在Python中查找两个日期时间之间的天数差异（不包括周末）？

Question

与查找两个日期之间的天数差异（不包括周末）相同的问题，但是这是针对JavaScript的。如何在Python中实现？

- Drake Guan

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这个问题似乎在请求将一种语言的源代码转写成另一种语言。 - Arafangion

Answer 1

试着使用scikits.timeseries：

import scikits.timeseries as ts
import datetime

a = datetime.datetime(2011,8,1)
b = datetime.datetime(2011,8,29)

diff_business_days = ts.Date('B', b) - ts.Date('B', a)
# returns 20

或使用dateutil：

import datetime
from dateutil import rrule

a = datetime.datetime(2011,8,1)
b = datetime.datetime(2011,8,29)

diff_business_days = len(list(rrule.rrule(rrule.DAILY,
                                          dtstart=a,
                                          until=b - datetime.timedelta(days=1),
                                          byweekday=(rrule.MO, rrule.TU, rrule.WE, rrule.TH, rrule.FR))))

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scikits.timeseries看起来已经过时了：http://pytseries.sourceforge.net/

用pandas代替可以做如下事情：

import pandas as pd

a = datetime.datetime(2015, 10, 1)
b = datetime.datetime(2015, 10, 29)

diff_calendar_days = pd.date_range(a, b).size
diff_business_days = pd.bdate_range(a, b).size

- eumiro

Answer 2

我不确定这是最好的解决方案，但它适合我：

from datetime import datetime, timedelta

startDate = datetime(2011, 7, 7)
endDate = datetime(2011, 10, 7)
dayDelta = timedelta(days=1)
diff = 0
while startDate != endDate:
    if startDate.weekday() not in [5,6]:
        diff += 1
    startDate += dayDelta

- Artsiom Rudzenka

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如果你希望更快地得到结果，可以减去 2 * (天数 // 7)，然后根据起始日和结束日在一周中的位置进行调整。请注意，调整后不要改变原有意思。 - agf

Answer 3

这里提供了一个O(1)复杂度的解决方案，仅使用Python内置库。它具有恒定的性能，无论时间间隔长度如何，也不关心参数顺序。

#
# by default, the last date is not inclusive
#
def workdaycount(first, second, inc = 0):
   if first == second:
      return 0
   import math
   if first > second:
      first, second = second, first
   if inc:
      from datetime import timedelta
      second += timedelta(days=1)
   interval = (second - first).days
   weekspan = int(math.ceil(interval / 7.0))
   if interval % 7 == 0:
      return interval - weekspan * 2
   else:
      wdf = first.weekday()
      if (wdf < 6) and ((interval + wdf) // 7 == weekspan):
         modifier = 0
      elif (wdf == 6) or ((interval + wdf + 1) // 7 == weekspan):
         modifier = 1
      else:
         modifier = 2
      return interval - (2 * weekspan - modifier)

#
# sample usage
#
print workdaycount(date(2011, 8, 15), date(2011, 8, 22)) # returns 5
print workdaycount(date(2011, 8, 15), date(2011, 8, 22), 1) # last date inclusive, returns 6

- Saul