我编写了一个函数,该函数相对于周末和节假日将日期移动指定的工作日数。当我将其与标量值一起使用时,它可以正常工作,但是尝试在pandas DataFrame上使用时遇到问题。 该函数本身如下所示:
def move_date_by_days(init_date, roll=1, nwd_key=None, hol_key=None):
'''
moves date by n-number of working days forward or backward
init_date: date, initial caluclation date
roll: integer, number of days to move forward (+) or backward (-)
nwd_key: string that stands for currency iso code, it is a key in non_working_days dictionary
hol_key: string that stands for currency iso code, it is a key in holidays dictonary
return: date
'''
nwd = non_working_days.get(nwd_key, [])
hol = holidays.get(hol_key,[])
moved_date = init_date + datetime.timedelta(days=roll)
if (moved_date.isoweekday() in nwd) or (moved_date in hol):
if roll >= 0:
moved_date=move_date_by_days(init_date + datetime.timedelta(days=1), roll=roll, nwd_key=nwd_key, hol_key=hol_key)
else:
moved_date=move_date_by_days(init_date + datetime.timedelta(days=-1), roll=roll, nwd_key=nwd_key, hol_key=hol_key)
return moved_date
现在,我的数据框为'df',列为'start_date'和'end_date'
start_date end_date
0 2020-01-31 2020-04-30
1 2020-04-30 2020-07-31
2 2020-07-31 2020-10-31
3 2020-10-31 2020-11-28
,我想创建名为“ fixing”的第三列,该列将在“ start_date”之前2个工作日。我正在尝试:
dates_table['fixing'] = move_date_by_days(self.dates_table['start_date'], -2, self.ccy, self.ccy)
但它返回AttributeError: 'Series' object has no attribute 'isoweekday'
请注意,例如,当我引用DataFrame的标量数据时,该功能可以正常工作
d1 = s1.dates_table.iat[0,0]
move_date_by_days(d1, -2, 'pln', 'pln')
它返回我所期望的:datetime.date(2020, 1, 29)
任何提示,请问如何在数据框的整个列上使用该功能?
答案 0 :(得分:1)
IIUC,为什么不使用offsets.BDay
或offsets.CustomBusinessDay
,它们使用holidays
的列表:
df['fixing'] = df['start_date'] - pd.offsets.BDay(2)
df['fixing'] = df['start_date'] - pd.offsets.CustomBusinessDay(2, holidays=holidays['pln'])
[出]
start_date end_date fixing
0 2020-01-31 2020-04-30 2020-01-29
1 2020-04-30 2020-07-31 2020-04-28
2 2020-07-31 2020-10-31 2020-07-29
3 2020-10-31 2020-11-28 2020-10-29