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.

Duplicate of http://stackoverflow.com/questions/13486479/how-to-get-an-array-of-unique-values-from-an-array-containing-duplicates-in-java and http://stackoverflow.com/questions/1960473/unique-values-in-an-array ? — mplungjan, Jan 09 '17 at 14:41
Possible duplicate of [Unique values in an array](http://stackoverflow.com/questions/1960473/unique-values-in-an-array) — RomanPerekhrest, Jan 09 '17 at 14:42
Unless he MUST use reduce and find - which is the reason I did not hammer close it — mplungjan, Jan 09 '17 at 14:42
@mplungjan the linked questions doesn't use reduce and find helper. — StandardNerd, Jan 09 '17 at 14:43
WTH is `find(array.number)`? Did you mean `array.find(number)`? And why do you pass the result of that to `indexOf`? It seems what you are looking for is `if (uniqueArray.includes(number))`. — Bergi, Jan 09 '17 at 15:07

Nina Scholz · Accepted Answer · 2017-01-09T15:09:41.077

4

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));

edited Jan 09 '17 at 15:09

answered Jan 09 '17 at 14:39

Nina Scholz

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includes is fine, but it does not work in edge, actually. – Nina Scholz Jan 09 '17 at 14:45
your `find` does not work on arrays that contain a `0`. You should use `some` or simply `includes`. – Bergi Jan 09 '17 at 15:05
@Bergi, then indexOf is better, at least as find. – Nina Scholz Jan 09 '17 at 15:11
Yes indeed. An alternative (as long as the array does only contain numbers) is `….find(…) != null` – Bergi Jan 09 '17 at 15:26
@Bergi, if not found, then [`find`](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/find) returns `undefined`, but if `undefined` is a value, then it does not work. – Nina Scholz Jan 09 '17 at 15:29
1

That's what I meant by "if the array does only contain numbers" (which `undefined` is not) – Bergi Jan 09 '17 at 15:38

Pranav C Balan · Answer 2 · 2017-01-10T04:39:41.130

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));

Nenad Vracar · Answer 3 · 2017-01-09T14:56:54.267

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));

score 0 · Answer 4 · answered Jan 09 '17 at 14:51

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;
}

Vanilla Javascript unique numbers in array with reduce and find

4 Answers4