Vanilla Javascript unique numbers in array with reduce and find - Stack Overflow

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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

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.

• javascript

Share Improve this question Follow edited Jan 9, 2017 at 14:45 StandardNerd asked Jan 9, 2017 at 14:37 StandardNerdStandardNerd 4,1831010 gold badges4848 silver badges7979 bronze badges 9

• Duplicate of stackoverflow./questions/13486479/… and stackoverflow./questions/1960473/unique-values-in-an-array ? – mplungjan Commented Jan 9, 2017 at 14:41
• Possible duplicate of Unique values in an array – RomanPerekhrest Commented Jan 9, 2017 at 14:42
• Unless he MUST use reduce and find - which is the reason I did not hammer close it – mplungjan Commented Jan 9, 2017 at 14:42
• @mplungjan the linked questions doesn't use reduce and find helper. – StandardNerd Commented Jan 9, 2017 at 14:43
• Which is why I did not actually close as duplicate – mplungjan Commented Jan 9, 2017 at 14:43

 |  Show 4 more ments

4 Answers 4

Sorted by: Reset to default 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));

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) {
    // pare 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));

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));

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;
}

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