Remove duplicates from arrays using reduce

Question

I am trying to remove duplicates from a list of arrays. The way I was trying to do this is by using reduce to create an empty array that pushes all undefined indexes onto that array. I am getting errors though that

if(acc[item]===undefined){
      ^
TypeError: Cannot read property '1' of undefined

my function below:

function noDuplicates(arrays) {
  var arrayed = Array.prototype.slice.call(arguments);

  return reduce(arrayed, function(acc, cur) {
    forEach(cur, function(item) {
      if (acc[item] === undefined) {
        acc.push(item);
      }
      return acc;
    });
  }, []);
}

console.log(noDuplicates([1, 2, 2, 4], [1, 1, 4, 5, 6]));

What should be the output of console.log(noDuplicates([1,2,2,4],[1,1,4,5,6]));? — Ved
– Ved, Commented May 4, 2017 at 4:41
Possible duplicate of Remove Duplicates from JavaScript Array — Preview
– Preview, Commented May 4, 2017 at 5:07

Shekhar Chikara · Accepted Answer · 2017-05-04 04:51:21Z

8

First concatenate the two arrays, next use filter() to filter out only the unique items-

var a = [1, 2, 2, 4], b = [1, 1, 4, 5, 6];
var c = a.concat(b);
var d = c.filter(function (item, pos) {return c.indexOf(item) == pos});
console.log(d);

answered May 4, 2017 at 4:51

Shekhar Chikara

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Comments

RobG · Accepted Answer · 2017-05-04 06:22:46Z

There are a number of issues with how you're calling methods and where you return acc from:

function noDuplicates(arrays) {
  var arrayed = Array.prototype.slice.call(arguments);

  // reduce is a method of an array, so call it as a method
  // return reduce(arrayed, function(acc, cur) {
  return arrayed.reduce(function(acc, cur) {
  
    // Same with forEach
    cur.forEach(function(item) {
      if (acc[item] === undefined) {
        acc.push(item);
      }
       // Return acc from the reduce callback, forEach returns undefined always
       // return acc;
    });
    return acc;
  }, []);
}

console.log(noDuplicates([1, 2, 2, 4], [1, 1, 4, 5, 6]));

You could also call reduce directly on arguments using call:

Array.prototype.reduce.call(arguments, function(acc, curr) {
  // ...
});

The above makes your code run, but it doesn't produce the correct output as the test:

  if (acc[item] === undefined)

doesn't do what you want. What you need to do is remember each value and only push it to acc if it's not been seen before:

function noDuplicates(arrays) {
  var arrayed = Array.prototype.slice.call(arguments);
  var seen = {};

  return arrayed.reduce(function(acc, cur) {
    cur.forEach(function(item) {
      if (!seen[item]) {
        acc.push(item);
        seen[item] = true;
      }
    });
    return acc;
  }, []);
}

console.log(noDuplicates([1, 2, 2, 4], [1, 1, 4, 5, 6]));

Some other approaches:

// A more concise version of the OP
function noDupes() {
  return [].reduce.call(arguments, function(acc, arr) {
    arr.forEach(function(value) {
      if (acc.indexOf(value) == -1) acc.push(value);
    });
    return acc;
   },[]);
}
 
console.log(noDupes([1, 2, 2, 4], [1, 1, 4, 5, 6]));

// Some ECMAScript 2017 goodness
function noDupes2(...args){
 return [].concat(...args).filter((v, i, arr) => arr.indexOf(v)==i);
}

console.log(noDupes2([1, 2, 2, 4], [1, 1, 4, 5, 6]));

Mayank Shukla · Accepted Answer · 2017-05-04 04:58:59Z

4

Any reason of using reduce? because we can do this easily by first merging these two arrays then by using Set to remove the duplicate keys.

Check this:

function noDuplicates(a, b){
    var k = a.concat(b);
    return [...new Set(k)];
}

console.log(noDuplicates([1,2,2,4],[1,1,4,5,6]));

Check the DOC, how Set works.

edited May 4, 2017 at 4:58

answered May 4, 2017 at 4:51

Mayank Shukla

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1 Comment

Igor Soloydenko Over a year ago

Using the array concatenation kinda diminishes the point of using Set which is a good idea.

Rahul Thakur · Accepted Answer · 2020-04-28 14:22:51Z

4

My Solution is -

var numbers = [1, 1, 2, 3, 4, 4];

function unique(array){
  return array.reduce(function(previous, current) {
     if(!previous.find(function(prevItem){
         return prevItem === current;
     })) {
        previous.push(current);
     }
     return previous;
 }, []);
}

unique(numbers);

answered Apr 28, 2020 at 14:22

Rahul Thakur

413 bronze badges

Comments

bla · Accepted Answer · 2019-01-16 21:36:29Z

2

looking for a smoother solution for MDN's exact same problem, I've came up with that solution, I find it simple and nice. I have also just updated it in MDN and wanted to share it here (I'm really new to that stuff, so sorry if did something wrong)

let myArray = ['a', 'b', 'a', 'b', 'c', 'e', 'e', 'c', 'd', 'd', 'd', 'd'];
var myOrderedArray = myArray.reduce(function (accumulator, currentValue) {
  if (accumulator.indexOf(currentValue) === -1) {
    accumulator.push(currentValue);
  }
  return accumulator
}, [])

console.log(myOrderedArray);

(I'm really new to this, hope it'd help)

edited Jan 16, 2019 at 21:36

answered Jan 16, 2019 at 21:21

bla

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Comments

Shivananda Sai · Accepted Answer · 2023-08-01 07:52:27Z

1

Remove duplicate elements from an array using reduce function in JS

const arr = [1,2,3,4,4,5,5,5,6];

const uniqueArray = (arr) => {
    return arr.reduce((acc,ele) => {
        return acc.includes(ele) ? acc : [...acc,ele]
    },[])
}

console.log(uniqueArray(arr)); // [1,2,3,4,5,6]

edited Aug 1, 2023 at 7:52

answered Aug 1, 2023 at 7:50

Shivananda Sai

114 bronze badges

Comments

Curt · Accepted Answer · 2022-04-29 03:37:53Z

0

The previous answers are not optimized for large arrays. The following allows for linear big O notation:

const dedupWithReduce = (arr) =>
  arr.reduce(
    (acc, cur) => {
      if (!acc.lookupObj[cur]) {
        return {
          lookupObj: {
            ...acc.lookupObj,
            [cur]: true
          },
          dedupedArray: acc.dedupedArray.concat(cur)
        };
      } else {
        return acc;
      }
    },
    { lookupObj: {}, dedupedArray: [] }
  ).dedupedArray;

answered Apr 29, 2022 at 3:37

Curt

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