Time difference in seconds from numpy.timedelta64

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(time1) - np.datetime64(time2)
print dt

0:01:05

I'd like to convert dt to number (int or float) representing time difference in seconds.

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

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

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.

You can simply cast the value to the desired time unit using np.astype, as shown in the example:

timedelta = np.datetime64('2011-07-18')-np.datetime64('2011-07-16')
seconds = timedelta.astype('timedelta64[s]').astype(np.int32)
hours = timedelta.astype('timedelta64[h]').astype(np.int32)