计算某年某月的第三个周五日期

一本正经胡说八道的猫

因为数据处理需要，有这个需求。
用两种方法，涉及“datetime.timedelta”和“dateutil.relativedelta.relativedelta”的操作。
查看了区别：

Definition:   datetime.timedelta([days[, seconds[, microseconds[, 
milliseconds[, minutes[, hours[, weeks]]]]]]])

Definition:   relativedelta.relativedelta(self, dt1=None, dt2=None,
years=0, months=0, days=0, leapdays=0, weeks=0, hours=0, minutes=0,
seconds=0, microseconds=0, year=None, month=None, day=None,
weekday=None, yearday=None, nlyearday=None, hour=None, minute=None,
second=None, microsecond=None)

方法一：
用datetime.weekday加 datetime.timedelta
先用for循环算出月头哪天为周五，然后在该日期基础上加两周得到日期

import datetime
def pay_day(y, m, weekday):
    for d in range(1, 8):
        if datetime.date(int(y), int(m), d).weekday() == weekday:
            date = datetime.date(int(y), int(m), d)    
    return  date + datetime.timedelta(weeks=2)
 #return  date + datetime.timedelta(days=14)加14天也可以

显示结果：

[In]
for i in range(1,13):
    print(pay_day(2022,i,4))
[Out]
2022-01-21
2022-02-18
2022-03-18
2022-04-15
2022-05-20
2022-06-17
2022-07-15
2022-08-19
2022-09-16
2022-10-21
2022-11-18
2022-12-16

方法二：
用dateutil.relativedelta.relativedelta

def pay_days(y, m, x):
    return datetime.date(y,m, 1)+relativedelta.relativedelta(day = 0, weekday = relativedelta.FR(x))

[In]
for i in range(1,13):
    print(pay_days(2022,i,3))
[Out]
2022-01-21
2022-02-18
2022-03-18
2022-04-15
2022-05-20
2022-06-17
2022-07-15
2022-08-19
2022-09-16
2022-10-21
2022-11-18
2022-12-16

两种方法都能快速算出，用时都是6ms。
不过方法二的代码更加简洁，我更喜欢用。