Extracting Days from a Numpy.Timedelta64 Value

extracting days from a numpy.timedelta64 value

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.

how to extract days as integers from a timedelta64[ns] object in python

This should convert your timedelta64[ns] type to float64 representing days:

data['difference'].astype('timedelta64[D]')

how to convert timedelta64 to number of days integer?

For a single value, you can use Timedelta.days

td = pd.to_timedelta('146 days 12:37:08.787241')
td.days
146

However, for a pandas Series of Timedeltas you'll need the dt accessor

my_pd_series.dt.days

How to get the number value from timedelta64

use parameter ms inside np.datetime64(datetime,'ms') as second then do subtraction

first_date = np.datetime64("2013-04-08 15:52:17.207",'ms')
last_date = np.datetime64("2013-05-29 10:44:44",'ms')
a=last_date-first_date
b=a.astype('int')
type(b)

Time difference in seconds from numpy.timedelta64

You can access it through the "wrapped" datetime item:

>>> dt.item().total_seconds()
65.0

Explanation: here dt is an array scalar in numpy, which is a zero rank array or 0-dimensional array. So you will find the dt here also has all the methods an ndarray possesses, and you can do for example dt.astype('float'). But it wraps a python object, in this case a datetime.timedelta object.

To get the original scalar you can use dt.item(). To index the array scalar you can use the somewhat bizarre syntax of getitem using an empty tuple:

>>> dt[()]
array(datetime.timedelta(0, 65), dtype='timedelta64[s]')

This should work in all versions of numpy, but if you are using numpy v1.7+ it may be better to use the newer numpy datetime API directly as explained in the answer from J.F. Sebastien here.

How to convert numpy.timedelta64 to minutes

Use array.astype() to convert the type of an array safely:

>>> import numpy as np
>>> a = np.timedelta64(1620000000000,'ns')
>>> a.astype('timedelta64[m]')
numpy.timedelta64(27,'m')

How to work with `numpy.timedelta64` outside of pandas/numpy?

When working with datetime64 dtype arrays, tolist() or item() do a good job of converting the array to base Python objects. Let's try that with your timedelta:

In [174]: x=np.timedelta64(-2700000000000,'ns')
In [175]: x.item()
Out[175]: -2700000000000

Not quite what we want. But if I first convert it to minutes:

In [176]: x.astype('timedelta64[m]')
Out[176]: numpy.timedelta64(-45,'m')
In [177]: x.astype('timedelta64[m]').item()
Out[177]: datetime.timedelta(days=-1, seconds=83700)

getting seconds from numpy timedelta64

Your own answer is correct and good. Slightly different way is to specify scale constants with timedelta expression.

For example, to scale to seconds:

>>> np.diff(index)/np.timedelta64(1, 's')
array([ 3.6139351 ,  3.39279693,  1.87199821])

To minutes:

>>> np.diff(index)/np.timedelta64(1, 'm')
array([ 0.06023225,  0.05654662,  0.03119997])

Get year, month or day from numpy datetime64

As datetime is not stable in numpy I would use pandas for this:

In [52]: import pandas as pd

In [53]: dates = pd.DatetimeIndex(['2010-10-17', '2011-05-13', "2012-01-15"])

In [54]: dates.year
Out[54]: array([2010, 2011, 2012], dtype=int32)

Pandas uses numpy datetime internally, but seems to avoid the shortages, that numpy has up to now.

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