我有一个带有 start_date 列和 add_days 列 (=10) 的数据框 (df)。我想创建不包括周末和假期(假期作为数据框)的 target_date (=start_date + add_days)

我做了一些研究,然后尝试这个。

from datetime import date,  timedelta
import datetime as dt

df["star_date"] = pd.to_datetime(df["star_date"])
Holidays['Date_holi'] = pd.to_datetime(Holidays['Date_holi'])


def date_by_adding_business_days(from_date, add_days, holidays):
    business_days_to_add = add_days
    current_date = from_date
    while business_days_to_add > 0:
        current_date += datetime.timedelta(days=1)
        weekday = current_date.weekday()
        if weekday >= 5: # sunday = 6
            continue
        if current_date in holidays:
            continue
        business_days_to_add -= 1
    return current_date


#demo:
base["Target_date"]=date_by_adding_business_days(df["start_date"], 10, Holidays['Date_holi'])

但我收到此错误:



谢谢你的帮助。

最佳答案

ALollz 的评论非常有效;在创建期间自定义您的日期以仅保留定义为您的问题的工作日将是最佳的。

但是,我假设您无法事先定义工作日,并且您需要使用按原样构建的数据框来解决问题。

这是一种可能的解决方案:

import pandas as pd
import numpy as np
from datetime import timedelta

# Goal is to offset a start date by N business days (weekday + not a holiday)

# Here we fake the dataset as it was not provided
num_row = 1000
df = pd.DataFrame()
df['start_date'] = pd.date_range(start='1/1/1979', periods=num_row, freq='D')
df['add_days'] = pd.Series([10]*num_row)

# Define what is a week day
week_day = [0,1,2,3,4] # Monday to Friday
# Define what is a holiday with month and day without year (you can add more)
holidays = ['10-30','12-24']

def add_days_to_business_day(df, week_day, holidays, increment=10):
    '''
       modify the dataframe to increment only the days that are part of a weekday
       and not part of a pre-defined holiday
       >>> add_days_to_business_day(df, [0,1,2,3,4], ['10-31','12-31'])
           this will increment by 10 the days from Monday to Friday excluding Halloween and new year-eve
    '''
    # Increment everything that is in a business day
    df.loc[df['start_date'].dt.dayofweek.isin(week_day),'target_date'] = df['start_date'] + timedelta(days=increment)
    # Remove every increment done on a holiday
    df.loc[df['start_date'].dt.strftime('%m-%d').isin(holidays), 'target_date'] = np.datetime64('NaT')


add_days_to_business_day(df, week_day, holidays)
df

注意: 我没有使用“add_days”列,因为它只是一个重复值。我改为使用我的函数增量参数,该参数将增加 N 天(默认为 N = 10)。

希望能帮助到你!

关于python - 如何在不包括节假日的日期中添加工作日,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/59636651/

10-12 21:54