Question

给定一个字符串reqDayOf，这是一个工作日名称，你如何计算工作日为十进制（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，然后将strptime与年份+周年+工作日名称一起使用，但感觉就像是黑客。

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'

Answer 1

weekday method：

返回星期几（整数）

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，您可以高层次地表达这个想法，而不必担心肮脏细节中的逐个错误，例如

DT.timedelta(days=(dow[reqDayOf.lower()]-weekday-1)%7+1)

Answer 2

这与@unutbu's anwser中的工作日方法类似，但实现方式并不要求将日期名称（或其缩写）硬编码到其中。它使用datetime.date＆＃39; toordinal()方法将日期转换为单个数字，然后计算添加到其中的天数以获得所需的工作日，最后转换新数字使用date.fromordinal()方法返回日期。

它创建并使用名为DAYNUMS的字典将reqDayOf工作日名称转换为1到7之间的数字，date.isoweekday()方法用于dateTimeObj方法。 1}}传递给它的参数。 DAYNUMS字典是在不使用硬编码日期名称的情况下创建的，因此在给定适当的reqDayOf参数的情况下，它也适用于非英语语言环境和语言，例如'lundi'而不是'Monday' }用于法语区域设置。

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

给定一个字符串是工作日名称，如何将工作日计算为十进制（以及它出现的下一个日期）？

2 个答案: