如何修正月末最后工作日计算Python函数并集成节假日校验？

我正在编写一个筛选月末最后工作日的函数，目前遇到以下问题：
当前函数获取的是上月最后工作日，但假设上月最后工作日为周五时，我需要将日期调整至下一个工作日（即周一），但实际结果却跳到了周六。请问如何修改函数使其仅返回工作日？同时如何集成我已有的节假日表进行校验？

现有代码 #

import calendar
from datetime import date
import pandas as pd
from datetime import timedelta

def BsDay():
    today = date.today()
    last_day = max(calendar.monthcalendar(today.year, today.month)[-1:][0][:5])
    validDay = (today.year, today.month-1,
                max(calendar.monthcalendar(today.year, today.month-1)[-1:][0][:5]))
    if today.month == 1 and today.day <= last_day:
        validDay = (today.year-1, today.month+11,
                    max(calendar.monthcalendar(today.year, today.month-1)[-1:][0][:5]))

    elif today.day <= last_day:
        validDay
    else:
        validDay = (today.year, today.month, max(
            calendar.monthcalendar(today.year, today.month)[-1:][0][:5]))

    return validDay

尝试的代码及输出 #

BusinessDay = ''.join(map(str, BsDay()))
BusinessDay = BusinessDay[0:4] + '-0' + BusinessDay[4]+'-'+BusinessDay[5:7]

dateD='2022-07-29'
dateD=pd.to_datetime(dateD)
dateF=pd.to_datetime(BusinessDay)

LastDayNotWeekend=dateD+timedelta(days=1)
print('LastDayNotWeekend not Weekend',LastDayNotWeekend)
LastDayInWeekend=dateF+timedelta(days=1)
print('LastDayInWeekend in Weekend->',LastDayInWeekend)

月末最后工作日不在周末时的输出： #

LastDayNotWeekend not Weekend 2022-07-30 00:00:00

月末最后工作日在周末时的输出： #

LastDayInWeekend in Weekend-> 2022-09-01 00:00:00

1. 重构函数，直接返回date对象 #

原函数返回元组，后续字符串拼接容易出格式bug（比如10月会被拆成010），改成直接返回date类型，逻辑更清晰：

import calendar
from datetime import date, timedelta
import pandas as pd

def get_last_business_day():
    today = date.today()
    # 确定目标月份：当前日 <= 当月最后工作日则取上月，否则取当月
    current_month_last_workday = max(calendar.monthcalendar(today.year, today.month)[-1][:5])
    if today.day <= current_month_last_workday:
        target_year = today.year - 1 if today.month == 1 else today.year
        target_month = 12 if today.month == 1 else today.month - 1
    else:
        target_year = today.year
        target_month = today.month
    
    # 取目标月份周一到周五的最后一天
    last_week = calendar.monthcalendar(target_year, target_month)[-1]
    last_workday = max(last_week[:5])
    return date(target_year, target_month, last_workday)

2. 处理周五转周一的逻辑 #

按需求，只要候选日是周五，直接跳转到下周一：

def get_adjusted_business_day():
    candidate = get_last_business_day()
    # weekday()返回0=周一，4=周五
    if candidate.weekday() == 4:
        candidate += timedelta(days=3)
    # 保险校验：防止调整后意外碰到周末（虽然加3天肯定是周一）
    while candidate.weekday() >= 5:
        candidate += timedelta(days=1)
    return candidate

3. 集成节假日表校验 #

假设你的节假日表是pandas.DataFrame，包含holiday_date列存储节假日日期。把节假日转成集合快速查询，循环向后找第一个非节假日的工作日：

def get_final_valid_workday(holiday_df):
    adjusted_date = get_adjusted_business_day()
    # 把节假日转成date集合，查询效率更高
    holiday_set = set(holiday_df['holiday_date'].dt.date)
    
    while True:
        # 检查是否是周末或节假日
        if adjusted_date.weekday() >= 5 or adjusted_date in holiday_set:
            adjusted_date += timedelta(days=1)
        else:
            break
    return adjusted_date

测试用例 #

# 模拟节假日表，比如2022-09-01是节假日
holiday_df = pd.DataFrame({'holiday_date': pd.to_datetime(['2022-09-01'])})
print(get_final_valid_workday(holiday_df))

关键修复点 #

• 删掉原函数里无效的elif today.day <= last_day: validDay代码
• 用date对象替代字符串拼接，彻底解决日期格式错误
• 节假日用集合存储，比DataFrame直接查询快数倍
• 循环校验确保最终返回的一定是合法工作日

内容的提问来源于stack exchange，提问作者Zan