extracting days from a numpy.timedelta64 value
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)
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
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.