Converting integer Seconds values into datetime. Python

Question

I'm trying to figure out the easiest way to automate the conversion of an array of seconds into datetime. I'm very familiar with converting the seconds from 1970 into datetime, but the values that I have here are for the seconds elapsed in a given day. For example, 14084 is the number if seconds that has passed on 2011,11,11, and I was able to generate the datetime below.

str(dt.timedelta(seconds = 14084))
Out[245]: '3:54:44'

dt.datetime.combine(date(2011,11,11),time(3,54,44))
Out[250]: datetime.datetime(2011, 11, 11, 3, 54, 44)

Is there a faster way of conversion for an array.

Why don't you just do datetime(2011,11,11) + timedelta(seconds=14084)? — Selcuk
– Selcuk, Commented Sep 2, 2020 at 22:59

Michael · Accepted Answer · 2020-09-02 23:08:43Z

1

numpy has support for arrays of datetimes with a timedelta type for manipulating them: https://numpy.org/doc/stable/reference/arrays.datetime.html

e.g. you can do this:

import numpy as np
date_array = np.arange('2005-02', '2005-03', dtype='datetime64[D]')
date_array += np.timedelta64(4, 's')  # Add 4 seconds

If you have an array of seconds, you could convert it into an array of timedeltas and add that to a fixed datetime

answered Sep 2, 2020 at 23:08

Michael

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RichieV · Accepted Answer · 2020-09-02 23:07:57Z

0

Say you have

seconds = [14084, 14085, 15003]

You can use pandas

import pandas as pd
series = pd.to_timedelta(seconds, unit='s') + pd.to_datetime('2011-11-11')
series = series.to_series().reset_index(drop=True)

print(series)
0   2011-11-11 03:54:44
1   2011-11-11 03:54:45
2   2011-11-11 04:10:03
dtype: datetime64[ns]

Or a list comprehension

list_comp = [datetime.datetime(2011, 11, 11) +
    datetime.timedelta(seconds=s) for s in seconds]

print(list_comp)
[datetime.datetime(2011, 11, 11, 3, 54, 44), datetime.datetime(2011, 11, 11, 3, 54, 45), datetime.datetime(2011, 11, 11, 4, 10, 3)]

answered Sep 2, 2020 at 23:07

RichieV

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