Vanilla Javascript unique numbers in array with reduce and find

Question

I'm trying to achieve to write an array function with the use of reduce and find helpers that returns an array of unique numbers.

 var numbers = [1, 1, 2, 3, 4, 4]; // function should return [1, 2, 3, 4] function unique(array) { array.reduce((uniqueArray, number) => { if (uniqueArray.indexOf(find(array.number))) { uniqueArray.push(array.number); } return uniqueArray; }, []); } console.log(unique(numbers)); // undefined // undefined 

When running this code I get

undefined

twice in Browser Javascript console.

Answer 1

The reasons for the errors are explained in previous answers. So I just adding an alternate method with Array#filter method.

 var numbers = [1, 1, 2, 3, 4, 4]; // function should return [1, 2, 3, 4] function unique(array) { return array.filter(function(v, i, arr) { // compare index with first element index return i == arr.indexOf(v); }) } console.log(unique(numbers)); 

---

With ES6 arrow function.

 var numbers = [1, 1, 2, 3, 4, 4]; // function should return [1, 2, 3, 4] function unique(array) { return array.filter((v, i, arr) => i == arr.indexOf(v)) } console.log(unique(numbers)); 

---

UPDATE : With a reference object instead of checking the index.

 var numbers = [1, 1, 2, 3, 4, 4], ref = {}; function unique(array) { return array.filter(function(v) { if (!(v in ref)) { ref[v] = true; return true; } return false; }) } console.log(unique(numbers)); 

Answer 2

You need a return statment.

return array.reduce((uniqueArray // ...
// ^^^

And some better find method with Array.indexOf

 function unique(array) { return array.reduce((uniqueArray, number) => { if (uniqueArray.indexOf(number) === -1) { uniqueArray.push(number); } return uniqueArray; }, []); } var numbers = [1, 1, 2, 3, 4, 4]; console.log(unique(numbers)); 

And now with Set and spread syntax ... for collecting the items in a new array.

 function unique(array) { return [... new Set(array)]; } var numbers = [1, 1, 2, 3, 4, 4]; console.log(unique(numbers)); 

Answer 3

You have few errors. First you need to return value from your function and also to check if element is already in uniqueArray you can use indexOf() == -1 .

 var numbers = [1, 1, 2, 3, 4, 4]; function unique(array) { return array.reduce((uniqueArray, number) => { if (uniqueArray.indexOf(number) == -1) uniqueArray.push(number) return uniqueArray; }, []); } console.log(unique(numbers)); 

With ES6/7 you can use includes() and arrow functions like this.

 var numbers = [1, 1, 2, 3, 4, 4]; function unique(arr) { return arr.reduce((r, n) => (!r.includes(n) ? r.push(n) : 1) && r , []); } console.log(unique(numbers)); 

Answer 4

You can always use Array.includes .

function SillyFunctionName(array) {
    "use strict";

    var uniqueArray = [];

    for (var i = 0; i < array.length; i++) {
        if (uniqueArray.includes(array[i])) {
            break;
        } else {
            uniqueArray.push(array[i]);
        }
    }

    return uniqueArray;
}