我想根据销售额最高的连续两个月来计算奖金。所以我可以每连续两个月迭代一次总数来找到最大值，即得到
value = Max[total_between_firstdayMonth1_and_lastDayMonth2, total_between_firstdayMonth2_and_lastDayMonth3, ... , total_between_firstdaySecondToLastMonth_andlastDayLastMonth]
所以我可能需要一组日期时间对象或类似的东西。
start= model.Order.order('created').get().created # get the oldest order
end = model.Order.order('-created').get().created # get the newest order
所以在开始和结束之间，我必须将时间分成连续 2 个月的重叠对，例如。如果第一个订单是在 2008 年 12 月，最后一个订单是在 2011 年 11 月，那么从中选择最大值的列表应该是 [total_december2008 + total_january2009, total_january2009 + total_february2009, ... , total_october2011 + total_november2011]
但是，如果我像上面那样知道开始时间，那么如何获得第二个月的最后一天呢？如何创建时间和总计列表？
我可能无法立即创建总计列表，但如果我可以创建开始和结束列表，那么我可以调用我们可以假设的辅助函数，例如。
total(start_datetime, end_datetime)
感谢您的帮助
更新
我想我找到了如何计算示例间隔的时间，其中时间线是从任何日期到下个月的最后一天:
>>> d = date(2007,12,18)
>>> print d
2007-12-18
>>> d + relativedelta(months=2) - timedelta(days=d.day)
datetime.date(2008, 1, 31)
更新2
我可以计算到第一级的第一个持续时间。现在我只需要概括它来遍历所有持续时间并检查哪个是最高级别:
def level(self):
startdate = model.Order.all().filter('status =', 'PAID').filter('distributor_id =' , self._key.id()).get().created.date()
last_day_nextmonth =startdate + relativedelta(months=2) - timedelta(days=1)
if self.personal_silver(startdate, last_day_nextmonth) + self.non_manager_silver(startdate, last_day_nextmonth) < 25:
maxlevel = _('New distributor')
elif self.personal_silver(startdate, last_day_nextmonth) + self.non_manager_silver(startdate, last_day_nextmonth) > 25:
maxlevel = _('Assistant Teamleader')
return maxlevel
更新3
更接近我的意思是从开始到现在取一些函数值的最大值。 Basecase 可以是下个月的最后一天是 future ，辅助函数可以是递归的，但我没有时间或帮助使其递归到它现在只适用于前 2 个时期，即从开始后的 4 个月:
def level(self):
startdate = model.Order.all().filter('status =', 'PAID'
).filter('distributor_id =',
self._key.id()).get().created.date()
last_day_nextmonth = startdate + relativedelta(months=2) \
- timedelta(days=1)
total = self.personal_silver(startdate, last_day_nextmonth) + self.non_manager_silver(startdate, last_day_nextmonth)
if total >= 125:
level = 5
elif total >= 75:
level = 4
elif total >= 25:
level = 3
elif total >= 2:
level = 2
else:
level = 1
return self.levelHelp(level, last_day_nextmonth + timedelta(days=1))
def levelHelp(self, level, startdate):
#if startdate in future return level
last_day_nextmonth = startdate + relativedelta(months=2) \
- timedelta(days=1)
total = self.personal_silver(startdate, last_day_nextmonth) + self.non_manager_silver(startdate, last_day_nextmonth)
if total >= 125:
newlevel = 5
elif total >= 75:
newlevel = 4
elif total >= 25:
newlevel = 3
elif total >= 2:
newlevel = 2
else:
newlevel = 1
return level if level > newlevel else newlevel
更新4
我添加了递归，其中基本情况是下一步是在未来，如果是这样它将返回最大级别:
def level(self):
startdate = model.Order.all().filter('status =', 'PAID'
).filter('distributor_id =',
self._key.id()).get().created.date()
last_day_nextmonth = startdate + relativedelta(months=2) \
- timedelta(days=1)
total = self.personal_silver(startdate, last_day_nextmonth) + self.non_manager_silver(startdate, last_day_nextmonth)
if total >= 125:
level = 5
elif total >= 75:
level = 4
elif total >= 25:
level = 3
elif total >= 2:
level = 2
else:
level = 1
return self.levelHelp(level, last_day_nextmonth + timedelta(days=1))
def levelHelp(self, level, startdate):
last_day_nextmonth = startdate + relativedelta(months=2) \
- timedelta(days=1)
total = self.personal_silver(startdate, last_day_nextmonth) + self.non_manager_silver(startdate, last_day_nextmonth)
if total >= 125:
newlevel = 5
elif total >= 75:
newlevel = 4
elif total >= 25:
newlevel = 3
elif total >= 2:
newlevel = 2
else:
newlevel = 1
maxlevel = level if level > newlevel else newlevel
nextstart = last_day_nextmonth + timedelta(days=1)
now = datetime.now().date()
if nextstart > now: #next start in is the future
return maxlevel
else: return self.levelHelp(maxlevel, nextstart)
请您参考如下方法:
这对于函数式方法来说听起来不错。最后有一个完整的工作示例，但我只想强调以 FP 风格编写的核心功能的优雅和简单:
def find_best_two_months(orders):
first = lambda x: x[0]
second = lambda x: x[1]
orders_by_year_and_month = [
("%04d-%02d" % (date.year, date.month), amount)
for date, amount in orders]
sorted_orders = sorted(orders_by_year_and_month, key=first)
totals_by_month = [
(ym, sum(map(second, groupped_orders)))
for ym, groupped_orders in groupby(sorted_orders, key=first)]
totals_two_months = [
( "%s - %s" % (m1[0], m2[0]), m1[1]+m2[1] )
for m1, m2 in zip(totals_by_month, totals_by_month[1:]) ]
return max(totals_two_months, key=second)
这是一个带有注释的完整工作示例:
#!/usr/bin/python
from random import randint
from datetime import date, timedelta
from itertools import groupby
""" finding best two months the functional way """
def find_best_two_months(orders):
"""
Expect a list of tuples of form (date_of_order, amount):
[ (date1, amount1), (date2, amount2), ...]
"""
" helper functions for extracting first or second from tuple "
first = lambda x: x[0]
second = lambda x: x[1]
" converts [(date, amount)] -> [(YYYY-MM, amount)] "
orders_by_year_and_month = [ ("%04d-%02d" % (date.year, date.month), amount) for date, amount in orders]
" Sorts by YYYY-MM. This step can be omitted if orders were already sorted by date"
sorted_orders = sorted(orders_by_year_and_month, key=first)
" Compresses orders from the same month, so ve get [(YYYY-MM), total_amount_of_orders]"
totals_by_month = [ (ym, sum(map(lambda x:x[1], groupped_orders)))
for ym, groupped_orders in groupby(sorted_orders, key=first)]
" Zips orders to two month periods"
totals_two_months = [ ("%s - %s" % (m1[0], m2[0]), m1[1]+m2[1]) for m1, m2 in zip(totals_by_month, totals_by_month[1:]) ]
" Returns two-month period with maximum total amount. If there were many periods with max amount, only the first is returned "
return max(totals_two_months, key=second)
"""
this code is for generating random list of orders
and is not a part of the solution
"""
MIN_AMOUNT=70
MAX_AMOUNT=500
MAX_DAY_SPREAD=5
def gen_order(last_date):
""" returns (order_date, amount) """
days = timedelta()
return (
last_date+timedelta(days=randint(0, MAX_DAY_SPREAD)), # new date
randint(MIN_AMOUNT, MAX_AMOUNT)) # amount
def gen_orders(total, start_date):
orders = []
last_date = start_date
for i in range(total):
order = gen_order(last_date)
orders.append(order)
last_date = order[0]
return orders
if __name__ == "__main__":
orders = gen_orders(300, date(2010,1,1))
print find_best_two_months(orders)
