conditionally reducing a JavaScript array of objects

Question

If I have

const foo = [
    { id: 1234, b: 'blah', v: 1 },
    { id: 1234, b: 'blahr', v: 3 },
    { id: 1234, b: 'blag', v: 2 }
]

const latest = foo.reduce((a, b) => (a.v > b.v ? a : b));
console.log(latest);
// { a: 1234, b: 'blahr', v: 3 }

update: fixed a typo in the code above

all is good so far. But I actually have

const foo = [
    { id: 1234, b: 'blah', v: 1 },
    { id: 1234, b: 'blahr', v: 3 },
    { id: 1234, b: 'blag', v: 2 },
    { id: 3424, b: 'gaaar', v: 1 },
    { id: 5324, b: 'moorg' },
]

and I want

[
    { id: 1234, b: 'blahr', v: 3 },
    { id: 3424, b: 'gaaar', v: 1 },
    { id: 5324, b: 'moorg' }
]

In other words, I want the reduce function to apply only if there are more than one objects with the same id. Objects with unique ids, and objects without any version v info should be returned as is.

Update2: Ok, using the answer to Most efficient method to groupby on an array of objects and combining my reduce function above, I am able to accomplish what I want with the following code:

const groupBy = function(xs, key) {
    return xs.reduce((rv, x) => {
        (rv[x[key]] = rv[x[key]] || []).push(x);
        return rv;
    }, {});
};

bar = groupBy(foo, 'id');
baz = Object.values(bar).map(e => e.reduce((a, b) => (a.v > b.v ? a : b)));
console.log(baz)
//[ { id: 1234, b: 'blahr', v: 3 },
//{ id: 3424, b: 'gaaar', v: 1 },
//{ id: 5324, b: 'moorg' } ]