Time difference in seconds (as a floating point)

Question

>>> from datetime import datetime
>>> t1 = datetime.now()
>>> t2 = datetime.now()
>>> delta = t2 - t1
>>> delta.seconds
7
>>> delta.microseconds
631000

Is there any way to get that as 7.631000 ? I can use time module, but I also need that t1 and t2 variables as DateTime objects. So if there is an easy way to do it with datettime, that would be great. Otherwise it'll look ugly:

t1 = datetime.now()
_t1 = time.time()
t2 = datetime.now()
diff = time.time() - _t1

score 50 · Answer 1 · edited Dec 18 '18 at 06:35

50

for newer version of Python (Python 2.7+ or Python 3+), you can also use the method total_seconds:

from datetime import datetime
t1 = datetime.now()
t2 = datetime.now()
delta = t2 - t1
print(delta.total_seconds())

edited Dec 18 '18 at 06:35

Miraj50

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answered Jun 17 '14 at 09:21

daniel kullmann

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score 24 · Accepted Answer · answered May 21 '10 at 08:58

24

combined = delta.seconds + delta.microseconds/1E6

answered May 21 '10 at 08:58

Edward Dale

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or combined = delta.seconds + (float(1) / delta.microseconds) – pocoa May 21 '10 at 16:22
1

@pocoa - this is actually an incorrect conversion. 1/time is a rate (Hz) which really doesn't make sense here. this also clearly provides a different result than the accepted answer on which you commented. – underrun Oct 07 '16 at 15:31

score 7 · Answer 3 · edited May 21 '10 at 09:23

7

I don't know if there is a better way, but:

((1000000 * delta.seconds + delta.microseconds) / 1000000.0)

or possibly:

"%d.%06d"%(delta.seconds,delta.microseconds)

edited May 21 '10 at 09:23

Mike Graham

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answered May 21 '10 at 08:59

Douglas Leeder

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Time difference in seconds (as a floating point)

3 Answers3

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