pandas：to_timedelta 工作日

Question

我想遍历一个只有工作日的日期范围，即没有周末。为此，我有以下命令，基本上选择一个月中的每个第三个星期五，并使列 window 等于从第三个星期五开始的 2、20 天。这段代码工作得很好。

for beg in pd.bdate_range("2000-01-01", "2017-05-01"):     
     beg= third_friday
     df["window"].loc[beg: beg + pd.to_timedelta(20,"D")] = 2
     if month==12:
         year=year+1
         month=0
     if year>=2017 and month>=3:
         break
     month = month +3
     monthcal = c.monthdatescalendar(year,month)
     third_friday = [day for week in monthcal for day in week if \
                day.weekday() == calendar.FRIDAY and \
               day.month == month][2]

然而，

中的 20

df["window"].loc[beg: beg + pd.to_timedelta(20,"D")] = 2

命令指的是 20 天，包括周末，但我希望它指的是 20 个工作日；例如像这样：

df["window"].loc[beg: beg + pd.to_timedelta(20, "Weekdays_only")] = 2

是否有简单的修复方法，以便我可以用其他东西替换 "D"，或者我是否必须重写所有内容？

此外，我还想用不同的值标记第三个星期五前后的日子，例如third_friday 之后的第 +1 天是 1，第 2 天是 2。为此，我写了第二个 for 循环。这里是完整的例子：

for beg in pd.bdate_range("2000-01-01", "2017-05-01"):     
 beg= third_friday
 lower_counter = 0
 for j in range(0,-21,-1):    
   df["window_counter"].loc[beg - pd.to_timedelta(j,"D"):beg] = lower_counter       
   lower_counter = j         

 df["window"].loc[beg: beg + pd.to_timedelta(20,"D")] = 2
 if month==12:
     year=year+1
     month=0
 if year>=2017 and month>=3:
     break
 month = month +3
 monthcal = c.monthdatescalendar(year,month)
 third_friday = [day for week in monthcal for day in week if \
            day.weekday() == calendar.FRIDAY and \
           day.month == month][2]

Answer 1

我相信您正在寻找 BDay 日期偏移

import pandas as pd
from pandas.tseries.offsets import *

new_date = beg + BDay(20)

http://pandas.pydata.org/pandas-docs/stable/timeseries.html#dateoffset-objects

Answer 2

20 个工作日恰好是 28 个日历日。

因此

df["window"].loc[beg: beg + pd.to_timedelta(28,"D")] = 2

应该可以

pandas：to_timedelta 工作日

pandas: to_timedelta with business days

python

datetime

timedelta

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