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 ;-)