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Alexander Belopolsky wrote:
> Alexander Belopolsky <belopolsky@users.sourceforge.net> added the comment:
> 
> This has nothing to do with set.update, the difference is due to the
> time to setup the generator:
> 
> $ python -m timeit -s 'x = set(range(10000)); y = []' 'x.update(y)'
> 1000000 loops, best of 3: 0.38 usec per loop
> $ python -m timeit -s 'x = set(range(10000)); y = (i for i in [])'
> 'x.update(y)'
> 1000000 loops, best of 3: 0.335 usec per loop
> 
> ----------
> nosy: +belopolsky
> 
> __________________________________
> Tracker <report@bugs.python.org>
> <http://bugs.python.org/issue2672>
> __________________________________
> 

That is true, though if I just force a generator overhead:

% python -m timeit -s 'x = set(range(10000)); y = []' 'x.update(y)'
1000000 loops, best of 3: 0.204 usec per loop
% python -m timeit -s 'x = set(range(10000)); y = (i for i in [])'
'x.update(y)'
10000000 loops, best of 3: 0.173 usec per loop
% python -m timeit -s 'x = set(range(10000)); l = []' 'x.update(i for i
in l)'
1000000 loops, best of 3: 0.662 usec per loop
 python -m timeit -s 'x = set(range(10000)); l = []; y = (i for i in l)'
'(i for i in l); x.update(y)'
1000000 loops, best of 3: 1.87 usec per loop

So if you compare consuming a generator multiple times to creating it
each time, it is 0.662 usec - 0.173 usec = 0.489 usec to create a generator.

So why does: "(i for i in l); x.update(y)" take an additional 1.208 usec.

(I'm certainly willing to believe that set.update() is generator/list
agnostic, but something weird is still happening.)

John
=:->

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