Named tuples existence in a hashset-CodePudding

In [1]: x = set()
In [2]: pos = collections.namedtuple('Position', ['x','y'])
In [4]: x.add(pos(1,1))
In [5]: x
Out[5]: {Position(x=1, y=1)}
In [6]: pos(1,1) in x
Out[6]: True
In [8]: pos(1,2) in x
Out[8]: False

I was not expecting Line 6 pos(1,1) in x to work. Since it does seem that pos(1,1) creates an object with a different object id every time.

In [9]: id(pos(1,1))
Out[9]: 140290954200696
In [10]: id(pos(1,1))
Out[10]: 140290954171016

How does the set in operator work on named tuples in this case? Does it check the contents of namedtuple?

CodePudding user response：

namedtuple is not special. The element should be exactly equal(__eq__) and must have similar hash to pass the containment.

>>> hash(pos(1, 1))
8389048192121911274
>>> pos(1, 1) == pos(1, 1)
True

If you are interested see the implementation here, set().__contains__(y) first have to compute the hash value of y.

static int
set_contains_key(PySetObject *so, PyObject *key)
{
    Py_hash_t hash;

    if (!PyUnicode_CheckExact(key) ||
        (hash = _PyASCIIObject_CAST(key)->hash) == -1) {
        hash = PyObject_Hash(key);
        if (hash == -1)
            return -1;
    }
    return set_contains_entry(so, key, hash);
}

but computing hash alone does not says whether the element are equal. For example

>>> hash(-1)
-2
>>> hash(-2)
-2

Which means that if the hash values are equal then __eq__ check is required to confirm if the elements are exactly equal.

Please note that my answer is purely based on Cpython implementation.

CodePudding user response：

A tuple of hashable objects is hashable, so it can be used as a member of a set and therefore pass the in check.
Since you are using numbers as tuple values, which are obviously hashable, there is no problem hashing the entire tuple.