Question

我有一个包含这些项目的清单：

hours = ['19:30', '20:10', '20:30', '21:00', '22:00']

假设现在是20:18，我如何从列表中获得'20：10'项目？我想用它来查找电视指南中当前正在播放的节目。

Answer 1

>>> import datetime
>>> hours = ['19:30', '20:10', '20:30', '21:00', '22:00']
>>> now = datetime.datetime.strptime("20:18", "%H:%M")
>>> min(hours, key=lambda t: abs(now - datetime.datetime.strptime(t, "%H:%M")))
'20:10'

Answer 2

简单但又脏的方式

max(t for t in sorted(hours) if t<=now)

Answer 3

我不是Python程序员，但我使用以下算法：

将所有内容转换为“午夜后的分钟”，例如hours = [1170 (= 19*60+30), 1210, ...]，currenttime = 1218 (= 20*60+18)。
然后循环搜索hours，找到最后一个小于currenttime的条目。

Answer 4

您可以使用时间模块中的功能; time.strptime()允许您将字符串解析为时间元组，然后time.mktime()将其转换为秒。然后，您可以在几秒钟内简单地比较所有项目，并找出最小的差异。

Answer 5

import bisect
# you can use the time module like katrielalex answer which a standard library 
# in python, but sadly for me i become an addict to dateutil :)
from dateutil import parser 

hour_to_get = parser.parse('20:18')

hours = ['19:30', '20:10', '20:30', '21:00', '22:00']
hours = map(parser.parse, hours) # Convert to datetime.

hours.sort() # In case the list of hours isn't sorted.

index = bisect.bisect(hours, hour_to_get)

if index in (0, len(hours) - 1):
    print "there is no show running at the moment"
else:
    print "running show started at %s " % hours[index-1]

希望这可以帮助你：）

Answer 6

@katrielalex＆amp;添

import itertools
[x for x in itertools.takewhile( lambda t: now > datetime.datetime.strptime(t, "%H:%M"), hours )][-1]

找到最近的小时

6 个答案: