距离上周三返回日期较短

Question

我想用Python返回上周三的日期。但是，如果今天是星期三，那么它仍然应该返回上星期三的日期，而不是今天的日期。现在，我使用以下代码：

today = date.today()
if today.weekday() == 2:
last_wednesday = today - timedelta(weeks=1)
else:
last_wednesday = today - timedelta(days=today.weekday()) + timedelta(days=2)
return last_wednesday

由于我来自Java，我习惯了一种更短的方式：return LocalDate.now().with(TemporalAdjusters.previous(DayOfWeek.WEDNESDAY));

所以，我的问题是：在Python中，是否也有类似的方法只需一行就可以实现这一点？

编辑：为了澄清我的问题，函数不应该返回上周星期三的日期，而应该返回"星期三"的日期；最新的"；星期三所以，也就是说，我想要的：

2021-03-22 (Monday) --> 2021-03-17 (Last week's wednesday)
2021-03-23 (Tuesday) --> 2021-03-17 (Last week's wednesday)
2021-03-24 (Wednesday) --> 2021-03-17 (Last week's wednesday)
2021-03-25 (Thursday) --> 2021-03-24 (This week's wednesday)
2021-03-26 (Friday) --> 2021-03-24 (This week's wednesday)

Answer 1

您可以这样做：

def last_wednesday(date):
return date - datetime.timedelta((4 + date.weekday())%7 + 1)

这将在任何日期返回前一个星期三，而不会返回日期本身。

Answer 2

略低于@mousetail

day - timedelta(days=5 + day.weekday())

测试功能。。

from datetime import date, timedelta
today = date.today()
for x in range(-4, 4):
day = today + timedelta(days=x)
print(day, day - timedelta(days=5 + day.weekday()))

2021-03-20 2021-03-10
2021-03-21 2021-03-10
2021-03-22 2021-03-17
2021-03-23 2021-03-17
2021-03-24 2021-03-17
2021-03-25 2021-03-17
2021-03-26 2021-03-17
2021-03-27 2021-03-17