Improve performance of set membership testing from set arguments

[HarryLHW]

Feature or enhancement

Proposal:

For set s and set elem, the operations elem in s, s.remove(elem), and s.discard(elem) are equivalent to frozenset(elem) in s, s.remove(frozenset(elem)), and s.discard(frozenset(elem)), respectively. This feature is documented in Built-in Types:

Note, the elem argument to the contains(), remove(), and discard() methods may be a set. To support searching for an equivalent frozenset, a temporary one is created from elem.

The implementation can be improved by calculating the hash value directly from the set object (as if it is a frozenset), instead of copying it into a new frozenset before calculating the hash value. The hash function for set is internal use only, and set is still unhashable.

Has this already been discussed elsewhere?

This is a minor feature, which does not need previous discussion elsewhere

Links to previous discussion of this feature:

No response

Linked PRs

• gh-121795: Improve performance of set membership testing from set arguments #121796

[HarryLHW]

Benchmark (on an M2 Macbook Air):

Script:

import timeit

for n in (1, 10, 100, 1000, 10000):
    setup = f'''
a = set(range({n}))
b = {{frozenset(a)}}
'''
    print(timeit.timeit("a in b", setup), f'N = {n}')

main branch result:

0.3688333749996673 N = 1
0.48004929200033075 N = 10
1.781020708000142 N = 100
11.256919416999153 N = 1000
152.03752795800028 N = 10000

my branch result:

0.30569541699878755 N = 1
0.3746081659992342 N = 10
1.2535620420021587 N = 100
7.2957435000025725 N = 1000
86.63279329200304 N = 10000

There will be no performance regression on the normal case.

[rhettinger]

Sorry, I haven't been feeling well. Please get another core dev to review this.

[erlend-aasland]

Thanks for the PR, @HarryLHW! I believe this can be closed, as the PR was recently merged. Good job!