extracting days from a numpy.timedelta64 value

I am using pandas/python and I have two date time series s1 and s2, that have been generated using the 'to_datetime' function on a field of the df containing dates/times.

When I subtract s1 from s2

s3 = s2 - s1

I get a series, s3, of type

timedelta64[ns]

    0    385 days, 04:10:36
    1     57 days, 22:54:00
    2    642 days, 21:15:23
    3    615 days, 00:55:44
    4    160 days, 22:13:35
    5    196 days, 23:06:49
    6     23 days, 22:57:17
    7      2 days, 22:17:31
    8    622 days, 01:29:25
    9     79 days, 20:15:14
    10    23 days, 22:46:51
    11   268 days, 19:23:04
    12                  NaT
    13                  NaT
    14   583 days, 03:40:39

How do I look at 1 element of the series:

s3[10]

I get something like this:

numpy.timedelta64(2069211000000000,'ns')

How do I extract days from s3 and maybe keep them as integers(not so interested in hours/mins etc.)?

Thanks in advance for any help.

You can convert it to a timedelta with a day precision. To extract the integer value of days you divide it with a timedelta of one day.

    >>> x = np.timedelta64(2069211000000000, 'ns')
    >>> days = x.astype('timedelta64[D]')
    >>> days / np.timedelta64(1, 'D')
    23

Or, as @PhillipCloud suggested, just days.astype(int) since the timedelta is just a 64bit integer that is interpreted in various ways depending on the second parameter you passed in ('D', 'ns', ...).

You can find more about it here.

From: stackoverflow.com/q/18215317