Why is “1000000000000000 in range(1000000000000001)” so fast in Python 3?

Question

It is my understanding that the range() function, which is actually an object type in Python 3, generates its contents on the fly, similar to a generator.

Answer 1

To add to Martijn’s answer, this is the relevant part of the source (in C, as the range object is written in native code):

static int
range_contains(rangeobject *r, PyObject *ob)
{
    if (PyLong_CheckExact(ob) || PyBool_Check(ob))
        return range_contains_long(r, ob);

    return (int)_PySequence_IterSearch((PyObject*)r, ob,
                                       PY_ITERSEARCH_CONTAINS);
}

So for PyLong objects (which is int in Python 3), it will use the range_contains_long function to determine the result. And that function essentially checks if ob is in the specified range (although it looks a bit more complex in C).

If it’s not an int object, it falls back to iterating until it finds the value (or not).

The whole logic could be translated to pseudo-Python like this:

def range_contains (rangeObj, obj):
    if isinstance(obj, int):
        return range_contains_long(rangeObj, obj)

    # default logic by iterating
    return any(obj == x for x in rangeObj)

def range_contains_long (r, num):
    if r.step > 0:
        # positive step: r.start <= num < r.stop
        cmp2 = r.start <= num
        cmp3 = num < r.stop
    else:
        # negative step: r.start >= num > r.stop
        cmp2 = num <= r.start
        cmp3 = r.stop < num

    # outside of the range boundaries
    if not cmp2 or not cmp3:
        return False

    # num must be on a valid step inside the boundaries
    return (num - r.start) % r.step == 0