If/then style question

[Steven D'Aprano]

On Sat, 18 Dec 2010 12:29:31 +0100, Francesco wrote:

[...]
> I agree to your point, but I'm afraid you chose a wrong example (AFAIK,
> and that's not much). Sure, the second version of function(arg) is much
> more readable, but why do you think the first one would do "*lots*  of
> unnecessary work"?
> All the overhead in that function would be:
>    if some_condition, three IF tests, and you know that's NOT a lot! 

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.



-- 
Steven