Question

我有一个tuple列表，每次间隔15分钟，持续24小时。现在我需要在工作时间之外删除时间。

我的时代：

times = []
for i in range(0, 24 * 4):
    times.append((datetime.combine(date.today(), time()) + timedelta(minutes=15) * i).time())

营业时间：

time_open = '08:00:00'
time_close = '17:00:00'

如何在营业时间之外从列表中删除时间？

Answer 1

如果我明白你想要什么，这是一个快速而肮脏的解决方案。您可以在追加到列表之前检查业务开放/关闭时间之间的时间。

from datetime import datetime as d, timedelta, time
import datetime as o_d

time_open = '08:00:00'
time_close = '17:00:00'
times = []
for i in range(0, 24 * 4):
    date = (d.combine(o_d.date.today(), time()) + timedelta(minutes=15) * i).time()
    date_str = date.strftime("%H:%M:%S")
    if time_open <= date_str <= time_close:
        times.append(date_str)
        #times.append(date)
print(times)

根据评论进行编辑：如果您想要一个通常称为“军事”时间的元组表示，而不是大多数人习惯的时间，即上午/下午，您可以执行以下操作

from datetime import datetime as d, timedelta, time
import datetime as o_d

time_open = '08:00:00'
time_close = '17:00:00'
times = []
for i in range(0, 24 * 4):
    date = (d.combine(o_d.date.today(), time()) + timedelta(minutes=15) * i).time()
    date_str = date.strftime("%H:%M:%S")
    if time_open <= date_str <= time_close:

        #If you want a list of just standard times
        #times.append(date_str)

        #If you want the time representation(s) here and no longer need the 
        #the datetime object

        times.append((date_str, date.strftime("%I:%M %p"))

        #If you still need the datetime object append this do list comprehension below in comment with #options = ....
        #times.append(date)
print(times)

#assuming you have appended date above see comment
#options = [( c.strftime("%H:%M:%S"), c.strftime("%I:%M %p")) for c in times]

Python根据营业时间从时间列表中删除

1 个答案: