Returns every element that exists in any of the two arrays at least once, after applying the provided function to each array element of both.

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 }]

Returns every element that exists in any of the two arrays at least once, after applying the provided function to each array element of both.
<ul>
<li>Create a <code class="notranslate">Set</code> by applying all <code class="notranslate">fn</code> to all values of <code class="notranslate">a</code>.</li>
<li>Create a <code class="notranslate">Set</code> from <code class="notranslate">a</code> and all elements in <code class="notranslate">b</code> whose value, after applying <code class="notranslate">fn</code> does not match a value in the previously created set.</li>
<li>Return the last set converted to an array.</li>
</ul>

Mapped array union

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