5

给定一个字符串，reqDayOf这是一个工作日名称，您如何将工作日计算为十进制（然后根据datetime对象返回它的下一个实例）？

获取datetime对象的年份和 iso 年份的第几周，然后使用strptime年份 + 年份中的星期 + 工作日名称，但感觉就像一个 hack。

import datetime

def getDateFromDayOf(dateTimeObj,reqDayOf):
  #reqDayOf is one of ['monday','tuesday','wednesday','thursday','friday','saturday','sunday']
  #return the next instance of reqDayOf
  #after dateTimeObj
  #as a datetime object 
  #Get the WeekOfYear from dateTimeObj and then
  #get the date based on Year + WeekOfYear + reqDayOf 
  (dtoYear,dtoIsoWeek,x)=dateTimeObj.isocalendar()
  checkDate= '{}-{}-{}'.format( dtoYear,dtoIsoWeek,reqDayOf)
  dateOfDay=datetime.datetime.strptime(checkDate,"%Y-%U-%A")
  #return dateOfDay if it's greater than the original date
  if dateOfDay > dateTimeObj:
    return dateOfDay
  else:
    #this is needed on Sundays
    #add a week
    return dateOfDay + datetime.timedelta(days=7)  


>>> datetime.datetime.now().strftime('%a %Y %b %d')
'Fri 2013 Apr 05'
>>> getDateFromDayOf(datetime.datetime.now(),'Monday').strftime('%a %Y %b %d')
'Mon 2013 Apr 08'
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2 回答 2

4

星期几（作为整数）由weekday 方法返回：

import datetime as DT
dow = dict(zip('monday tuesday wednesday thursday friday saturday sunday'.split(),
           range(7)))

def getDateFromDayOf(dateTimeObj, reqDayOf):
    weekday = dateTimeObj.weekday()        
    return dateTimeObj + DT.timedelta(days=(dow[reqDayOf.lower()]-weekday-1)%7+1)

In [90]: getDateFromDayOf(DT.datetime.now(), 'Monday').date()
Out[90]: datetime.date(2013, 4, 8)

In [91]: getDateFromDayOf(DT.datetime.now(), 'Tuesday').date()
Out[91]: datetime.date(2013, 4, 9)

In [92]: getDateFromDayOf(DT.datetime.now(), 'Friday').date()
Out[92]: datetime.date(2013, 4, 12)

或者，使用dateutil，

import datetime as DT
import dateutil
import dateutil.relativedelta as rdelta
import dateutil.rrule as rrule

dow = dict(zip('monday tuesday wednesday thursday friday saturday sunday'.split(),
               (getattr(rdelta, d) for d in 'MO TU WE TH FR SA SU'.split())))
def getDateFromDayOf(dateTimeObj, reqDayOf):
    rr = rrule.rrule(
        rrule.DAILY,                       # step by days
        byweekday = dow[reqDayOf.lower()], # return only this day of the week
        dtstart = dateTimeObj)             # start on this day 
    res = rr.after(dateTimeObj, inc=False) # inc=False means don't include the dtstart day
    return res

使用 dateutil 您可以以高级方式表达这个想法，而不必担心肮脏的细节中的错误，例如

DT.timedelta(days=(dow[reqDayOf.lower()]-weekday-1)%7+1)
于 2013-04-05T13:17:43.747 回答
2

这类似于@unutbu 的 anwser中的 weekday 方法，但以不需要将日期名称（或其缩写）硬编码到其中的方式实现。它使用datetime.date'toordinal()方法将日期转换为单个数字，然后计算要添加多少天才能获得所需的工作日，最后使用该date.fromordinal()方法将新数字转换回日期。

它创建并使用一个名为的字典DAYNUMS将reqDayOf工作日名称转换为从 1 到 7 的数字，并且该date.isoweekday()方法用于对dateTimeObj传递给它的参数做同样的事情。该DAYNUMS字典是在不使用硬编码的日期名称的情况下创建的，因此它也可以在非英语语言环境和给定适当reqDayOf参数的语言中工作，例如'lundi'代替'Monday'法语语言环境。

import datetime

# Create dictionary based on fact that 2013-April-01 was a Monday.
DAYNUMS = {datetime.date(2013, 4, i).strftime('%A').lower(): i
               for i in range(1, 8)}

def getDateFromDayOf(dateTimeObj, reqDayOf):
    daysDiff = (DAYNUMS[reqDayOf.lower()] - dateTimeObj.isoweekday() - 1) % 7 + 1
    return datetime.date.fromordinal(dateTimeObj.toordinal() + daysDiff)

a_date = datetime.datetime.strptime('2013-04-05', '%Y-%m-%d')
print(a_date.strftime('%a %Y %b %d'))                              # -> Fri 2013 Apr 05
print(getDateFromDayOf(a_date, 'Monday').strftime('%a %Y %b %d'))  # -> Mon 2013 Apr 08
于 2013-04-05T16:34:56.530 回答