Time difference in seconds from numpy.timedelta64

Question

How to get time difference in seconds from numpy.timedelta64 variable?

time1 = \'2012-10-05 04:45:18\'
time2 = \'2012-10-05 04:44:13\'
dt = np.datetime64(tim         


        

Answer 1

To get number of seconds from numpy.timedelta64() object using numpy 1.7 experimental datetime API:

seconds = dt / np.timedelta64(1, 's')

Answer 2

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.