Basically there are two functions that deal with an iterable and return True or False depending on which boolean values elements of the sequence evaluate to.

1. all(iterable) returns True if all elements of the iterable are considered as true values (like reduce(operator.and_, iterable)).

2. any(iterable) returns True if at least one element of the iterable is a true value (again, using functional stuff, reduce(operator.or_, iterable)).

Using the all function, you can map operator.not_ over your list or just build a new sequence with negated values and check that all the elements of the new sequence are true:

>>> all(not element for element in data)

With the any function, you can check that at least one element is true and then negate the result since you need to return False if there's a true element:

>>> not any(data)

According to De Morgan's law, these two variants will return the same result, but I would prefer the last one (which uses any) because it is shorter, more readable (and can be intuitively understood as "there isn't a true value in data") and more efficient (since you don't build any extra sequences).

How do Python's any and all functions work?

any and all take iterables and return True if any and all (respectively) of the elements are True.

>>> any([0, 0.0, False, (), '0']), all([1, 0.0001, True, (False,)])
(True, True)            #   ^^^-- truthy non-empty string
>>> any([0, 0.0, False, (), '']), all([1, 0.0001, True, (False,), {}])
(False, False)                                                #   ^^-- falsey

If the iterables are empty, any returns False, and all returns True.

>>> any([]), all([])
(False, True)

I was demonstrating all and any for students in class today. They were mostly confused about the return values for empty iterables. Explaining it this way caused a lot of lightbulbs to turn on.

Shortcutting behavior

They, any and all, both look for a condition that allows them to stop evaluating. The first examples I gave required them to evaluate the boolean for each element in the entire list.

(Note that list literal is not itself lazily evaluated - you could get that with an Iterator - but this is just for illustrative purposes.)

Here's a Python implementation of any and all:

def any(iterable):
    for i in iterable:
        if i:
            return True
    return False # for an empty iterable, any returns False!

def all(iterable):
    for i in iterable:
        if not i:
            return False
    return True  # for an empty iterable, all returns True!

Of course, the real implementations are written in C and are much more performant, but you could substitute the above and get the same results for the code in this (or any other) answer.

all

all checks for elements to be False (so it can return False), then it returns True if none of them were False.

>>> all([1, 2, 3, 4])                 # has to test to the end!
True
>>> all([0, 1, 2, 3, 4])              # 0 is False in a boolean context!
False  # ^--stops here!
>>> all([])
True   # gets to end, so True!

any

The way any works is that it checks for elements to be True (so it can return True), then it returnsFalseif none of them wereTrue`.

>>> any([0, 0.0, '', (), [], {}])     # has to test to the end!
False
>>> any([1, 0, 0.0, '', (), [], {}])  # 1 is True in a boolean context!
True   # ^--stops here!
>>> any([])
False   # gets to end, so False!

I think if you keep in mind the short-cutting behavior, you will intuitively understand how they work without having to reference a Truth Table.

Evidence of all and any shortcutting:

First, create a noisy_iterator:

def noisy_iterator(iterable):
    for i in iterable:
        print('yielding ' + repr(i))
        yield i

and now let's just iterate over the lists noisily, using our examples:

>>> all(noisy_iterator([1, 2, 3, 4]))
yielding 1
yielding 2
yielding 3
yielding 4
True
>>> all(noisy_iterator([0, 1, 2, 3, 4]))
yielding 0
False

We can see all stops on the first False boolean check.

And any stops on the first True boolean check:

>>> any(noisy_iterator([0, 0.0, '', (), [], {}]))
yielding 0
yielding 0.0
yielding ''
yielding ()
yielding []
yielding {}
False
>>> any(noisy_iterator([1, 0, 0.0, '', (), [], {}]))
yielding 1
True

The source

Let's look at the source to confirm the above.

Here's the source for any:

static PyObject *
builtin_any(PyObject *module, PyObject *iterable)
{
    PyObject *it, *item;
    PyObject *(*iternext)(PyObject *);
    int cmp;

    it = PyObject_GetIter(iterable);
    if (it == NULL)
        return NULL;
    iternext = *Py_TYPE(it)->tp_iternext;

    for (;;) {
        item = iternext(it);
        if (item == NULL)
            break;
        cmp = PyObject_IsTrue(item);
        Py_DECREF(item);
        if (cmp < 0) {
            Py_DECREF(it);
            return NULL;
        }
        if (cmp > 0) {
            Py_DECREF(it);
            Py_RETURN_TRUE;
        }
    }
    Py_DECREF(it);
    if (PyErr_Occurred()) {
        if (PyErr_ExceptionMatches(PyExc_StopIteration))
            PyErr_Clear();
        else
            return NULL;
    }
    Py_RETURN_FALSE;
}

And here's the source for all:

static PyObject *
builtin_all(PyObject *module, PyObject *iterable)
{
    PyObject *it, *item;
    PyObject *(*iternext)(PyObject *);
    int cmp;

    it = PyObject_GetIter(iterable);
    if (it == NULL)
        return NULL;
    iternext = *Py_TYPE(it)->tp_iternext;

    for (;;) {
        item = iternext(it);
        if (item == NULL)
            break;
        cmp = PyObject_IsTrue(item);
        Py_DECREF(item);
        if (cmp < 0) {
            Py_DECREF(it);
            return NULL;
        }
        if (cmp == 0) {
            Py_DECREF(it);
            Py_RETURN_FALSE;
        }
    }
    Py_DECREF(it);
    if (PyErr_Occurred()) {
        if (PyErr_ExceptionMatches(PyExc_StopIteration))
            PyErr_Clear();
        else
            return NULL;
    }
    Py_RETURN_TRUE;
}

For the question in the title:

if a list contains one set of values or another

it might be more natural to use set operations. In other words, instead of

if any(x==playerOne for x in board) or any(x==playerTwo for x in board):

# or 
if playerOne in board or playerTwo in board:

use set.issubset (or set.intersection¹):

if {playerOne, playerTwo}.issubset(board):

# or
if {playerOne, playerTwo} & set(board):

If playerOne and playerTwo are set/list/tuple of values, then compute their union and test if it's a subset of board:

if {*playerOne,*playerTwo}.issubset(board):

Also if the question is

if every item on the board is either playerOne marker or playerTwo marker

then instead of

if all(x == playerOne or x == playerTwo for x in board):

test set equality:¹

if {playerOne, playerTwo} == set(board):

¹ You can obviously assign set(board) to some variable beforehand so that you don't have to cast board to a set every time you need to test this condition.

Cool question. Best way to approach this, is to try it out for oneself in a Python REPL:

$ python
[Clang 14.0.0 (clang-1400.0.29.202)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> bool([False])
True
>>> bool(False)
False
>>> all([[]])
False
>>> all([[False]])
True

But hey, who am I but a humble raven, standing atop the shoulders of giants here. Check out this answer here for more details and clarification.