Round datetime64 array to nearest second

Question

I have an array of type datetime64[ns]. Each element looks something like '2019-08-30T14:02:03.684000000'. How do I round the values to the nearest second such that I would obtain '2019-08-30T14:02:04' in this example?

I know I can truncate the values by

t = t.astype('datetime64[s]')

but I specifically need to round the values and not truncate them. And the numpy ’round’ function doesn’t seem to like the datetime64[ns] data type.

Asked By: Hypermale

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Source

Answer 1

You can do it by converting np.datetime64 to datetime.datetime.

import numpy as np
from datetime import datetime, timedelta

dt64 = np.datetime64('2019-08-30T14:02:03.684000000')

# np to datetime object
ts = (dt64 - np.datetime64('1970-01-01T00:00:00Z')) / np.timedelta64(1, 's')
dt = datetime.utcfromtimestamp(ts)

# Rounding
if dt.microsecond/1000000 > 0.5:
  date = (dt + timedelta(seconds=1)).replace(microsecond=0)
else:
  date = dt.replace(microsecond=0)

# Datetime to np 
date_rounded = np.datetime64(date).astype('datetime64[s]')

Output:

numpy.datetime64(‘2019-08-30T14:02:04’)

Answered By: Nischal Sanil

Answer 2

Another handy utility for rounding datetime64 by specified interval (‘accuracy’):

from datetime import datetime
from typing import Union
import numpy as np

UNIX_T0 = np.datetime64('1970-01-01T00:00:00')

def round_t64(time: Union[np.datetime64, datetime], dt: Union[np.timedelta64, int]):
    """
    Round timestamp by dt
    """
    if isinstance(dt, int):
        dt = np.timedelta64(dt, 's')

    if isinstance(time, datetime):
        time = np.datetime64(time)

    return time - (time - UNIX_T0) % dt

And use case:

# floor to 1 sec
round_t64(np.datetime64(datetime.now()), np.timedelta64(1, 's'))

# ceil to 1 sec
round_t64(np.datetime64(datetime.now()), -np.timedelta64(1, 's'))

# floor to 5 min
round_t64(np.datetime64(datetime.now()), np.timedelta64(5, 'm'))

Answered By: mde

Round datetime64 array to nearest second

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