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

在过去的两天里学到了python,并且想知道是否有更干净的方法来做这个.这个方法有效...它只是看起来很难看而且为每个日期创建一个新的时间变量似乎很愚蠢是一种通过一个简单的调用将给定日期偏移到一周的开始和结束但我无法在互联网上找到任何东西的方法,或者看起来它会起作用的文档.

Answer 1

使用该datetime模块.

这将产生一周的开始和结束(从星期一到星期日):

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

如果要保留标准时间格式并参考当天，则略有变化：

from datetime import datetime, timedelta

today = datetime.now().date()
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

您还可以使用Arrow：

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

Answer 4

使用摆锤模块

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