On Mon, Jun 18, 2018 at 09:20:25PM -0400, Juancarlo Añez wrote:
> >> For all practical purpose, it would be enough to define that the
> >> expression:
> >>
> >> mylist += [item]
> >>
> >> gets optimized to mylist.append(item).
What if mylist doesn't have an append method?
Just because the variable is *called* "mylist" doesn't mean it actually
is a list. Since Python is dynamically typed, the compiler has no clue
ahead of time whether mylist is a list, so the best it could do is
compile the equivalent of:
if builtins.type(mylist) is builtins.list:
call mylist.append(item)
else:
call mylist.__iadd__([item])
which I suppose it possible, but that's the sort of special-case
optimization which CPython has avoided. (And when it has been tried, has
often been very disappointing.)
Better to move the optimization into list.__iadd__, but doing that still
pays the cost of making a one-item list. Now that's likely to be fast,
but if the aim is to avoid making that one-item list (can you say
"premature optimization"?) then it is a failure.
Besides, it's hardly worthwhile: += is already virtually as fast as
calling append.
[steve@ando ~]$ python3.5 -m timeit -s "L = []" "L.append(1)"
1000000 loops, best of 3: 0.21 usec per loop
[steve@ando ~]$ python3.5 -m timeit -s "L = []" "L += [1]"
1000000 loops, best of 3: 0.294 usec per loop
[...]
> It seems that the optimization is already in place:
Not in 3.5 it isn't.
[steve@ando ~]$ python3.5 -c "import dis; dis.dis('L.append(1)')"
1 0 LOAD_NAME 0 (L)
3 LOAD_ATTR 1 (append)
6 LOAD_CONST 0 (1)
9 CALL_FUNCTION 1 (1 positional, 0 keyword pair)
12 RETURN_VALUE
[steve@ando ~]$ python3.5 -c "import dis; dis.dis('L += [1]')"
1 0 LOAD_NAME 0 (L)
3 LOAD_CONST 0 (1)
6 BUILD_LIST 1
9 INPLACE_ADD
10 STORE_NAME 0 (L)
13 LOAD_CONST 1 (None)
16 RETURN_VALUE
> import timeit
[...]
That times a large amount of irrelevant code that has nothing to do with
either += or append.
--
Steve
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