[Python-ideas] Re: Revisiting a frozenset display literal

Jan. 17, 2022

      On Mon, Jan 17, 2022 at 3:50 AM Steven D'Aprano <steve@pearwood.info> wrote:
...
If you saw this code in a review:
t = tuple([1, 2, 3, 4, 5])
would you say "that is okay, because the name lookup is smaller than the
cost of building the list"?
I wouldn't. I would change the code to `(1, 2, 3, 4, 5)`.
Of course, everyone would -- because tuple displays already exist. I'd
suggest refactoring that code even if the compiler could completely
optimize it away.

Would you let:

 l = list([1, 2, 3, 4, 5])

pass code review either? even if there were no performance penalty? I
wouldn't, because it's redundant, not because it's slower.

Also that pattern is actually very common for types that aren't built-in
(or even in the stdlib).

It's always kind of bugged me that I need to write:

arr = np.array([1, 2, 3, 4])

And I'm creating a list just so I can pass it to the array constructor. But
in practice, it's not a performance problem at all.

And in code in the wild, I'll bet numpy arrays are used orders of
magnitude more than frozen sets ;-)

Sometimes, now, the compiler *pessimizes* the construction of the frozen
...
set. See b.p.o #46393.
yup. Using a 'constant' frozenset is slower than 'constant' set, when doing
not much else:

In [29]: def setfun():
    ...:     s = {1, 3, 5, 2}
    ...:     i = 3
    ...:     if i in s:
    ...:         return 'yes'
    ...:

In [30]: def fsetfun():
    ...:     s = frozenset((1, 3, 5, 2))
    ...:     i = 3
    ...:     if i in s:
    ...:         return 'yes'
    ...:

In [31]: %timeit setfun()
194 ns ± 10.6 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

In [32]: %timeit fsetfun()
286 ns ± 2.72 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

But: would you notice if that function did any real work?

And I think we could call this one of the many micro-optimizations we have
in Python:

Don't use a frozenset as a constant when a regular set will do.

So it comes down to how often frozen sets as constants are required.

-CHB

-- 
Christopher Barker, PhD (Chris)

Python Language Consulting
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