并将tulp of tulp（间隔）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（默认情况下禁用）。这是从模块加载的。这适用于python3（测试3.6-3.8）
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)
下面是课堂：
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))
粒度=1-5，四舍五入=真/假，转换=真/假
一些测试显示差异：
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
Full - 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
秒：563440
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
秒：604780
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
秒：2419180
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
秒：113478160
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
我已经准备好了法语翻译。但是翻译速度很快…只有几个字。
希望这能对我有所帮助，因为另一个答案对我很有帮助。提示：让你这样做：divmod（36603600）#（1,60）
和divmod（60,60）#（1,0）
。另外，你到底在问什么？在纸上写下你会怎么做，然后把它转换成代码。根据你的描述，你的“如果”语句应该是“>=”，而不是“的可能副本不能完全满足要求。我相信有最大秒数。因此这不是建议的方法。请参阅文档：0到86399之间的秒数，包括0到86399之间的秒数，一天有86400秒，这就是为什么只能有0-86399秒。换句话说，timedelta（秒=86400）将解析为天=1，秒=0。因此，86399不是秒的最大输入值。似乎我误解了文档，如果你这么说的话。那么你的意思是，如果我使用以下timedelta:timedelta（秒=86450），它将自动解析为timedelta（天=1，秒=50）？正确吗？谢谢你的评论。如果持续时间太长，以致于它改变了月份，那么此建议将不起作用。尝试5184000（60*24*3600）秒。使用3.0样式的格式字符串，可以使用：t=timedelta（seconds=long（valu））；打印（“保存时间：{}m-{}d{}（h:mm:ss）”。格式（t.days/30，t.days%30，timedelta（seconds=t.seconds））
仅仅发布代码作为答案是不好的。即使你认为这可能是不言而喻的，你也应该包括一些解释。在我的Python 2.6.6系统上，我不得不使用result.append（“%s%s”%”（值，名称））
我得到一个类型错误，以修复它：sec=timedelta（seconds=（int（输入（'Enter number of seconds:'））)
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))
myformater = FormatTimestamp()
myconverter = myformater.convert(seconds)
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()
Full - 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