如何基于日期范围创建二进制变量 标记节日前后一周的行

完整可运行代码 #

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

# 构建测试数据
df = pd.DataFrame({'Date': ['1/1/2019', '1/5/2019', '12/28/2018'],
                   'Event': ['Music', 'Poetry', 'Theatre'],
                   'Cost': [10000, 5000, 15000]})
df['newDate'] = pd.to_datetime(df['Date'], format='%m/%d/%Y')

# 通用节日窗口标记函数
def mark_holiday_window(df, date_col, holiday_obj, window_days=7):
    # 转换所有节日日期为datetime格式
    holiday_dates = pd.to_datetime(list(holiday_obj.keys()))
    # 为每个节日生成前后N天的时间窗口
    holiday_windows = [(d - timedelta(days=window_days), d + timedelta(days=window_days)) for d in holiday_dates]
    # 批量判断日期是否落在任意节日窗口内
    is_in_window = df[date_col].apply(lambda x: any(start <= x <= end for start, end in holiday_windows))
    return np.where(is_in_window, 1, 0)

# ------------------- 元旦标记示例 -------------------
new_year_holidays = holidays.HolidayBase()
new_year_holidays.append({
    "2018-01-01": "New Year's Day",
    "2019-01-01": "New Year's Day"
})
df['hol_new_year'] = mark_holiday_window(df, 'newDate', new_year_holidays, window_days=7)

# ------------------- 感恩节标记示例（可替换为任意节日） -------------------
us_holidays = holidays.US(years=[2018,2019])
thanksgiving_holidays = holidays.HolidayBase()
for date, name in us_holidays.items():
    if 'Thanksgiving' in name:
        thanksgiving_holidays.append({date: name})
df['hol_thanksgiving'] = mark_holiday_window(df, 'newDate', thanksgiving_holidays, window_days=7)

print(df)

核心逻辑说明 #

• 通用适配：只要传入holidays库支持的任意节日对象，就能实现复活节、圣诞节等不同节日的窗口标记，支持自定义国家/地区的节日规则
• 窗口可调整：修改window_days参数即可自定义前后覆盖的天数，设置为0即仅标记节日当天
• 修正了原有代码的逻辑错误：原代码直接对整个日期列用in判断，无法实现逐行匹配，改为向量式批量判断后运行效率更高

运行上述代码后，测试数据里的3行日期均落在2019年元旦前后7天范围内，hol_new_year字段值全部为1，符合预期效果。

内容的提问来源于stack exchange，提问作者Ketty