On Thu, Oct 14, 2021 at 7:37 PM Jeremiah Vivian
<nohackingofkrow...@gmail.com> wrote:
>
> So I implemented these functions as operators in a downloaded source of
> CPython... the differences are insane! (Sorry if this produces nested quotes)
> > >>> import timeit
> > # d + 1 vs list(d.values())[0]: 2133x speedup
> > >>> timeit.main(['-s', "d = {x: x+1 for x in range(10000)}", "d + 1"])
> > 2000000 loops, best of 5: 165 nsec per loop
> > >>> timeit.main(['-s', "d = {x: x+1 for x in range(10000)}",
> > >>> "list(d.values())[0]"])
> > 1000 loops, best of 5: 352 usec per loop
Insane speedup but what does it prove? That if you handcraft C code to
do the precise and narrow feature you need, it's faster than a naive
listification?
Try the actually-recommended way: next(iter(d)). See how much
difference you get. Not having your hacked-on interpreter I can't
fully test this, but in my test of vanilla, d[next(iter(d))] was three
orders of magnitude faster than list(d.values())[0] - bringing it
within cooee of your "d + 1" version. Neither form copes with the
possibility of an empty dictionary, so for proper comparisons, you'd
need to try/except it; although recent CPythons have very very low
cost to a try block that doesn't end up catching anything, so it
probably won't make a material difference.
You've created some *incredibly* arbitrary operators which have no
justification other than "this operator isn't being used". Even if you
*do* manage to see a two thousand to one speedup, that is a
microoptimization that comes at an extremely steep readability
penalty. In no way does this make mathematical sense as "adding 1 to
the dictionary", nor does it parallel the way that other types behave,
nor is it internally consistent (multiplying isn't repeated addition),
nor is it particularly apt use of a symbol (like Path/str). And I
suspect that your speedups are really in the order of 2:1, not 2000:1.
ChrisA
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