python - Python：给出给定日期的开始和结束周数据

Question

day = "13/Oct/2013"
print("Parsing :",day)
day, mon, yr= day.split("/")
sday = yr+" "+day+" "+mon
myday = time.strptime(sday, '%Y %d %b')
Sstart = yr+" "+time.strftime("%U",myday )+" 0"
Send = yr+" "+time.strftime("%U",myday )+" 6"
startweek = time.strptime(Sstart, '%Y %U %w')
endweek = time.strptime(Send, '%Y %U %w')
print("Start of week:",time.strftime("%a, %d %b %Y",startweek))
print("End of week:",time.strftime("%a, %d %b %Y",endweek))
print("Data entered:",time.strftime("%a, %d %b %Y",myday))

out:
Parsing : 13/Oct/2013
Start of week: Sun, 13 Oct 2013
End of week: Sat, 19 Oct 2013
Sun, 13 Oct 2013

在过去的 2 天里学习了 python，想知道是否有更清洁的方法来做到这一点。这种方法有效......它看起来很丑，而且必须为每个日期创建一个新的时间变量似乎很愚蠢，而且应该是一种通过简单的电话将给定日期偏移到一周开始和结束的方法，但我无法在互联网上找到任何看起来可以工作的文档或文档。

Answer 1

使用datetime模块。

这将产生一周的开始和结束（从周一到周日）：

from datetime import datetime, timedelta

day = '12/Oct/2013'
dt = datetime.strptime(day, '%d/%b/%Y')
start = dt - timedelta(days=dt.weekday())
end = start + timedelta(days=6)
print(start)
print(end)

编辑：

print(start.strftime('%d/%b/%Y'))
print(end.strftime('%d/%b/%Y'))

Answer 2

如果您想保持标准时间格式并参考当天，则略有不同：

from datetime import date, timedelta

today = date.today()
start = today - timedelta(days=today.weekday())
end = start + timedelta(days=6)
print("Today: " + str(today))
print("Start: " + str(start))
print("End: " + str(end))

Answer 3

使用摆模块：

today = pendulum.now()
start = today.start_of('week')
end = today.end_of('week')

Answer 4

你也可以使用Arrow：

import arrow
now = arrow.now()
start_of_week = now.floor('week')
end_of_week = now.ceil('week')