timeit.timeit的返回值是平均值还是最佳值？

Question

the doc tells us clearly that timeit Command-Line Interface outputs the best.

python -m timeit '"-".join(map(str, range(100)))'
10000 loops, best of 3: 25.2 usec per loop

how about the Python Interface?

>>> import timeit
>>> timeit.timeit('char in text', setup='text = "sample string"; char = "g"')
0.41440500499993504

is it still the best?

Answer 1

The command line sets the repeat option to 3:

 -r N, --repeat=N 

how many times to repeat the timer (default 3)

and 'the best' is the best out of those 3. That's different from the number parameter, which is determined automatically for you when you don't set the -n / --number parameter.

The timeit.timeit() function, on the other hand, doesn't repeat . It runs the statement number of times, and gives you the total time . From the timeit.timeit() documentation :

Create a Timer instance with the given statement, setup code and timer function and run its timeit() method with number executions.

and from timeit.Timer.timeit() :

Time number executions of the main statement. This executes the setup statement once, and then returns the time it takes to execute the main statement a number of times , measured in seconds as a float.

If you want to get a best out of result, use the timeit.repeat() function :

  timeit.repeat(stmt='pass', setup='pass', timer=<default timer>, repeat=3, number=1000000) 

Create a Timer instance with the given statement, setup code and timer function and run its repeat() method with the given repeat count and number executions.

Using timeit.repeat() will not auto-range the number for you however. For that you'd have to create your own Timer() instance.

The documentation links to the implementation , so you can see how this is done in the main() function; simplifying this down to the code that is executed when you use the default options:

t = Timer(stmt, setup, timer)
repeat = 3
for i in range(1, 10):
    number = 10**i
    x = t.timeit(number)
    if x >= 0.2:
        break
r = t.repeat(repeat, number)
best = min(r)
print "%d loops," % number,
usec = best * 1e6 / number
if usec < 1000:
    print "best of %d: %.*g usec per loop" % (repeat, 3, usec)
else:
    msec = usec / 1000
    if msec < 1000:
        print "best of %d: %.*g msec per loop" % (repeat, 3, msec)
    else:
        sec = msec / 1000
        print "best of %d: %.*g sec per loop" % (repeat, 3, sec)

In Python 3, the above has been much improved with the new Timer.autorange() method and a better handling of scales.

Demo using your statement and setup:

>>> import timeit
>>> t = timeit.Timer('char in text', setup='text = "sample string"; char = "g"')
>>> repeat = 3
>>> for i in range(1, 10):
...     number = 10**i
...     x = t.timeit(number)
...     if x >= 0.2:
...         break
...
>>> r = t.repeat(repeat, number)
>>> best = min(r)
>>> print "%d loops," % number,
10000000 loops,
>>> usec = best * 1e6 / number
>>> if usec < 1000:
...     print "best of %d: %.*g usec per loop" % (repeat, 3, usec)
... else:
...     msec = usec / 1000
...     if msec < 1000:
...         print "best of %d: %.*g msec per loop" % (repeat, 3, msec)
...     else:
...         sec = msec / 1000
...         print "best of %d: %.*g sec per loop" % (repeat, 3, sec)
...
best of 3: 0.0305 usec per loop

Answer 2

From the manual : It runs the main statement number times (default is 1000000 ), and returns the sum of time it took to execute all number times.

you can add number=10**8 to your call and see how the result changes:

>>> import timeit
>>> timeit.timeit('char in text', setup='text = "sample string"; char = "g"')
0.03136014938354492
>>> timeit.timeit('char in text', setup='text = "sample string"; char = "g"', number=10**7)
0.22713899612426758
>>> timeit.timeit('char in text', setup='text = "sample string"; char = "g"', number=10**8)
2.130625009536743
>>>