CODEWITHSUNDEEP
pythonpython-3.xpandasnumpy
Benus13
Benus13

Reputation: 91

Generating a list of hour:minute in Python

Im trying to generate a numpy array of minutes from 9:30 to 16:00. (6.5 hours*60 minutes=390 elements)

Is there something similar to:

import pandas as pd
pd.date_range("09:30", "16:00", freq="1min")

Of course after it to convert to numpy is easy...

Upvotes: 3

Views: 1755

Answers (4)

Paul Panzer
Paul Panzer

Reputation: 53089

You could use np.datetime64. But it will insist on being given a year-month-day, as well.

r = np.arange('0000-01-01T09:30', '0000-01-01T16:00', dtype='M8')
r
# array(['0000-01-01T09:30', '0000-01-01T09:31', '0000-01-01T09:32',
#        '0000-01-01T09:33', '0000-01-01T09:34', '0000-01-01T09:35',
#         ...
#        '0000-01-01T15:54', '0000-01-01T15:55', '0000-01-01T15:56',
#        '0000-01-01T15:57', '0000-01-01T15:58', '0000-01-01T15:59'],
#       dtype='datetime64[m]')

You could get rid of the date by subtracting it but the resulting np.timedelta64 does display as minutes.

d = r - r[0].astype('M8[D]')
d
# array([570, 571, 572, 573, 574, 575, 576, 577, 578, 579, 580, 581, 582,
#        583, 584, 585, 586, 587, 588, 589, 590, 591, 592, 593, 594, 595,
#        ...
#        934, 935, 936, 937, 938, 939, 940, 941, 942, 943, 944, 945, 946,
#        947, 948, 949, 950, 951, 952, 953, 954, 955, 956, 957, 958, 959],
#       dtype='timedelta64[m]')

Hours can be extracted, though:

>> d.astype('m8[h]')
# array([ 9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,
#         9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9, 10, 10, 10, 10,
#         ...

#        15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15,
#        15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15],
#       dtype='timedelta64[h]')

Upvotes: 0

norok2
norok2

Reputation: 26906

There is nothing that high-level for date/time management in numpy, but it is not too difficult to have your own with the help of the datetime standard module. One way of doing this would be:

import datetime

def date_range(
        begin_time,
        end_time,
        step_time,
        in_date_fmt='%H:%M',
        out_date_fmt=None,
        upper_bound=False):
    if out_date_fmt is None:
        out_date_fmt = in_date_fmt 

    begin_time = datetime.datetime.strptime(begin_time, in_date_fmt)
    end_time = datetime.datetime.strptime(end_time, in_date_fmt)

    delta_time = (end_time - begin_time)
    origin_time = datetime.datetime.strptime('0', '%S')
    step_time = (
        datetime.datetime.strptime(step_time, in_date_fmt) - origin_time)

    if upper_bound:
        upper_bound = step_time.seconds

    for i in range(0, delta_time.seconds + upper_bound, step_time.seconds):
       yield (
           begin_time + 
           datetime.timedelta(seconds=i)).strftime(out_date_fmt)

Which can be used this way:

my_date_range = np.array(list(date_range('09:30', '16:00', '00:01')))

The docstring is left as an exercise for the reader ;-)

Of course, if you can use pandas you should really use it and not invent Yet Another Wheel™.

Upvotes: 0

jezrael
jezrael

Reputation: 863291

I believe you need DatetimeIndex.strftime:

a = pd.date_range("09:30", "16:00", freq="1min").strftime('%H:%M')
print (a[:10])
['09:30' '09:31' '09:32' '09:33' '09:34' '09:35' '09:36' '09:37' '09:38'
 '09:39']

Upvotes: 4

Connor John
Connor John

Reputation: 433

Try this:

t = np.arange(datetime.datetime(2018,3,12,9,30), datetime.datetime(2018,3,12,16,0), datetime.timedelta(hours=1)).astype(datetime.datetime)

Upvotes: 0

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