python subtracting timedeltas in seconds

Question

I need to subtract timedeltas in python and return the number in just a second form.

I currently have:

from datetime import datetime, timedelta 
q1_start_time = datetime.now()
q1_end_time = datetime.now()
total_time = q1_end_time - q1_start_time
print(total_time)

This returns:

0:00:01.549000

How can I get it to just return 1, or best case scenario it would return 2 ( rounding up ). I don't need 0:00:01 just need 1.

Answer 1

Use:

print(total_time.total_seconds())

It's available for objects of type datetime.timedelta since Python 2.7.

math.ceil() is used to round floats to the next greater/equal integer number. Other rounding functions are math.floor() , and round() . If it's just about beautifying the printed representation of the number, you could also use Python's string formatting functionality, eg,

print "elapsed: %.2f seconds" % (t_end-t_start).total_seconds()

For reference:

• http://docs.python.org/2/library/datetime.html#datetime.timedelta.total_seconds
• http://docs.python.org/2.7/library/functions.html#round
• http://docs.python.org/2/library/math.html#math.ceil
• http://docs.python.org/2/library/math.html#math.floor
• http://docs.python.org/2/library/stdtypes.html#string-formatting-operations