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.

Use dt.days to obtain the days attribute as integers.

For eg:

In [14]: s = pd.Series(pd.timedelta_range(start='1 days', end='12 days', freq='3000T'))

In [15]: s
Out[15]: 
0    1 days 00:00:00
1    3 days 02:00:00
2    5 days 04:00:00
3    7 days 06:00:00
4    9 days 08:00:00
5   11 days 10:00:00
dtype: timedelta64[ns]

In [16]: s.dt.days
Out[16]: 
0     1
1     3
2     5
3     7
4     9
5    11
dtype: int64

More generally - You can use the .components property to access a reduced form of timedelta.

In [17]: s.dt.components
Out[17]: 
   days  hours  minutes  seconds  milliseconds  microseconds  nanoseconds
0     1      0        0        0             0             0            0
1     3      2        0        0             0             0            0
2     5      4        0        0             0             0            0
3     7      6        0        0             0             0            0
4     9      8        0        0             0             0            0
5    11     10        0        0             0             0            0

Now, to get the hours attribute:

In [23]: s.dt.components.hours
Out[23]: 
0     0
1     2
2     4
3     6
4     8
5    10
Name: hours, dtype: int64

Suppose you have a timedelta series:

import pandas as pd
from datetime import datetime
z = pd.DataFrame({'a':[datetime.strptime('20150101', '%Y%m%d')],'b':[datetime.strptime('20140601', '%Y%m%d')]})

td_series = (z['a'] - z['b'])

One way to convert this timedelta column or series is to cast it to a Timedelta object (pandas 0.15.0+) and then extract the days from the object:

td_series.astype(pd.Timedelta).apply(lambda l: l.days)

Another way is to cast the series as a timedelta64 in days, and then cast it as an int:

td_series.astype('timedelta64[D]').astype(int)