我正在尝试实现一个功能，以计算给定日期的下一个和第三个工作日（最好考虑到某些给定的假期）
def day_of_week(year, month, day):
t = [0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4]
year -= month < 3
return (year + int(year/4) - int(year/100) + int(year/400) + t[month-1] + day) % 7
输入的格式为YYYYMMDD，2018年3月21日输入为20180321，输出日期的格式应相同。
我正在尝试做这样的事情，但我意识到这不是最佳实践
def leap_year(year):
if(year%4==0 and year%100!=0):
return True
elif (year%4==0 and year%100==0 and year%400==0):
return True
else:
return False
def business_day(year, month, day):
if (month==12):
if(day_of_week(year, month, day)<5 and day_of_week(year, month, day)>0 and day==31):
return str(year+1)+str(0)+str(1)+str(0)+str(1)
elif (day_of_week(year, month, day)<5 and day_of_week(year, month, day)>0 and day!=31):
newDay="0"
if(day<10):
newDay = newDay + str(day+1)
else:
newDay = str(day+1)
return str(year) + str(month) + newDay
elif (day_of_week(year, month, day)>=5 and day==31):
if(day_of_week(year, month, day)==5):
return str(year+1)+"01"+"03"
if (day_of_week(year, month, day) == 6):
return str(year + 1) + "01" + "02"
if (day_of_week(year, month, day) == 0):
return str(year + 1) + "01" + "01"
elif (day_of_week(year, month, day)>=5 and day==30):
if((day_of_week(year, month, day)==5)):
return str(year + 1) + "01" + "02"
if ((day_of_week(year, month, day) == 6)):
return str(year + 1) + "01" + "01"
if ((day_of_week(year, month, day) == 0)):
return str(year + 1) + str(month) + str(day+1)
不能使用该解决方案中的任何库。感谢您的帮助
答案 0 :(得分：1)
import datetime
example = '20180321'
# you can parse the time string directly to a datetime object
next_buisness_day = datetime.datetime.strptime(example, '%Y%m%d')
# specify the increment based on the day of the week or any
#other condition
increment = 1
print('day day is', next_buisness_day.weekday())
# if friday
if next_buisness_day.weekday() == 4:
increment = 3
# if saturday
elif next_buisness_day.weekday() == 5:
increment = 2
next_buisness_day += datetime.timedelta(days=increment)
# and convert back to whatever format you like
print('{:%Y%m%d}'.format(next_buisness_day))
看看datetime模块，您可以使用它进行各种操作。 https://docs.python.org/3/library/datetime.html
答案 1 :(得分：1)
没有图书馆！我学习Python很有趣。你是否？ ：-）
def day_of_week(year, month, day):
t = [0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4]
year -= month < 3
dw = (year + year // 4 - year // 100 + year // 400 + t[month-1] + day) % 7
return [6, 0, 1, 2, 3, 4, 5][dw] # To be consistent with 'datetime' library
def leap_year(year):
leap = year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)
return True if leap else False
def valid_day(year, month, day):
month_list = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
if year < 1 or year > 9999 or month < 1 or month > 12:
return False
m = month_list[month-1] if month != 2 or not leap_year(year) else 29
return True if 1 <= day <= m else False
class Career(Exception):
def __str__(self): return 'So I became a waiter...'
MAX_DATE_AND_DAYS_INT = 365 * 100
class Date:
# raise ValueError
def __init__(self, year, month, day):
if not valid_day(year, month, day):
raise Career()
self.y, self.m, self.d = year, month, day
@classmethod
def fromstring(cls, s):
s1, s2, s3 = int(s[0:4]), int(s[4:6]), int(s[6:8])
return cls(s1, s2, s3)
def __repr__(self) -> str:
return '%04d%02d%02d' % (self.y, self.m, self.d)
def weekday_date(self) -> str:
names = ['Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat', 'Sun']
return names[self.weekday()] + ' ' + str(self)
def next_day(self):
if valid_day(self.y, self.m, self.d + 1):
return Date(self.y, self.m, self.d + 1)
elif valid_day(self.y, self.m + 1, 1):
return Date(self.y, self.m + 1, 1)
elif valid_day(self.y + 1, 1, 1):
return Date(self.y + 1, 1, 1)
else:
raise Career
def weekday(self):
return day_of_week(self.y, self.m, self.d)
def __add__(self, other):
"Add a Date to an int."
if isinstance(other, int):
if other < 1 or other > MAX_DATE_AND_DAYS_INT:
raise OverflowError("int > MAX_DATE_AND_DAYS_INT")
new_date = Date(self.y, self.m, self.d)
while other >= 1:
new_date = new_date.next_day()
other -= 1
return new_date
return NotImplemented
def next_working_day(self):
day = self.next_day()
while True:
while day.weekday() >= 5:
day = day.next_day()
holidays_list = year_holidays(day.y)
for str_day in holidays_list:
s2 = str(day)
if str_day == s2:
day = day.next_day()
break # for
if day.weekday() < 5:
break # while True
return day
def year_holidays(year):
holidays = [
["New Year's Day", 1, 1], # Fixed: January 1
["Birthday of Martin Luther King, Jr.", 1, 0, 0, 3], # Floating
["Washington's Birthday", 2, 0, 0, 3], # Third Monday in February
["Memorial Day", 5, 0, 0, 5], # Last Monday
["Independence Day", 7, 4],
["Labor Day", 9, 0, 0, 1],
["Columbus Day", 10, 0, 0, 2],
["Veterans Day", 11, 11],
["Thanksgiving Day", 11, 0, 3, 4],
["Christmas Day", 12, 25]
]
year_list = []
for h in holidays:
if h[2] > 0:
day = Date(year, h[1], h[2]) # Fixed day
else:
day = Date(year, h[1], 1) # Floating day
while h[3] != day.weekday(): # Advance to match the weekday
day = day.next_day()
count = 1
while count != h[4]: # Match the repetition of this day
next_week = day + 7
if next_week.m == day.m:
day = next_week
count += 1
year_list.append(str(day))
return year_list # return the holidays as list of strings
if __name__ == '__main__':
dates = [
['20190308', '20190311', '20190313'],
['20190309', '20190311', '20190313'],
['20190310', '20190311', '20190313'],
['20190311', '20190312', '20190314'],
['20190329', '20190401', '20190403'],
['20181231', '20190102', '20190104'],
['20190118', '20190122', '20190124'],
['20190216', '20190219', '20190221'],
['20190526', '20190528', '20190530'],
['20190703', '20190705', '20190709'],
['20190828', '20190829', '20190903'],
['20191010', '20191011', '20191016'],
['20191108', '20191112', '20191114'],
['20191125', '20191126', '20191129'],
['20191224', '20191226', '20191230'],
['20191227', '20191230', '20200102']]
print('\n Today Next and 3rd business day')
for days in dates:
today = Date.fromstring(days[0])
next_day = today.next_working_day()
third_day = next_day.next_working_day().next_working_day()
if str(next_day) != days[1] or str(third_day) != days[2]:
print('*** ERROR *** ', end='')
else:
print(' ', end='')
print(today.weekday_date(), next_day.weekday_date(), third_day.weekday_date())
输出：
Today Next and 3rd business day
Fri 20190308 Mon 20190311 Wed 20190313
Sat 20190309 Mon 20190311 Wed 20190313
Sun 20190310 Mon 20190311 Wed 20190313
Mon 20190311 Tue 20190312 Thu 20190314
Fri 20190329 Mon 20190401 Wed 20190403
Mon 20181231 Wed 20190102 Fri 20190104
Fri 20190118 Tue 20190122 Thu 20190124
Sat 20190216 Tue 20190219 Thu 20190221
Sun 20190526 Tue 20190528 Thu 20190530
Wed 20190703 Fri 20190705 Tue 20190709
Wed 20190828 Thu 20190829 Tue 20190903
Thu 20191010 Fri 20191011 Wed 20191016
Fri 20191108 Tue 20191112 Thu 20191114
Mon 20191125 Tue 20191126 Fri 20191129
Tue 20191224 Thu 20191226 Mon 20191230
Fri 20191227 Mon 20191230 Thu 20200102
答案 2 :(得分：0)
我从'datetime'库使用了几个函数。您可以很有趣地编写它们：date（y，m，d），timedelta（days = 7），day，weekday（），'{：％Y％m％d}'。format（day），strptime（input ，'％Y％m％d'），strftime（datetime，'％a％x'）。好的主意是为日期创建一个类，并摆脱所有格式转换。因此，只有日期（y，m，d），timedelta（days = 7），day，weekday（）可以锻炼。
import datetime
from datetime import date, timedelta
def day2string(day):
return '{:%Y%m%d}'.format(day)
def year_holidays(year):
holidays = [
["New Year's Day", 1, 1], # Fixed: January 1
["Birthday of Martin Luther King, Jr.", 1, 0, 0, 3], # Floating
["Washington's Birthday", 2, 0, 0, 3], # Third Monday in February
["Memorial Day", 5, 0, 0, 5], # Last Monday
["Independence Day", 7, 4],
["Labor Day", 9, 0, 0, 1],
["Columbus Day", 10, 0, 0, 2],
["Veterans Day", 11, 11],
["Thanksgiving Day", 11, 0, 3, 4],
["Christmas Day", 12, 25]
]
year_list = []
for h in holidays:
if h[2] > 0:
day = date(year, h[1], h[2]) # Fixed day
else:
day = date(year, h[1], 1) # Floating day
while h[3] != day.weekday(): # Advance to match the weekday
day += timedelta(days=1)
count = 1
while count != h[4]: # Match the repetition of this day
next_week = day + timedelta(days=7)
if next_week.month == day.month:
day = next_week
count += 1
year_list.append(day2string(day))
return year_list # return the holidays as list of strings
def str2datetime(string):
return datetime.datetime.strptime(string, '%Y%m%d')
def next_working_day(string):
day = str2datetime(string)
day += timedelta(days=1)
while True:
while day.weekday() >= 5:
day += timedelta(days=1)
holidays_list = year_holidays(day.year)
for str_day in holidays_list:
s2 = day2string(day)
if str_day == s2:
day += timedelta(days=1)
break # for
if day.weekday() < 5:
break # while True
return day2string(day)
if __name__ == '__main__':
dates = [
['20190308', '20190311', '20190313'],
['20190309', '20190311', '20190313'],
['20190310', '20190311', '20190313'],
['20190311', '20190312', '20190314'],
['20190329', '20190401', '20190403'],
['20181231', '20190102', '20190104'],
['20190118', '20190122', '20190124'],
['20190216', '20190219', '20190221'],
['20190526', '20190528', '20190530'],
['20190703', '20190705', '20190709'],
['20190828', '20190829', '20190903'],
['20191010', '20191011', '20191016'],
['20191108', '20191112', '20191114'],
['20191125', '20191126', '20191129'],
['20191224', '20191226', '20191230'],
['20191227', '20191230', '20200102']]
print('\n Today Next and 3rd business day')
for days in dates:
next_day = next_working_day(days[0])
third_day = next_working_day(next_working_day(next_day))
if next_day != days[1] or third_day != days[2]:
print('*** ERROR *** ', end='')
else:
print(' ', end='')
def f(x): return datetime.datetime.strftime(str2datetime(x), '%a %x')
print(f(days[0]), f(next_day), f(third_day))
它应该创建下一个输出：
Today Next and 3rd business day
Fri 03/08/19 Mon 03/11/19 Wed 03/13/19
Sat 03/09/19 Mon 03/11/19 Wed 03/13/19
Sun 03/10/19 Mon 03/11/19 Wed 03/13/19
Mon 03/11/19 Tue 03/12/19 Thu 03/14/19
Fri 03/29/19 Mon 04/01/19 Wed 04/03/19
Mon 12/31/18 Wed 01/02/19 Fri 01/04/19
Fri 01/18/19 Tue 01/22/19 Thu 01/24/19
Sat 02/16/19 Tue 02/19/19 Thu 02/21/19
Sun 05/26/19 Tue 05/28/19 Thu 05/30/19
Wed 07/03/19 Fri 07/05/19 Tue 07/09/19
Wed 08/28/19 Thu 08/29/19 Tue 09/03/19
Thu 10/10/19 Fri 10/11/19 Wed 10/16/19
Fri 11/08/19 Tue 11/12/19 Thu 11/14/19
Mon 11/25/19 Tue 11/26/19 Fri 11/29/19
Tue 12/24/19 Thu 12/26/19 Mon 12/30/19
Fri 12/27/19 Mon 12/30/19 Thu 01/02/20