Run results for: Intersection of multiple arrays | Lodash _.intersection() vs various custom optimized methods

Run details:

User agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/129.0.0.0 Safari/537.36

Browser: Chrome 129

Operating system: Windows

Device Platform: Desktop

Date tested: one year ago

Custom optimized function 1 (Best)

577/s

Lodash _.intersection()

386/s

Custom optimized function 2

54/s

View exact numbers

Test name	Executions per second
✓ Custom optimized function 1 (Best)	577 Ops/sec
Lodash _.intersection()	386 Ops/sec
Custom optimized function 2	54 Ops/sec

HTML Preparation code:


<script src='https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.5/lodash.min.js'></script>

Script Preparation code:

var arr1 = [];
var arr2 = [];
var arr3 = [];
var arr4 = [];

for (i = 0; i < 1000; i++) {
    arr1.push(getRandom());
    arr2.push(getRandom());
    arr3.push(getRandom());
    arr4.push(getRandom());
}

function getRandom() {
    const minimum = 1;
    const maximum = 1000;
    var randomnumber = Math.floor(Math.random() * (maximum - minimum + 1)) + minimum;
    return randomnumber;
}

Tests:

Lodash _.intersection()
const results = new Set(_.intersection(arr1, arr2, arr3, arr4));
Custom optimized function 1 (Best)
function intersectMultipleArrays(...arrays) { if (arrays.length === 0) return new Set(); // Sort arrays by length to optimize operations arrays.sort((a, b) => a.length - b.length); // Start with the smallest array as the base intersection set let intersection = new Set(arrays[0]); // Iterate over the remaining arrays and filter the intersection for (let i = 1; i < arrays.length; i++) { intersection = new Set(arrays[i].filter(item => intersection.has(item))); // Early exit if intersection is empty if (intersection.size === 0) return new Set(); } // Return the final Set return intersection; } const results = intersectMultipleArrays(arr1, arr2, arr3, arr4);
Custom optimized function 2
const results = new Set( [arr1, arr2, arr3, arr4] .sort((a, b) => b.length - a.length) .reduce((a, b) => a.filter(aValue => b.includes(aValue))) );