获取 DataFrame 中的前一个工作日

Question

我有一个包含两列、日期和类别的 DataFrame。我想根据规则创建一个新的日期列：如果类别是B那么值应该是最接近日期的工作日（仅来自过去或当天本身），否则它是日期列本身的值。

我将工作日定义为不在周末的任何一天，也不出现在holidays下面最小示例中定义的列表中。

请考虑以下 DataFrame df：

import datetime as dt

import pandas as pd

from IPython.display import display

holidays = [dt.datetime(2018, 10, 11)]

df = pd.DataFrame({"day": ["2018-10-10", "2018-10-11", "2018-10-12",

                       "2018-10-13", "2018-10-14", "2018-10-15"

                      ],

               "category":["A", "B", "C", "B", "C", "A"]

              }

)

df["day"] = pd.to_datetime(df.day, format="%Y-%m-%d")

display(df)

         day category

0 2018-10-10        A

1 2018-10-11        B

2 2018-10-12        C

3 2018-10-13        B

4 2018-10-14        C

5 2018-10-15        A

如何获得第三列，其值为下面列出的值？

2018-10-10

2018-10-10

2018-10-12

2018-10-12

2018-10-14

2018-10-15

我创建了一个函数，可以在处理列表时查找最后一个工作日，如果有帮助的话。

# creates a list whose elements are all days in the years 2017, 2018 and 2019

days = [dt.datetime(2017, 1 , 1) + dt.timedelta(k) for k in range(365*3)]

def lastt_bus_day(date):

    return max(

        [d for d in days if d.weekday() not in [5, 6]

                            and d not in holidays

                            and d <= date

        ]

    )

for d in df.day:

    print(last_bus_day(d))

#prints

2018-10-10 00:00:00

2018-10-10 00:00:00

2018-10-12 00:00:00

2018-10-12 00:00:00

2018-10-12 00:00:00

2018-10-15 00:00:00

Answer 1

Pandas 支持通过自定义工作日提供您自己的假期。该解决方案的好处是无缝支持相邻的假期；例如，某些地区的节礼日和圣诞节。# define custom business daysweekmask = 'Mon Tue Wed Thu Fri'holidays = ['2018-10-11']bday = pd.tseries.offsets.CustomBusinessDay(holidays=holidays, weekmask=weekmask)# construct mask to identify when days must be sutractedm1 = df['category'] == 'B'm2 = df['day'].dt.weekday.isin([5, 6]) | df['day'].isin(holidays)# apply conditional logicdf['day'] = np.where(m1 & m2, df['day'] - bday, df['day'])print(df)&nbsp; category&nbsp; &nbsp; &nbsp; &nbsp; day0&nbsp; &nbsp; &nbsp; &nbsp; A 2018-10-101&nbsp; &nbsp; &nbsp; &nbsp; B 2018-10-102&nbsp; &nbsp; &nbsp; &nbsp; C 2018-10-123&nbsp; &nbsp; &nbsp; &nbsp; B 2018-10-124&nbsp; &nbsp; &nbsp; &nbsp; C 2018-10-145&nbsp; &nbsp; &nbsp; &nbsp; A 2018-10-15编辑：根据您的评论，“我刚刚意识到我没有问清楚我想要什么。我想找到前一个工作日”，您可以简单地使用：df['day'] -= bday