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time in notebook

asked 2010-12-06 08:35:27 +0100


Could you help me? I need to know the time of a sage program written in the notebook, but when I use the timeit command, it doesn't work if I use "for" iterators in the program.

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answered 2010-12-06 08:51:08 +0100

kcrisman gravatar image

Did you try %timeit as the first line in the cell? I'm not sure whether that would work.

Another option is to write a dummy program that calls your other program, like def foo(): return my_program(n) and then do timeit on foo.

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answered 2010-12-08 11:20:08 +0100

v_2e gravatar image

In my case the following syntax works:

for i in range(0,10):

The working example:

time print("The first line took:")
time print("\nThe second line took:")
for i in range(0,10):
print("\nAltogether took:")

The output should look like this:

The first line took:

Time: CPU 0.00 s, Wall: 0.00 s

The second line took:

Time: CPU 0.00 s, Wall: 0.00 s

Altogether took:

CPU time: 0.01 s, Wall time: 10.04 s

Hope, that helps. :)

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However, this doesn't do what timeit does, which is to do it over several iterations. For instance, in the first line we really want `sage: timeit('print("The first line took:")') 625 loops, best of 3: 734 ns per loop`

kcrisman gravatar imagekcrisman ( 2010-12-08 15:43:24 +0100 )edit

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Asked: 2010-12-06 08:35:27 +0100

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Last updated: Dec 08 '10