Consider conditionally applying rule PLR6201

[ofek]

When the sequences are small, the cost of instantiation of a set in fact negatively impacts performance:

❯ python -m timeit "{0, '0', 9, '9', 8, '8', 7, '7', 6, '6'}"
2000000 loops, best of 5: 105 nsec per loop
❯ python -m timeit "[0, '0', 9, '9', 8, '8', 7, '7', 6, '6']"
5000000 loops, best of 5: 43.6 nsec per loop
❯ python -m timeit "(0, '0', 9, '9', 8, '8', 7, '7', 6, '6')"
50000000 loops, best of 5: 8.45 nsec per loop
❯ python -m timeit -s "seq = {0, '0', 9, '9', 8, '8', 7, '7', 6, '6'}" "0 in seq"
10000000 loops, best of 5: 20.6 nsec per loop
❯ python -m timeit -s "seq = [0, '0', 9, '9', 8, '8', 7, '7', 6, '6']" "0 in seq"
20000000 loops, best of 5: 17.9 nsec per loop
❯ python -m timeit -s "seq = (0, '0', 9, '9', 8, '8', 7, '7', 6, '6')" "0 in seq"
20000000 loops, best of 5: 17.5 nsec per loop
❯ python -m timeit -s "seq = {0, '0', 9, '9', 8, '8', 7, '7', 6, '6'}" "6 in seq"
10000000 loops, best of 5: 19.8 nsec per loop
❯ python -m timeit -s "seq = [0, '0', 9, '9', 8, '8', 7, '7', 6, '6']" "6 in seq"
5000000 loops, best of 5: 85.5 nsec per loop
❯ python -m timeit -s "seq = (0, '0', 9, '9', 8, '8', 7, '7', 6, '6')" "6 in seq"
5000000 loops, best of 5: 90.1 nsec per loop

[tdulcet]

There are optimizations here that are not accounted for in your benchmarks. See What’s New In Python 3.2 for example. Also see #8322 and pylint-dev/pylint#4776.

[flying-sheep]

Yup, with these, the x in {...} pattern wins out against list and tuple literals unless it’s the first element:

$ python -m timeit '0 in {0, "0", 9, "9", 8, "8", 7, "7", 6, "6"}'
20000000 loops, best of 5: 13.6 nsec per loop
$ python -m timeit '0 in [0, "0", 9, "9", 8, "8", 7, "7", 6, "6"]'
20000000 loops, best of 5: 11.5 nsec per loop
$ python -m timeit '0 in (0, "0", 9, "9", 8, "8", 7, "7", 6, "6")'
20000000 loops, best of 5: 11.4 nsec per loop
$ python -m timeit '6 in {0, "0", 9, "9", 8, "8", 7, "7", 6, "6"}'
20000000 loops, best of 5: 12.9 nsec per loop
$ python -m timeit '6 in [0, "0", 9, "9", 8, "8", 7, "7", 6, "6"]'
5000000 loops, best of 5: 52.3 nsec per loop
$ python -m timeit '6 in (0, "0", 9, "9", 8, "8", 7, "7", 6, "6")'
5000000 loops, best of 5: 56.5 nsec per loop

The manual version is maybe even a tiny bit faster, but this is noise level.

$ python -m timeit -s 'seq = {0, "0", 9, "9", 8, "8", 7, "7", 6, "6"}' '0 in seq'
20000000 loops, best of 5: 12.5 nsec per loop
$ python -m timeit -s 'seq = [0, "0", 9, "9", 8, "8", 7, "7", 6, "6"]' '0 in seq'
20000000 loops, best of 5: 11.2 nsec per loop
$ python -m timeit -s 'seq = (0, "0", 9, "9", 8, "8", 7, "7", 6, "6")' '0 in seq'
20000000 loops, best of 5: 10.5 nsec per loop
$ python -m timeit -s 'seq = {0, "0", 9, "9", 8, "8", 7, "7", 6, "6"}' '6 in seq'
20000000 loops, best of 5: 15.3 nsec per loop
$ python -m timeit -s 'seq = [0, "0", 9, "9", 8, "8", 7, "7", 6, "6"]' '6 in seq'
5000000 loops, best of 5: 53.8 nsec per loop
$ python -m timeit -s 'seq = (0, "0", 9, "9", 8, "8", 7, "7", 6, "6")' '6 in seq'
5000000 loops, best of 5: 49.6 nsec per loop

[oprypin]

@flying-sheep Thanks! Could you also please run an equivalent benchmark for a 3-item collection? 🙏

[flying-sheep]

Microoptimizations like this are rarely indicative of real life performance. I’m simply pointing out that the optimization mentioned by @tdulcet is indeed applied, so using a inline set literal is preferable, as it’s equivalent to creating a frozenset ahead of time.

But sure, here you go. Finding the 3rd item in a sequence is slower than finding it in a set, only finding the first item in a sequence is very very slightly faster than finding that item in a set.

So if x in {...} is best. Ruff does it right.

---

Peephole optimized:

$ python -m timeit '0 in {0, "0", 9}'
20000000 loops, best of 5: 13.7 nsec per loop
$ python -m timeit '0 in [0, "0", 9]'
20000000 loops, best of 5: 11.4 nsec per loop
$ python -m timeit '0 in (0, "0", 9)'
20000000 loops, best of 5: 11.3 nsec per loop
$ python -m timeit '9 in {0, "0", 9}'
20000000 loops, best of 5: 12.8 nsec per loop
$ python -m timeit '9 in [0, "0", 9]'
10000000 loops, best of 5: 21.4 nsec per loop
$ python -m timeit '9 in (0, "0", 9)'
10000000 loops, best of 5: 21.5 nsec per loop

… has the same performance as manually optimized:

$ python -m timeit -s 'seq = {0, "0", 9}' '0 in seq'
20000000 loops, best of 5: 12.7 nsec per loop
$ python -m timeit -s 'seq = [0, "0", 9]' '0 in seq'
20000000 loops, best of 5: 11.3 nsec per loop
$ python -m timeit -s 'seq = (0, "0", 9)' '0 in seq'
20000000 loops, best of 5: 10.5 nsec per loop
$ python -m timeit -s 'seq = {0, "0", 9}' '9 in seq'
20000000 loops, best of 5: 12.2 nsec per loop
$ python -m timeit -s 'seq = [0, "0", 9]' '9 in seq'
10000000 loops, best of 5: 21.7 nsec per loop
$ python -m timeit -s 'seq = (0, "0", 9)' '9 in seq'
20000000 loops, best of 5: 20 nsec per loop

[zanieb]

Thanks for the details @flying-sheep