Find the closest hour

I have a list with these items:

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

Assuming that now it's 20:18, how can I get the '20:10' item from list? I want to use this to find the current r开发者_JAVA技巧unning show in a TV Guide.

>>> 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'

easy but dirty way

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

I'm not a Python programmer, but I'd use the following algorithm:

1. Convert everything to "minutes after midnight", e.g. hours = [1170 (= 19*60+30), 1210, ...], currenttime = 1218 (= 20*60+18).

2. Then just loop thorugh hours and find the last entry which is smaller than currenttime.

You can use functions in the time module; time.strptime() allows you to parse a string into a time-tuple, then time.mktime() converts this to seconds. You can then simply compare all items in seconds, and find the smallest difference.

@katrielalex & Tim

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

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] 

Hope this can help you :)