Vanilla Javascript数组中的唯一数字，具有reduce和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;
}