[issue2672] speed of set.update(

[Alexander Belopolsky]

Alexander Belopolsky <belopolsky at users.sourceforge.net> added the comment:

On Thu, Apr 24, 2008 at 2:23 PM, John Arbash Meinel
<report at bugs.python.org> wrote:
..
>  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.)

I've seen a similar strangeness in timings:

$ python -m timeit '(i for i in [])'
100000 loops, best of 3: 4.16 usec per loop

but

$ python -m timeit -s 'x = set()' 'x.update(i for i in [])'
1000000 loops, best of 3: 1.31 usec per loop

on the other hand,

$ python -m timeit -s 'x = []' 'x.extend(i for i in [])'
100000 loops, best of 3: 4.54 usec per loop

How can x.update(i for i in []) take *less* time than simply creating a genexp?

Note that there is no apparent bytecode tricks here:

  1           0 LOAD_CONST               0 (<code object <genexpr> at
0xf7e88920, file "<stdin>", line 1>)
              3 MAKE_FUNCTION            0
              6 BUILD_LIST               0
              9 GET_ITER
             10 CALL_FUNCTION            1
             13 RETURN_VALUE
>>> dis(lambda:x.update(i for i in []))
  1           0 LOAD_GLOBAL              0 (x)
              3 LOAD_ATTR                1 (update)
              6 LOAD_CONST               0 (<code object <genexpr> at
0xf7e88920, file "<stdin>", line 1>)
              9 MAKE_FUNCTION            0
             12 BUILD_LIST               0
             15 GET_ITER
             16 CALL_FUNCTION            1
             19 CALL_FUNCTION            1
             22 RETURN_VALUE

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