How to return only values that are unique in an array?

Question

Given

[1,3,1,6,22] ,

it should return

[3,6,22]

So far I have this:

const returnUniques = (array) {
  if (array.length == 1) return array[0]
  for (let i = 0; i < array.length; i++) {
    for (let j = 0; j < array.length; j++) {
      if (i !== j) {
        if (array[i] === array[j]) {
        }
      }
    }
  }
  return array
}

I tried

const returnUniques = (array) {
  if (array.length == 1) return array[0]
  for (let i = 0; i < array.length; i++) {
    for (let j = 0; j < array.length; j++) {
      if (i !== j) {
        if (array[i] === array[j]) {
          array.splice(i, 1)
        }
      }
    }
  }
  return array
}

But it just removes duplicate and I could have this by one line with either new Set() or other techniques

So I added

const returnUniques = (array) => {
        if (array.length == 1) return array[0]
        
        for (let i = 0; i < array.length; i++) {
            for (let j = 0; j < array.length; j++) {
                if (i !== j) {
                    if(array[i] === array[j]) {
                       array.splice(i,1)
                       array.splice(j-1, 1)
                    } 
                }
            }
        }
        
        return array
    }

console.log(returnUniques([1,3,1,6,22]))

because I thought that if I removed array[i] so array[j] would be at index of itself -1. But this method doesn't work if there is more than twice the same number and if many other reasons..

Maybe this nested loop isn't the best way to go ? Thanks for helps !

Your approach is not that bad. However I would create a new empty array in your returnUniques function, fill it with the entries that pass your test and then return the new array. Don't modify the original array. — Krisztián Balla
– Krisztián Balla, Commented Nov 7, 2020 at 14:21
But how wil I fill the newArray with values that doesn't have duplicates ? — Yoël Zerbib
– Yoël Zerbib, Commented Nov 7, 2020 at 14:23
First I tried to change if (array[i] == array[j]) to (array[i] !== array[j]) but that doesn't do the trick lol. Thanks for you indications I'm gonna find the solution by trying more :) — Yoël Zerbib
– Yoël Zerbib, Commented Nov 7, 2020 at 14:28

charlietfl · Accepted Answer · 2020-11-07 14:25:28Z

4

Using a Set and filter to check if first and last index are the same

const returnUniques = (arr) => {
   return [...new Set(arr)].filter(n => arr.indexOf(n) === arr.lastIndexOf(n))    
}

console.log(returnUniques([1,3,1,6,22]))

edited Nov 7, 2020 at 14:25

answered Nov 7, 2020 at 14:17

charlietfl

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5 Comments

Yoël Zerbib Over a year ago

No I need to return [3,6,22]

Yoël Zerbib Over a year ago

You didn't read that I said that I could have used set method ?

Yoël Zerbib Over a year ago

Thanks, with the filter it's better, but why are we comparing the first index and last index only ? I mean, how your filter work in this case ?

Yoël Zerbib Over a year ago

Oh I see, if arr.indexOf(n) !== arr.lastIndexOf(n) it means that there is more than one like this so don't add it, am I right ?

charlietfl Over a year ago

Yes... for a very large data set this may not be the most efficient but for smaller sets should work fine

Derek Wang · Accepted Answer · 2020-11-07 14:19:17Z

2

Using Array.prototype.reduce, you can get the counts for each item.

And based on that content, you can filter the array items which are appeared only one time.

const input = [ 1, 3, 1, 6, 22 ];

const uniques = (arr) => {
  const groupBy = arr.reduce((acc, cur) => {
    acc[cur] ? acc[cur] ++ : acc[cur] = 1;
    return acc;
  }, {});
  return Object.entries(groupBy).filter(([key, count]) => count === 1).map(([key, count]) => Number(key));
};

console.log(uniques(input));

answered Nov 7, 2020 at 14:19

Derek Wang

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Comments

Rvfvl · Accepted Answer · 2020-11-07 14:22:57Z

1

You can do something like this.

const removeDuplicates = (arr) => {
  const results = [];

  for (const v of arr) {
    const occurences = arr.filter((x) => x === v);

    if (occurences.length === 1) results.push(v);
  }

  return results;
};

console.log(removeDuplicates([1, 3, 1, 6, 22]));

answered Nov 7, 2020 at 14:22

Rvfvl

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1 Comment

Krisztián Balla Over a year ago

Nice. I like this solution the most as it is solid clean code that is easy to understand.

Alan Omar · Accepted Answer · 2020-11-07 16:05:06Z

1

if the first index is equal to last index of the element then it is unique:

const unique = (array) =>
array.filter((e,idx,arr) => arr.indexOf(e) === arr.lastIndexOf(e))
console.log(unique([1,3,1,6,22]))

edited Nov 7, 2020 at 16:05

answered Nov 7, 2020 at 15:50

Alan Omar

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Comments

GMKHussain · Accepted Answer · 2020-11-07 14:38:23Z

0

Something like this

let arrayData = [1, 3, 1, 6, 22];

let getUniqueValuesFunc = (arrayParam) => {
  let unqiueValues = [];

  for (let unqiueValue of arrayParam ) {
    let compare = arrayParam.filter((x) => x === unqiueValue );

    if (compare.length === 1) unqiueValues.push(unqiueValue);
  }
  return unqiueValues;
};

console.log(getUniqueValuesFunc( arrayData ));

answered Nov 7, 2020 at 14:38

GMKHussain

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2 Comments

Yoël Zerbib Over a year ago

Love this one ! easily understandable :)

Krisztián Balla Over a year ago

This is just the same as the solution of @Rvfvl above. :-)

Abishek Kumar · Accepted Answer · 2020-11-07 14:38:32Z

0

Solution in single loop:

const input = [1, 3, 1, 1, 6, 22];

const returnUniques = (array) => {
    let output = [];
    let filter = [];
    for (let j = 0; j < array.length; j++) {
        const index = output.indexOf(array[j])
        if (index==-1) {
            output.push(array[j])
        } else {
            filter.push(array[j]);
        }
    }
    output = output.filter((o)=>(filter.indexOf(o)==-1))
    return output;
}

console.log(returnUniques(input));

edited Nov 7, 2020 at 14:38

answered Nov 7, 2020 at 14:25

Abishek Kumar

5395 silver badges15 bronze badges

2 Comments

Krisztián Balla Over a year ago

This won't work if for example 1 is within input three times.

Yoël Zerbib Over a year ago

@AbishekKumar Good reflection though !

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