new Set()
by applying all fn
to all values of a
.new Set()
from a
and all elements in b
whose value, after applying fn
does not match a value in the previously created set.const unionBy = (a, b, fn) => {
const s = new Set(a.map(fn));
return Array.from(new Set([...a, ...b.filter(x => !s.has(fn(x)))]));
};
unionBy([2.1], [1.2, 2.3], Math.floor); // [2.1, 1.2] unionBy([{ id: 1 }, { id: 2 }], [{ id: 2 }, { id: 3 }], x => x.id) // [{ id: 1 }, { id: 2 }, { id: 3 }]
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