I'd bet you would stress your point Steven! But you don't need to persuade me,
I do already agree.
I just meant to say that, when the advantage is little, there's no need to
rewrite a working function.
And that with modern CPUs, if tests take so little time, that even some
redundant one is not so much of a nuisance.
in your working example, the "payload" is just a couple of integer calculations, that take very little time too. So the overhead due
to redundant if tests does show clearly. And also in that not-really-real situation, 60% overhead just meant less than 3 seconds.
Just for the sake of discussion, I tried to give both functions some plough to pull, and a worst-case situation too:
>>> t1 = Timer('for x in range(100): print func1(0),',
... 'from __main__ import func1')
>>>
>>> t2 = Timer('for x in range(100): print func2(0),',
... 'from __main__ import func2')
>>>
>>> min(t1.repeat(number=1, repeat=1))
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
-1 -1 -1 -1 -1 -1 -1 -1
53.011015366479114
>>> min(t2.repeat(number=1, repeat=1))
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
-1 -1 -1 -1 -1 -1 -1 -1
47.55442856564332
that accounts for a scant 11% overhead, on more than one million tests per
cycle.
That said, let's make really clear that I would heartily prefer func2 to func1, based both on readability and speed. Thank you for
having spent some time playing with me!
Francesco
On 19/12/2010 1.05, Steven D'Aprano wrote:
Well, let's try it with a working (albeit contrived) example. This is
just an example -- obviously I wouldn't write the function like this in
real life, I'd use a while loop, but to illustrate the issue it will do.
def func1(n):
result = -1
done = False
n = (n+1)//2
if n%2 == 1:
result = n
done = True
if not done:
n = (n+1)//2
if n%2 == 1:
result = n
done = True
if not done:
n = (n+1)//2
if n%2 == 1:
result = n
done = True
if not done:
for i in range(1000000):
if not done:
n = (n+1)//2
if n%2 == 1:
result = n
done = True
return result
def func2(n):
n = (n+1)//2
if n%2 == 1:
return n
n = (n+1)//2
if n%2 == 1:
return n
n = (n+1)//2
if n%2 == 1:
return n
for i in range(1000000):
n = (n+1)//2
if n%2 == 1:
return n
return -1
Not only is the second far more readable that the first, but it's also
significantly faster:
from timeit import Timer
t1 = Timer('for i in range(20): x = func1(i)',
... 'from __main__ import func1')
t2 = Timer('for i in range(20): x = func2(i)',
... 'from __main__ import func2')
min(t1.repeat(number=10, repeat=5))
7.3219029903411865
min(t2.repeat(number=10, repeat=5))
4.530779838562012
The first function does approximately 60% more work than the first, all
of it unnecessary overhead.
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