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 stackoverflow.com/questions/13486479/… and stackoverflow.com/questions/1960473/unique-values-in-an-array ? — mplungjan
– mplungjan, Commented Jan 9, 2017 at 14:41
Unless he MUST use reduce and find - which is the reason I did not hammer close it — mplungjan
– mplungjan, Commented Jan 9, 2017 at 14:42
@mplungjan the linked questions doesn't use reduce and find helper. — StandardNerd
– StandardNerd, Commented Jan 9, 2017 at 14:43

Nina Scholz · Accepted Answer · 2017-01-09 15:09:41Z

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 9, 2017 at 15:09

answered Jan 9, 2017 at 14:39

Nina Scholz

388k26 gold badges367 silver badges417 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

Nina Scholz Over a year ago

includes is fine, but it does not work in edge, actually.

Bergi Over a year ago

your find does not work on arrays that contain a 0. You should use some or simply includes.

Nina Scholz Over a year ago

@Bergi, then indexOf is better, at least as find.

Bergi Over a year ago

Yes indeed. An alternative (as long as the array does only contain numbers) is ….find(…) != null

Bergi Over a year ago

That's what I meant by "if the array does only contain numbers" (which undefined is not)

|

Pranav C Balan · Accepted Answer · 2017-01-10 04:39:41Z

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 · Accepted Answer · 2017-01-09 14:56:54Z

2

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));

edited Jan 9, 2017 at 14:56

answered Jan 9, 2017 at 14:42

Nenad Vracar

122k16 gold badges160 silver badges184 bronze badges

Comments

Joshua · Accepted Answer · 2017-01-09 14:51:48Z

0

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;
}

answered Jan 9, 2017 at 14:51

Joshua

6497 silver badges18 bronze badges

Collectives™ on Stack Overflow

Vanilla Javascript unique numbers in array with reduce and find

4 Answers 4

6 Comments

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

6 Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related