在python中找到上一个/最后一个星期的星期五

如何解决在python中找到上一个/最后一个星期的星期五

| 例如1。假设我有一天4/30/07。那么我需要得到4/27/07。 例如2。假设我有一天6/29/07，那么我需要得到6/22/07。     

解决方法

        假设

day

是

datetime.date

或

datetime.datetime

对象，则此代码为上周的星期五创建

datetime

/

date

对象：

friday = day - timedelta(days=day.weekday()) + timedelta(days=4,weeks=-1)

说明：

timedelta(days=day.weekday())

是星期一和

day

之间的偏移量，因此加4天并减去一周将获得上周的星期五。 当然，您可以简化此过程（

+4d -1w = -3d

）：

friday = day - timedelta(days=day.weekday() + 3)

注意：要获得

timedelta

，请使用

from datetime import timedelta

或仅使用

import datetime

并使用

datetime.timedelta

    ,        另一种简便的方法是使用python-dateutil。 要获取上一个星期五：

>>> from dateutil.relativedelta import relativedelta,FR
>>> from datetime import datetime
>>> datetime(2015,7,8) + relativedelta(weekday=FR(-1))
datetime.datetime(2015,3,0)

下一个星期五：

>>> datetime(2015,8) + relativedelta(weekday=FR(+1))
datetime.datetime(2015,10,0)

    ,        不特定于星期五，而是指定为一天“ taget_dayofweek \”（其中星期一为0，星期日为6）

from datetime import datetime

target_dayofweek = 4  # Friday
current_dayofweek = datetime.now().weekday() # Today


if target_dayofweek <= current_dayofweek:
    # target is in the current week
    endDate = datetime.now() - timedelta(current_dayofweek - target_dayofweek)

else: 
    # target is in the previous week
    endDate = datetime.now() - timedelta(weeks=1) + timedelta(target_dayofweek - current_dayofweek)

    ,        

date_object = datetime.date.today()
from dateutil.relativedelta import relativedelta
previousWeekLastDay = date_object + relativedelta(weekday=SU(-1))
previousWeekFirstDay = previousWeekLastDay + relativedelta(weekday=MO(-1))
print(previousWeekFirstDay)
print(previousWeekLastDay)

previoustopreviousWeekLastDay1= date_object + relativedelta(weekday=SU(-2))
previoustopreviousWeekFirstDay1= previousWeekLastDay1 + relativedelta(weekday=MO(-1))
print(previoustopreviousWeekFirstDay1)
print(previoustopreviousWeekLastDay1)

    ,        pandas.tseries.offsets中有很多选项 这是前一周的星期五。

from pandas.tseries.offsets import Week
f_dates = required_df.index - Week(1,weekday=4)