python判断2018/2019年的某天是否是节假日

2019-01-25 09:15:00 by wst

python小组件

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网上找了半天也没找到可用的，没一个省心的。
无奈之下只好自己动手，丰衣足食！

hol:节日，work：被调休的工作日

 

返回结果：

• -1:报错
• 0: 工作日
• 1: 周末
• 2: 节日

"""
Author: deepinwst
Email: movingheart000@gmail.com
Date: 19-01-25 下午1:56
"""

import datetime
import time

def holiday(d, s="2018-01-01", e="2019-12-31"):
    # 节假日
    hol = {"2018-01-01", "2018-02-15", "2018-02-16", "2018-02-17", "2018-02-18", "2018-02-19", "2018-02-20",
           "2018-02-21", "2018-04-05", "2018-04-06", "2018-04-07", "2018-04-29", "2018-04-30", "2018-05-01"
           "2018-06-16", "2018-06-17", "2018-06-18", "2018-09-22", "2018-09-23", "2018-09-24", "2018-10-01",
           "2018-10-02", "2018-10-03", "2018-10-04", "2018-10-05", "2018-10-06", "2018-10-07", "2018-12-30",
           "2018-12-31", "2019-01-01", "2019-02-04", "2019-02-05", "2019-02-06", "2019-02-07", "2019-02-08",
           "2019-02-09", "2019-02-10", "2019-04-05", "2019-04-06", "2019-04-07", "2019-05-01", "2019-06-07",
           "2019-06-08", "2019-06-09", "2019-09-13", "2019-09-14", "2019-09-15", "2019-10-01", "2019-10-02",
           "2019-10-03", "2019-10-04", "2019-10-05", "2019-10-06", "2019-10-07"}
    # 调休日
    work = {"2018-02-11", "2018-02-24", "2018-04-08", "2018-04-28", "2018-09-29", "2018-09-30", "2018-12-29",
            "2019-02-02", "2019-02-03", "2019-09-29"}
    s1 = datetime.datetime.strptime(s, '%Y-%m-%d')
    e1 = datetime.datetime.strptime(e, '%Y-%m-%d')
    if not isinstance(d, str):
        print("Please input string date")
        return -1
    else:
        d1 = datetime.datetime.strptime(d, '%Y-%m-%d')
        if d1 > e1 or d1 < s1:
            print("not in 2018/2019 year")
            return -1
        elif d in hol:
            return 2
        elif d in work:
            return 0
        elif d1.weekday() in (5, 6):
            return 1
        else:
            return 0


if __name__ == "__main__":
    day = "2018-10-14"
    print(holiday(day))

 

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