./python.exe -m pyperf timeit -s 'a = [1]*100'  'a.count(1)'

Current Master: Mean +- std dev: 1.05 us +- 0.03 us
My patch: Mean +- std dev: 423 ns +- 11 ns

This is the side-effect of pr 17022.

New changeset 14d80d0b605d8b148e14458e4c1853a940071462 by Miss Islington (bot) (Dong-hee Na) in branch 'master':
bpo-39425: Fix list.count performance regression (GH-18119)
https://github.com/python/cpython/commit/14d80d0b605d8b148e14458e4c1853a940071462

New changeset fdb21609d944941f0732df72dc3d07a7a9a7efea by Pablo Galindo (Miss Islington (bot)) in branch '3.8':
bpo-39425: Fix list.count performance regression (GH-18119) (GH-18120)
https://github.com/python/cpython/commit/fdb21609d944941f0732df72dc3d07a7a9a7efea

New changeset 9e06d61af30bac4fcacb7973f826147ccc010392 by Pablo Galindo (Miss Islington (bot)) in branch '3.7':
bpo-39425: Fix list.count performance regression (GH-18119) (GH-18121)
https://github.com/python/cpython/commit/9e06d61af30bac4fcacb7973f826147ccc010392

Thanks for the PR, Dong-hee Na :)

The following result is a little bit surprising:

>>> nan=float("nan"); ([nan]*5).count(nan)
5
>>> nan == nan
False

But in fact, the optimization doesn't change the value. It was already 5 previously.

In fact, PyObject_RichCompareBool() has a fast path if the two object pointers are equal:

    /* Quick result when objects are the same.
       Guarantees that identity implies equality. */
    if (v == w) {
        if (op == Py_EQ)
            return 1;
        else if (op == Py_NE)
            return 0;
    }

In short, the optimization is good: thank you ;-)

> In fact, PyObject_RichCompareBool() has a fast path if the two object pointers are equal

It's documented:

https://docs.python.org/dev/c-api/object.html?highlight=pyobject_richcomparebool#c.PyObject_RichCompareBool

> The following result is a little bit surprising:

haha, I went to exactly the same journey when reviewing this PR :)