Microsecond accurate timestamp in python?

Question

Is there a way to maintain microsecond accuracy when converting a Python datetime to a timestamp?

>>> import datetime
>>> d1 = datetime.datetime(2013,7,31,9,13,8,829)
>>> import time
>>> d1_ts = time.mktime(d1.timetuple())
>>> d1
datetime.datetime(2013, 7, 31, 9, 13, 8, 829)
>>> d1_ts
1375279988.0
>>> d1.fromtimestamp(d1_ts)
datetime.datetime(2013, 7, 31, 9, 13, 8)

I lose that .829 on the conversion. This is rather important, because I have start and end timestamps that I need to step through at set intervals (with sub-second steps) to gather data from some sensors.

Eventually it will be used in a function similar to this:

from scipy import arange
sample_time = 0.02
for i in arange(d1_ts, d2_ts, sample_time):
    # do stuff

With a sample_time that small, the mircoseconds are important.

Answer 1

import datetime as DT
d1 = DT.datetime(2013, 7, 31, 9, 13, 8, 829)
epoch = DT.datetime(1970, 1, 1)

print(d1.timestamp()) #  python3
# 1375276388.000829
print((d1 - epoch).total_seconds())  # python2
# 1375261988.000829

Also note that if you are using NumPy 1.7 or newer, you could use np.datetime64:

In [23]: x = np.datetime64(d1)

In [24]: x.view('<i8')/1e6
Out[24]: 1375261988.000829

In [38]: x.astype('<i8').view('<M8[us]')
Out[38]: numpy.datetime64('2013-07-31T05:13:08.000829-0400')

In [40]: x.astype('<i8').view('<M8[us]') == x
Out[40]: True

Sinces the np.datetime64 provides an easy way to convert from dates to 8-byte ints, they can be very convenient for doing arithmetic as ints and then converting to dates.