使用python datetime在早上9点到晚上6点之间划分时间到3小时的时间间隔

Question

Basically, what I am trying to do is, divide the time between morning 9 AM and evening 6 PM into specified time hours interval

For example, if I specify the time gap interval as 3 hours then I should get the result by dividing the time between 9 to 6 as

['2017-05-03 09:00', '2017-05-03 12:00', '2017-05-03 15:00', '2017-05-03 18:00']

If I specify the time gap interval as 2 hours then I should get

['2017-05-03 09:00', '2017-05-03 11:00', '2017-05-03 13:00', '2017-05-03 15:00', '2017-05-03 17:00']

Below is my code for doing it

from datetime import datetime
from datetime import timedelta
start_time = datetime.now().replace(hour=9, minute=0) #(Morning 9 AM)
end_time = start_time + timedelta(hours=9) # (Evening 6 PM)

result = []
time_interval = 3
while start_time <=end_time:
    result.append(start_time)
    start_time += timedelta(hours=time_interval)

print(result)

Basically, I used while in the above code to check condition, so is it possible to do the same above logic without using while statement and by using some other logic using list comprehension etc .,?

Also if possible I need the output list as

[
'2017-05-03 09:00 to 2017-05-03 12:00', 
'2017-05-03 12:00 to 2017-05-03 15:00', 
'2017-05-03 15:00 to 2017-05-03 18:00'
]

Answer 1

Alternatively, you can solve the problem with rrule() from dateutil to generate the datetimes in a specified range hourly and then iterating over the results with a specified step :

from datetime import datetime
from datetime import timedelta
from itertools import islice

from dateutil.rrule import HOURLY, rrule

interval = 3
start_time = datetime.now().replace(hour=9, minute=0) #(Morning 9 AM)
end_time = start_time + timedelta(hours=9) # (Evening 6 PM)

stops = islice(rrule(freq=HOURLY, dtstart=start_time, until=end_time), None, None, interval)
print([datetime.strftime(stop, "%Y-%d-%m %H:%M") for stop in stops])

Prints:

['2017-05-03 09:00', '2017-05-03 12:00', '2017-05-03 15:00', '2017-05-03 18:00']

Note that the list comprehension and the strftime() calls here are for demonstration purposes only - stops is an iterator that "contains" the desired datetimes .

And, applying the same pairwise helper as @NiL used, we can get to your next desired output:

from datetime import datetime
from datetime import timedelta
from itertools import islice, tee

from dateutil.rrule import HOURLY, rrule


def pairwise(iterable):
    """s -> (s0,s1), (s1,s2), (s2, s3), ...

    https://docs.python.org/2/library/itertools.html
    """
    a, b = tee(iterable)
    next(b, None)
    return zip(a, b)


interval = 3
start_time = datetime.now().replace(hour=9, minute=0) #(Morning 9 AM)
end_time = start_time + timedelta(hours=9) # (Evening 6 PM)

stops = islice(rrule(freq=HOURLY, dtstart=start_time, until=end_time), None, None, interval)

print(['{0} to {1}'.format(datetime.strftime(start, "%Y-%d-%m %H:%M"),
                           datetime.strftime(stop, "%Y-%d-%m %H:%M"))
       for start, stop in pairwise(stops)])

Prints:

['2017-03-05 09:00 to 2017-03-05 12:00', '2017-03-05 12:00 to 2017-03-05 15:00', '2017-03-05 15:00 to 2017-03-05 18:00']

Answer 2

something like this might do what you want

from datetime import datetime
from datetime import timedelta
from itertools import tee
start_time = datetime.now().replace(hour=9, minute=0)  # (Morning 9 AM)
end_time = start_time + timedelta(hours=9)  # (Evening 6 PM)

time_interval = 3
delta = 9  # hours


def pairwise(iterable):
    """s -> (s0,s1), (s1,s2), (s2, s3), ...

    https://docs.python.org/2/library/itertools.html
    """
    a, b = tee(iterable)
    next(b, None)
    return zip(a, b)


iterations = (delta // time_interval) + 1
result = [start_time+i*timedelta(hours=time_interval)
          for i in range(iterations)]
print(result)

formatted=['{} to {}'.format(
    start.strftime('%Y-%m-%d %H:%M'),
    end.strftime('%Y-%m-%d %H:%M'))
    for start, end in pairwise(result)]

from pprint import pprint
pprint(formatted)