On Mar 28, 11:16 am, Patrick Maupin <pmau...@gmail.com> wrote:
> On Mar 28, 12:34 pm, Steve Howell <showel...@yahoo.com> wrote:
>
> > FYI I later obtained similar results with the more general:
> > accum += sublist
>
> Yeah, but you still have to create an object of the correct type for
> accum.
I think you overlooked the surrounding code.
Here is the code again:
def in_place(
start = [],
sublists = ([[None] * M]) * N
):
# only macro-optimized
i = 0
for sublist in sublists:
if i == 0:
accum = start + sublist
i += 1
else:
accum += sublist
if i == 0:
return 'whatever' # semantics here?
return accum
No need to infer the type.
> And for summing small lists, that will actually increase the
> runtime. (Worst case, a list of a single item: you have to create the
> accum object based on the type of the start value, then do two += into
> it, for the start value and the first item in the list, vs just doing
> a single + which automatically creates an object.)
>
For small lists, the performance of any sane implementation would
probably be so fast as to be negligible, unless you are in a really
tight loop. If you are in a tight loop, then your use case probably
eventually sums large lists. Even if it doesn't, the slowdown is just
a constant.
For M=5, I get these results on my machine:
M N t1 t2 (t2/t1)
5 1 3.50475311279e-05 3.2901763916e-05 0.938775510204
5 2 2.00271606445e-05 1.59740447998e-05 0.797619047619
5 4 6.79492950439e-05 6.31809234619e-05 0.929824561404
5 8 0.000124931335449 0.000126123428345 1.00954198473
5 64 0.000530958175659 0.00226187705994 4.25999101931
5 1024 0.00262904167175 0.27246594429 103.636981953
t1 = time with add only first element
t2 = time with add all elements
Very small penalty for small lists, very large gain for large lists.
> Personally, I think the most general approach, if sum were open to
> enhancements, would be to automatically apply Alf's suggestion of "+"
> on summing the first item to the start value, and "+=" on subsequent
> items.
See above. That's the approach I would use as well.
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