Is there a way to return a custom value for min and max in Python?

Yes. When min takes one arguments it assumes it to be an iterable, iterates over it and takes the minimum value. So,

class A:
    def __init__(self, a, b):
        self.a = a
        self.b = b
    def __iter__(self):
        yield self.a
        yield self.b

Should work.

Additional Note: If you don't want to use __iter__, I don't know of way to do that. You probably want to create your own min function, that calls some _min_ method if there is one in the argument it is passed to and calls the old min else.

oldmin = min
def min(*args):
    if len(args) == 1 and hasattr(args[0], '_min_'):
        return args[0]._min_()
    else:
        return oldmin(*args)

Since range is considered to be a sequence type by the very same docs, I was thinking that there must be some sort of optimization that is possible for range, and that perhaps I could take advantage of it.

There's no optimization going on for ranges and there are no specialized magic methods for min/max.

If you peek at the implementation for min/max you'll see that after some argument parsing is done, a call to iter(obj) (i.e obj.__iter__()) is made to grab an iterator:

it = PyObject_GetIter(v);
if (it == NULL) {
    return NULL;
}

then calls to next(it) (i.e it.__next__) are performed in a loop to grab values for comparisons:

while (( item = PyIter_Next(it) )) {
    /* Find min/max  */

Is it possible to have something like the following work?

No, if you want to use the built-in min* the only option you have is implementing the iterator protocol.

---

*By patching min, you can of-course, make it do anything you want. Obviously at the cost of operating in Pythonland. If, though, you think you can utilize some optimizations, I'd suggest you create a min method rather than re-defining the built-in min.

In addition, if you only have ints as instance variables and you don't mind a different call, you can always use vars to grab the instance.__dict__ and then supply it's .values() to min:

>>> x = A(20, 4)
>>> min(vars(x).values())
4

There are no __min__ and __max__ special methods*. This is kind of a shame since range has seen some pretty nice optimizations in Python 3. You can do this:

>>> 1000000000000 in range(1000000000000)
False

But don't try this unless you want to wait a long time:

>>> max(range(1000000000000))

However creating your own min/max functions is a pretty good idea, as suggested by Lærne.

Here is how I would do it. UPDATE: removed the dunder name __min__ in favor of _min, as recommended by PEP 8:

Never invent such names; only use them as documented

Code:

from functools import wraps

oldmin = min

@wraps(oldmin)
def min(*args, **kwargs)
    try:
        v = oldmin(*args, **kwargs)
    except Exception as err:
        err = err
    try:
        arg, = args
        v = arg._min()
    except (AttributeError, ValueError):
        raise err
    try:
        return v
    except NameError:
        raise ValueError('Something weird happened.')

I think this way is maybe a little bit better because it handles some corner cases the other answer hasn't considered.

Note that an iterable object with a _min method will still be consumed by oldmin as per usual, but the return value is overridden by the special method.

HOWEVER, if the _min method requires the iterator to still be available for consumption, this will need to be tweaked because the iterator is getting consumed by oldmin first.

Note also that if the __min method is simply implemented by calling oldmin, things will still work fine (even though the iterator was consumed; this is because oldmin raises a ValueError in this case).

* Such methods are often called "magic", but this is not the preferred terminology.