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问题:
What is the proper way to compare 2 times in Python in order to speed test a section of code? I tried reading the API docs. I'm not sure I understand the timedelta thing.
So far I have this code:
from datetime import datetime
tstart = datetime.now()
print t1
# code to speed test
tend = datetime.now()
print t2
# what am I missing?
# I'd like to print the time diff here
回答1:
datetime.timedelta
is just the difference between two datetimes ... so it's like a period of time, in days / seconds / microseconds
>>> a = datetime.datetime.now()
>>> b = datetime.datetime.now()
>>> c = b - a
>>> c
datetime.timedelta(0, 4, 316543)
>>> c.days
0
>>> c.seconds
4
>>> c.microseconds
316543
Be aware that c.microseconds
only returns the microseconds portion of the timedelta! For timing purposes always use c.total_seconds()
.
You can do all sorts of maths with datetime.timedelta, eg:
>>> c / 10
datetime.timedelta(0, 0, 431654)
It might be more useful to look at CPU time instead of wallclock time though ... that's operating system dependant though ... under Unix-like systems, check out the 'time' command.
回答2:
Since Python 2.7 there's the timedelta.total_seconds() method. So, to get the elapsed milliseconds:
>>> import datetime
>>> a = datetime.datetime.now()
>>> b = datetime.datetime.now()
>>> delta = b - a
>>> print delta
0:00:05.077263
>>> int(delta.total_seconds() * 1000) # milliseconds
5077
回答3:
You might want to use the timeit module instead.
回答4:
You could also use:
import time
start = time.clock()
do_something()
end = time.clock()
print "%.2gs" % (end-start)
Or you could use the python profilers.
回答5:
I know this is late, but I actually really like using:
start = time.time()
##### your timed code here ... #####
print "Process time: " + (time.time() - start)
time.time()
gives you seconds since the epoch. Because this is a standardized time in seconds, you can simply subtract the start time from the end time to get the process time (in seconds). time.clock()
is good for benchmarking, but I have found it kind of useless if you want to know how long your process took. For example, it's much more intuitive to say "my process takes 10 seconds" than it is to say "my process takes 10 processor clock units"
>>> start = time.time(); sum([each**8.3 for each in range(1,100000)]) ; print (time.time() - start)
3.4001404476250935e+45
0.0637760162354
>>> start = time.clock(); sum([each**8.3 for each in range(1,100000)]) ; print (time.clock() - start)
3.4001404476250935e+45
0.05
In the first example above, you are shown a time of 0.05 for time.clock() vs 0.06377 for time.time()
>>> start = time.clock(); time.sleep(1) ; print "process time: " + (time.clock() - start)
process time: 0.0
>>> start = time.time(); time.sleep(1) ; print "process time: " + (time.time() - start)
process time: 1.00111794472
In the second example, somehow the processor time shows "0" even though the process slept for a second. time.time()
correctly shows a little more than 1 second.
回答6:
You could simply print the difference:
print tend - tstart
回答7:
The following code should display the time detla...
from datetime import datetime
tstart = datetime.now()
# code to speed test
tend = datetime.now()
print tend - tstart
回答8:
time.time() / datetime is good for quick use, but is not always 100% precise. For that reason, I like to use one of the std lib profilers (especially hotshot) to find out what's what.
回答9:
You may want to look into the profile modules. You'll get a better read out of where your slowdowns are, and much of your work will be full-on automated.
回答10:
Here is a custom function that mimic's Matlab's/Octave's tic
toc
functions.
Example of use:
time_var = time_me(); # get a variable with the current timestamp
... run operation ...
time_me(time_var); # print the time difference (e.g. '5 seconds 821.12314 ms')
Function :
def time_me(*arg):
if len(arg) != 0:
elapsedTime = time.time() - arg[0];
#print(elapsedTime);
hours = math.floor(elapsedTime / (60*60))
elapsedTime = elapsedTime - hours * (60*60);
minutes = math.floor(elapsedTime / 60)
elapsedTime = elapsedTime - minutes * (60);
seconds = math.floor(elapsedTime);
elapsedTime = elapsedTime - seconds;
ms = elapsedTime * 1000;
if(hours != 0):
print ("%d hours %d minutes %d seconds" % (hours, minutes, seconds))
elif(minutes != 0):
print ("%d minutes %d seconds" % (minutes, seconds))
else :
print ("%d seconds %f ms" % (seconds, ms))
else:
#print ('does not exist. here you go.');
return time.time()
回答11:
I am not a Python programmer, but I do know how to use Google and here's what I found: you use the "-" operator. To complete your code:
from datetime import datetime
tstart = datetime.now()
# code to speed test
tend = datetime.now()
print tend - tstart
Additionally, it looks like you can use the strftime() function to format the timespan calculation in order to render the time however makes you happy.
回答12:
You could use timeit like this to test a script named module.py
$ python -mtimeit -s 'import module'