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简洁的方法来做这个。这个方法有效......它看起来很丑陋，为每个日期创建一个新的时间变量似乎很愚蠢，应该有一种方法可以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，但我无法在互联网或文档上找到任何看起来可行的文件。

Answer 1

使用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

您也可以使用Arrow：

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

Answer 4

使用钟摆模块

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

Python：给出给定日期的开始和结束周数据

4 个答案: