Remove all of the duplicate numbers in an array of numbers [duplicate]

This question already has an answer here:

Get all unique values in a JavaScript array (remove duplicates)

66 answers

I received this question for practice and the wording confused me, as I see 2 results that it might want.

And either way, I'd like to see both solutions.

For example, if I have an array:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

I'm taking this as wanting the final result as either:

let finalResult = [1, 2, 3, 4, 5, 8, 9, 10];

OR:

let finalResult = [1, 9, 10];

The difference between the two being, one just removes any duplicate numbers and leaves the rest and the second just wants any number that isn't a duplicate.

Either way, I'd like to write two functions that does one of each.

This, given by someone else gives my second solution.

let elems = {},



arr2 = arr.filter(function (e) {

   if (elems[e] === undefined) {

       elems[e] = true;

    return true;

  }

  return false;

});

console.log(arr2);

I'm not sure about a function for the first one (remove all duplicates).

edited Mar 20 at 17:43

pushkin

4,613112954

asked Mar 20 at 7:09

mph85

1279

marked as duplicate by Jared Smith, pushkin, BlueRaja - Danny Pflughoeft, the_lotus, Moira Mar 20 at 18:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

If you're using lodash, you can use _.uniq()

– BlueRaja - Danny Pflughoeft
Mar 20 at 9:29

1

Further, this is asking for the inverse of Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array. Finally, this post is asking two separate questions and both have good answers elsewhere already.

– Søren D. Ptæus
Mar 20 at 9:51

To answer the question "which one is it" in a comment-answer: if you're asked to remove duplicates, I believe you should understand the first variant. The second variant removes all element that have duplicates, meaning the "original" value AND its duplicates.

– Pierre Arlaud
Mar 20 at 13:18

add a comment |

This question already has an answer here:

Get all unique values in a JavaScript array (remove duplicates)

66 answers

I received this question for practice and the wording confused me, as I see 2 results that it might want.

And either way, I'd like to see both solutions.

For example, if I have an array:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

I'm taking this as wanting the final result as either:

let finalResult = [1, 2, 3, 4, 5, 8, 9, 10];

OR:

let finalResult = [1, 9, 10];

The difference between the two being, one just removes any duplicate numbers and leaves the rest and the second just wants any number that isn't a duplicate.

Either way, I'd like to write two functions that does one of each.

This, given by someone else gives my second solution.

let elems = {},



arr2 = arr.filter(function (e) {

   if (elems[e] === undefined) {

       elems[e] = true;

    return true;

  }

  return false;

});

console.log(arr2);

I'm not sure about a function for the first one (remove all duplicates).

edited Mar 20 at 17:43

pushkin

4,613112954

asked Mar 20 at 7:09

mph85

1279

marked as duplicate by Jared Smith, pushkin, BlueRaja - Danny Pflughoeft, the_lotus, Moira Mar 20 at 18:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

If you're using lodash, you can use _.uniq()

– BlueRaja - Danny Pflughoeft
Mar 20 at 9:29

1

Further, this is asking for the inverse of Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array. Finally, this post is asking two separate questions and both have good answers elsewhere already.

– Søren D. Ptæus
Mar 20 at 9:51

To answer the question "which one is it" in a comment-answer: if you're asked to remove duplicates, I believe you should understand the first variant. The second variant removes all element that have duplicates, meaning the "original" value AND its duplicates.

– Pierre Arlaud
Mar 20 at 13:18

add a comment |

This question already has an answer here:

Get all unique values in a JavaScript array (remove duplicates)

66 answers

I received this question for practice and the wording confused me, as I see 2 results that it might want.

And either way, I'd like to see both solutions.

For example, if I have an array:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

I'm taking this as wanting the final result as either:

let finalResult = [1, 2, 3, 4, 5, 8, 9, 10];

OR:

let finalResult = [1, 9, 10];

The difference between the two being, one just removes any duplicate numbers and leaves the rest and the second just wants any number that isn't a duplicate.

Either way, I'd like to write two functions that does one of each.

This, given by someone else gives my second solution.

let elems = {},



arr2 = arr.filter(function (e) {

   if (elems[e] === undefined) {

       elems[e] = true;

    return true;

  }

  return false;

});

console.log(arr2);

I'm not sure about a function for the first one (remove all duplicates).

edited Mar 20 at 17:43

pushkin

4,613112954

asked Mar 20 at 7:09

mph85

1279

This question already has an answer here:

Get all unique values in a JavaScript array (remove duplicates)

66 answers

I received this question for practice and the wording confused me, as I see 2 results that it might want.

And either way, I'd like to see both solutions.

For example, if I have an array:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

I'm taking this as wanting the final result as either:

let finalResult = [1, 2, 3, 4, 5, 8, 9, 10];

OR:

let finalResult = [1, 9, 10];

The difference between the two being, one just removes any duplicate numbers and leaves the rest and the second just wants any number that isn't a duplicate.

Either way, I'd like to write two functions that does one of each.

This, given by someone else gives my second solution.

let elems = {},



arr2 = arr.filter(function (e) {

   if (elems[e] === undefined) {

       elems[e] = true;

    return true;

  }

  return false;

});

console.log(arr2);

I'm not sure about a function for the first one (remove all duplicates).

This question already has an answer here:

Get all unique values in a JavaScript array (remove duplicates)

66 answers

javascript arrays duplicates

edited Mar 20 at 17:43

pushkin

4,613112954

asked Mar 20 at 7:09

mph85

1279

edited Mar 20 at 17:43

pushkin

4,613112954

asked Mar 20 at 7:09

mph85

1279

edited Mar 20 at 17:43

pushkin

4,613112954

edited Mar 20 at 17:43

pushkin

4,613112954

edited Mar 20 at 17:43

pushkin

4,613112954

asked Mar 20 at 7:09

mph85

1279

asked Mar 20 at 7:09

mph85

1279

asked Mar 20 at 7:09

mph85

1279

marked as duplicate by Jared Smith, pushkin, BlueRaja - Danny Pflughoeft, the_lotus, Moira Mar 20 at 18:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Jared Smith, pushkin, BlueRaja - Danny Pflughoeft, the_lotus, Moira Mar 20 at 18:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

If you're using lodash, you can use _.uniq()

– BlueRaja - Danny Pflughoeft
Mar 20 at 9:29

1

Further, this is asking for the inverse of Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array. Finally, this post is asking two separate questions and both have good answers elsewhere already.

– Søren D. Ptæus
Mar 20 at 9:51

To answer the question "which one is it" in a comment-answer: if you're asked to remove duplicates, I believe you should understand the first variant. The second variant removes all element that have duplicates, meaning the "original" value AND its duplicates.

– Pierre Arlaud
Mar 20 at 13:18

add a comment |

If you're using lodash, you can use _.uniq()

– BlueRaja - Danny Pflughoeft
Mar 20 at 9:29

1

Further, this is asking for the inverse of Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array. Finally, this post is asking two separate questions and both have good answers elsewhere already.

– Søren D. Ptæus
Mar 20 at 9:51

To answer the question "which one is it" in a comment-answer: if you're asked to remove duplicates, I believe you should understand the first variant. The second variant removes all element that have duplicates, meaning the "original" value AND its duplicates.

– Pierre Arlaud
Mar 20 at 13:18

If you're using lodash, you can use _.uniq()

– BlueRaja - Danny Pflughoeft
Mar 20 at 9:29

Further, this is asking for the inverse of Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array. Finally, this post is asking two separate questions and both have good answers elsewhere already.

– Søren D. Ptæus
Mar 20 at 9:51

To answer the question "which one is it" in a comment-answer: if you're asked to remove duplicates, I believe you should understand the first variant. The second variant removes all element that have duplicates, meaning the "original" value AND its duplicates.

– Pierre Arlaud
Mar 20 at 13:18

add a comment |

9 Answers
9

active

oldest

votes

Using Set and Array.from()

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



console.log(Array.from(new Set(arr)));

Alternate using regex

regex explanation here

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let res = arr

  .join(',')

  .replace(/(b,w+b)(?=.*1)/ig, '')

  .split(',')

  .map(Number);



console.log(res);

Alternate using objects

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let obj = arr.reduce((acc, val) => Object.assign(acc, {

  [val]: val

}), {});



console.log(Object.values(obj));

edited Mar 20 at 8:01

answered Mar 20 at 7:11

Aswin Kumar

999115

add a comment |

Just use a simple array.filter one-liner:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b){return a - b});

console.log(finalResult);

You could use another filter statement if you wanted the second result:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b){return a - b});

console.log(finalResult);

edited Mar 20 at 7:20

Mukyuu

1,82431123

answered Mar 20 at 7:11

Jack Bashford

13.1k31847

You could also add .sort() to sort them by numerical order: .sort(function(a, b){return a - b}) on finalresult

– Mukyuu
Mar 20 at 7:13

Yes @Mukyuu, that would also be useful

– Jack Bashford
Mar 20 at 7:16

5

Worth pointing out that the run time of this approach will be quadratic on the size of the input, which is probably not great unless the input arrays are known to be always fairly small.

– Joe Lee-Moyet
Mar 20 at 11:55

1

Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant

– Yosvel Quintero
Mar 20 at 13:08

Do note that the .sort() is not necessary - the end result without the sort is still an array without any duplicate items, just in the same order as the original array. (It does make it exactly match the finalResult variable in the question though.)

– Florrie
Mar 20 at 14:53

add a comment |

For the first part you can use Set() and Spread Syntax to remove duplicates.

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let res = [...new Set(arr)]

console.log(res)

For the second part you can use reduce()

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

//to get the object with count of each number in array.

let obj = arr.reduce((ac,a) => {

  //check if number doesnot occur before then set its count to 1

  if(!ac[a]) ac[a] = 1;

  //if number is already in object increase its count

  else ac[a]++;

  return ac;

},{})

//Using reduce on all the keys of object means all numbers.

let res = Object.keys(obj).reduce((ac,a) => {

  //check if count of current number 'a' is `1` in the above object then add it into array

  if(obj[a] === 1) ac.push(+a)

  return ac;

},)

console.log(res)

edited Mar 20 at 15:53

answered Mar 20 at 7:18

Maheer Ali

7,335619

nice appreciate that, that 2nd one looks crazy. I'm assuming the time complexity for would be less than ideal compared to other results?

– mph85
Mar 20 at 7:24

@mph85 Yes its a little complex because it doesnot go through the array again and again instead it just store all the result obj and then filter it.Its better regarding performance

– Maheer Ali
Mar 20 at 7:26

could you remind me why we need the spread operator? what happens if we don't have it? @Maheer Ali

– mph85
Mar 20 at 7:28

Is it because if we don't have it, it'll just log an object?

– mph85
Mar 20 at 7:29

@mph85 No if we will not have it. We will have a Set() inside array. We use it convert Set() to Array

– Maheer Ali
Mar 20 at 7:30

|
show 1 more comment

You could sort the array before and filter the array by checking only one side for duplicates or both sides.

var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],

    result1,

    result2;



array.sort((a, b) => a - b);



result1 = array.filter((v, i, a) => a[i - 1] !== v);

result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);



console.log(...result1);

console.log(...result2)

answered Mar 20 at 7:34

Nina Scholz

193k15106177

thank for that, for the result1 what is going on with the a[i - 1] !== v?

– mph85
Mar 20 at 22:01

a is the array, i is the actual index, and v is the actual value. it takes the value from the index before of the actual index and looks if the values are not equal.

– Nina Scholz
Mar 20 at 22:03

ah ok, and sorry, it takes the value from the index before of the actual index? Not sure what you mean by that

– mph85
Mar 20 at 22:10

for example if you have the value 9 of the array as v, then the value before is 8.

– Nina Scholz
Mar 20 at 22:13

ah ok, thank you for that

– mph85
Mar 20 at 22:15

add a comment |

You can create both arrays in One Go

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let unique = new Set();

let repeated = Array.from(arr.reduce((acc, curr) => {

	acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);

	return acc;

}, new Set()));



console.log(Array.from(unique))

console.log(repeated)

answered Mar 20 at 7:35

AZ_

656210

1

Nice one mate +1

– Maheer Ali
Mar 20 at 7:47

I wonder if the ternary plus && could be unclear though (x?y:z&&w)? It's not obvious to me how JS's order of operations would handle that, and I wonder if you could get across the same logic and reasoning by using if/else?

– Florrie
Mar 20 at 14:56

add a comment |

You can use Array.prototype.reduce() create a hash object where the keys are the numbers in the array and the values are going to be the the repeated occurrence of numbers in the arr array variable..

Then using Object.keys():

Remove all duplicates Object.keys(hash)

Remove all duplicates but filtering with Array.prototype.filter() to get the numbers with only one occurrence

Code:

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), {});



// [1, 2, 3, 4, 5, 8, 9, 10];

const finalResultOne = Object.keys(hash);



// [1, 9, 10];

const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);



console.log('finalResultOne:', ...finalResultOne);

console.log('finalResultTwo:', ...finalResultTwo);

edited Mar 21 at 3:23

answered Mar 20 at 7:47

Yosvel Quintero

11.9k42531

add a comment |

You can use closure and Map

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const build = ar => {

  const mapObj = ar.reduce((acc, e) => {

    acc.has(e) ? acc.set(e, true) : acc.set(e, false)

    return acc

  }, new Map())

  

  return function(hasDup = true) {

    if(hasDup) return [...mapObj.keys()]

    else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)

  }

}



const getArr = build(arr)



console.log(getArr())

console.log(getArr(false))

answered Mar 20 at 7:30

bird

997620

add a comment |

As many other have said, the first one is just [...new Set(arr)]

For the second, just filter out those that occur more than once:

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const count = (arr, e) => arr.filter(n => n == e).length



const unique = arr => arr.filter(e => count(arr, e) < 2)



console.log(unique(arr));

answered Mar 20 at 11:37

JollyJoker

21115

add a comment |

var arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

var map = {};

var finalResult = ;

for (var i = 0; i < arr.length; i++) {

  if (!map.hasOwnProperty(arr[i])) {

    map[arr[i]] = true;

    finalResult.push(arr[i]);

  }

}



//if you need it sorted otherwise it will be in order

finalResult.sort(function(a, b) {

  return a - b

});



console.log(finalResult);

edited 2 days ago

demo

2,38433984

answered Mar 20 at 13:54

Janspeed

1,3381315

add a comment |

9 Answers
9

active

oldest

votes

9 Answers
9

active

oldest

votes

Using Set and Array.from()

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



console.log(Array.from(new Set(arr)));

Alternate using regex

regex explanation here

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let res = arr

  .join(',')

  .replace(/(b,w+b)(?=.*1)/ig, '')

  .split(',')

  .map(Number);



console.log(res);

Alternate using objects

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let obj = arr.reduce((acc, val) => Object.assign(acc, {

  [val]: val

}), {});



console.log(Object.values(obj));

edited Mar 20 at 8:01

answered Mar 20 at 7:11

Aswin Kumar

999115

add a comment |

Using Set and Array.from()

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



console.log(Array.from(new Set(arr)));

Alternate using regex

regex explanation here

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let res = arr

  .join(',')

  .replace(/(b,w+b)(?=.*1)/ig, '')

  .split(',')

  .map(Number);



console.log(res);

Alternate using objects

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let obj = arr.reduce((acc, val) => Object.assign(acc, {

  [val]: val

}), {});



console.log(Object.values(obj));

edited Mar 20 at 8:01

answered Mar 20 at 7:11

Aswin Kumar

999115

add a comment |

Using Set and Array.from()

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



console.log(Array.from(new Set(arr)));

Alternate using regex

regex explanation here

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let res = arr

  .join(',')

  .replace(/(b,w+b)(?=.*1)/ig, '')

  .split(',')

  .map(Number);



console.log(res);

Alternate using objects

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let obj = arr.reduce((acc, val) => Object.assign(acc, {

  [val]: val

}), {});



console.log(Object.values(obj));

edited Mar 20 at 8:01

answered Mar 20 at 7:11

Aswin Kumar

999115

Using Set and Array.from()

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



console.log(Array.from(new Set(arr)));

Alternate using regex

regex explanation here

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let res = arr

  .join(',')

  .replace(/(b,w+b)(?=.*1)/ig, '')

  .split(',')

  .map(Number);



console.log(res);

Alternate using objects

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let obj = arr.reduce((acc, val) => Object.assign(acc, {

  [val]: val

}), {});



console.log(Object.values(obj));

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



console.log(Array.from(new Set(arr)));

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



console.log(Array.from(new Set(arr)));

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let res = arr

  .join(',')

  .replace(/(b,w+b)(?=.*1)/ig, '')

  .split(',')

  .map(Number);



console.log(res);

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let res = arr

  .join(',')

  .replace(/(b,w+b)(?=.*1)/ig, '')

  .split(',')

  .map(Number);



console.log(res);

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let obj = arr.reduce((acc, val) => Object.assign(acc, {

  [val]: val

}), {});



console.log(Object.values(obj));

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



let obj = arr.reduce((acc, val) => Object.assign(acc, {

  [val]: val

}), {});



console.log(Object.values(obj));

edited Mar 20 at 8:01

answered Mar 20 at 7:11

Aswin Kumar

999115

edited Mar 20 at 8:01

answered Mar 20 at 7:11

Aswin Kumar

999115

answered Mar 20 at 7:11

Aswin Kumar

999115

answered Mar 20 at 7:11

Aswin Kumar

999115

add a comment |

Just use a simple array.filter one-liner:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b){return a - b});

console.log(finalResult);

You could use another filter statement if you wanted the second result:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b){return a - b});

console.log(finalResult);

edited Mar 20 at 7:20

Mukyuu

1,82431123

answered Mar 20 at 7:11

Jack Bashford

13.1k31847

You could also add .sort() to sort them by numerical order: .sort(function(a, b){return a - b}) on finalresult

– Mukyuu
Mar 20 at 7:13

Yes @Mukyuu, that would also be useful

– Jack Bashford
Mar 20 at 7:16

5

Worth pointing out that the run time of this approach will be quadratic on the size of the input, which is probably not great unless the input arrays are known to be always fairly small.

– Joe Lee-Moyet
Mar 20 at 11:55

1

Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant

– Yosvel Quintero
Mar 20 at 13:08

Do note that the .sort() is not necessary - the end result without the sort is still an array without any duplicate items, just in the same order as the original array. (It does make it exactly match the finalResult variable in the question though.)

– Florrie
Mar 20 at 14:53

add a comment |

Just use a simple array.filter one-liner:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b){return a - b});

console.log(finalResult);

You could use another filter statement if you wanted the second result:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b){return a - b});

console.log(finalResult);

edited Mar 20 at 7:20

Mukyuu

1,82431123

answered Mar 20 at 7:11

Jack Bashford

13.1k31847

You could also add .sort() to sort them by numerical order: .sort(function(a, b){return a - b}) on finalresult

– Mukyuu
Mar 20 at 7:13

Yes @Mukyuu, that would also be useful

– Jack Bashford
Mar 20 at 7:16

5

Worth pointing out that the run time of this approach will be quadratic on the size of the input, which is probably not great unless the input arrays are known to be always fairly small.

– Joe Lee-Moyet
Mar 20 at 11:55

1

Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant

– Yosvel Quintero
Mar 20 at 13:08

Do note that the .sort() is not necessary - the end result without the sort is still an array without any duplicate items, just in the same order as the original array. (It does make it exactly match the finalResult variable in the question though.)

– Florrie
Mar 20 at 14:53

add a comment |

Just use a simple array.filter one-liner:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b){return a - b});

console.log(finalResult);

You could use another filter statement if you wanted the second result:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b){return a - b});

console.log(finalResult);

edited Mar 20 at 7:20

Mukyuu

1,82431123

answered Mar 20 at 7:11

Jack Bashford

13.1k31847

Just use a simple array.filter one-liner:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b){return a - b});

console.log(finalResult);

You could use another filter statement if you wanted the second result:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b){return a - b});

console.log(finalResult);

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b){return a - b});

console.log(finalResult);

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b){return a - b});

console.log(finalResult);

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b){return a - b});

console.log(finalResult);

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b){return a - b});

console.log(finalResult);

edited Mar 20 at 7:20

Mukyuu

1,82431123

answered Mar 20 at 7:11

Jack Bashford

13.1k31847

edited Mar 20 at 7:20

Mukyuu

1,82431123

edited Mar 20 at 7:20

Mukyuu

1,82431123

edited Mar 20 at 7:20

Mukyuu

1,82431123

answered Mar 20 at 7:11

Jack Bashford

13.1k31847

answered Mar 20 at 7:11

Jack Bashford

13.1k31847

answered Mar 20 at 7:11

Jack Bashford

13.1k31847

You could also add .sort() to sort them by numerical order: .sort(function(a, b){return a - b}) on finalresult

– Mukyuu
Mar 20 at 7:13

Yes @Mukyuu, that would also be useful

– Jack Bashford
Mar 20 at 7:16

5

Worth pointing out that the run time of this approach will be quadratic on the size of the input, which is probably not great unless the input arrays are known to be always fairly small.

– Joe Lee-Moyet
Mar 20 at 11:55

1

Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant

– Yosvel Quintero
Mar 20 at 13:08

Do note that the .sort() is not necessary - the end result without the sort is still an array without any duplicate items, just in the same order as the original array. (It does make it exactly match the finalResult variable in the question though.)

– Florrie
Mar 20 at 14:53

add a comment |

You could also add .sort() to sort them by numerical order: .sort(function(a, b){return a - b}) on finalresult

– Mukyuu
Mar 20 at 7:13

Yes @Mukyuu, that would also be useful

– Jack Bashford
Mar 20 at 7:16

5

Worth pointing out that the run time of this approach will be quadratic on the size of the input, which is probably not great unless the input arrays are known to be always fairly small.

– Joe Lee-Moyet
Mar 20 at 11:55

1

Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant

– Yosvel Quintero
Mar 20 at 13:08

Do note that the .sort() is not necessary - the end result without the sort is still an array without any duplicate items, just in the same order as the original array. (It does make it exactly match the finalResult variable in the question though.)

– Florrie
Mar 20 at 14:53

You could also add .sort() to sort them by numerical order: .sort(function(a, b){return a - b}) on finalresult

– Mukyuu
Mar 20 at 7:13

Yes @Mukyuu, that would also be useful

– Jack Bashford
Mar 20 at 7:16

Worth pointing out that the run time of this approach will be quadratic on the size of the input, which is probably not great unless the input arrays are known to be always fairly small.

– Joe Lee-Moyet
Mar 20 at 11:55

Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant

– Yosvel Quintero
Mar 20 at 13:08

Do note that the .sort() is not necessary - the end result without the sort is still an array without any duplicate items, just in the same order as the original array. (It does make it exactly match the finalResult variable in the question though.)

– Florrie
Mar 20 at 14:53

add a comment |

For the first part you can use Set() and Spread Syntax to remove duplicates.

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let res = [...new Set(arr)]

console.log(res)

For the second part you can use reduce()

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

//to get the object with count of each number in array.

let obj = arr.reduce((ac,a) => {

  //check if number doesnot occur before then set its count to 1

  if(!ac[a]) ac[a] = 1;

  //if number is already in object increase its count

  else ac[a]++;

  return ac;

},{})

//Using reduce on all the keys of object means all numbers.

let res = Object.keys(obj).reduce((ac,a) => {

  //check if count of current number 'a' is `1` in the above object then add it into array

  if(obj[a] === 1) ac.push(+a)

  return ac;

},)

console.log(res)

edited Mar 20 at 15:53

answered Mar 20 at 7:18

Maheer Ali

7,335619

nice appreciate that, that 2nd one looks crazy. I'm assuming the time complexity for would be less than ideal compared to other results?

– mph85
Mar 20 at 7:24

@mph85 Yes its a little complex because it doesnot go through the array again and again instead it just store all the result obj and then filter it.Its better regarding performance

– Maheer Ali
Mar 20 at 7:26

could you remind me why we need the spread operator? what happens if we don't have it? @Maheer Ali

– mph85
Mar 20 at 7:28

Is it because if we don't have it, it'll just log an object?

– mph85
Mar 20 at 7:29

@mph85 No if we will not have it. We will have a Set() inside array. We use it convert Set() to Array

– Maheer Ali
Mar 20 at 7:30

|
show 1 more comment

For the first part you can use Set() and Spread Syntax to remove duplicates.

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let res = [...new Set(arr)]

console.log(res)

For the second part you can use reduce()

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

//to get the object with count of each number in array.

let obj = arr.reduce((ac,a) => {

  //check if number doesnot occur before then set its count to 1

  if(!ac[a]) ac[a] = 1;

  //if number is already in object increase its count

  else ac[a]++;

  return ac;

},{})

//Using reduce on all the keys of object means all numbers.

let res = Object.keys(obj).reduce((ac,a) => {

  //check if count of current number 'a' is `1` in the above object then add it into array

  if(obj[a] === 1) ac.push(+a)

  return ac;

},)

console.log(res)

edited Mar 20 at 15:53

answered Mar 20 at 7:18

Maheer Ali

7,335619

nice appreciate that, that 2nd one looks crazy. I'm assuming the time complexity for would be less than ideal compared to other results?

– mph85
Mar 20 at 7:24

@mph85 Yes its a little complex because it doesnot go through the array again and again instead it just store all the result obj and then filter it.Its better regarding performance

– Maheer Ali
Mar 20 at 7:26

could you remind me why we need the spread operator? what happens if we don't have it? @Maheer Ali

– mph85
Mar 20 at 7:28

Is it because if we don't have it, it'll just log an object?

– mph85
Mar 20 at 7:29

@mph85 No if we will not have it. We will have a Set() inside array. We use it convert Set() to Array

– Maheer Ali
Mar 20 at 7:30

|
show 1 more comment

For the first part you can use Set() and Spread Syntax to remove duplicates.

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let res = [...new Set(arr)]

console.log(res)

For the second part you can use reduce()

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

//to get the object with count of each number in array.

let obj = arr.reduce((ac,a) => {

  //check if number doesnot occur before then set its count to 1

  if(!ac[a]) ac[a] = 1;

  //if number is already in object increase its count

  else ac[a]++;

  return ac;

},{})

//Using reduce on all the keys of object means all numbers.

let res = Object.keys(obj).reduce((ac,a) => {

  //check if count of current number 'a' is `1` in the above object then add it into array

  if(obj[a] === 1) ac.push(+a)

  return ac;

},)

console.log(res)

edited Mar 20 at 15:53

answered Mar 20 at 7:18

Maheer Ali

7,335619

For the first part you can use Set() and Spread Syntax to remove duplicates.

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let res = [...new Set(arr)]

console.log(res)

For the second part you can use reduce()

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

//to get the object with count of each number in array.

let obj = arr.reduce((ac,a) => {

  //check if number doesnot occur before then set its count to 1

  if(!ac[a]) ac[a] = 1;

  //if number is already in object increase its count

  else ac[a]++;

  return ac;

},{})

//Using reduce on all the keys of object means all numbers.

let res = Object.keys(obj).reduce((ac,a) => {

  //check if count of current number 'a' is `1` in the above object then add it into array

  if(obj[a] === 1) ac.push(+a)

  return ac;

},)

console.log(res)

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let res = [...new Set(arr)]

console.log(res)

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let res = [...new Set(arr)]

console.log(res)

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

//to get the object with count of each number in array.

let obj = arr.reduce((ac,a) => {

  //check if number doesnot occur before then set its count to 1

  if(!ac[a]) ac[a] = 1;

  //if number is already in object increase its count

  else ac[a]++;

  return ac;

},{})

//Using reduce on all the keys of object means all numbers.

let res = Object.keys(obj).reduce((ac,a) => {

  //check if count of current number 'a' is `1` in the above object then add it into array

  if(obj[a] === 1) ac.push(+a)

  return ac;

},)

console.log(res)

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

//to get the object with count of each number in array.

let obj = arr.reduce((ac,a) => {

  //check if number doesnot occur before then set its count to 1

  if(!ac[a]) ac[a] = 1;

  //if number is already in object increase its count

  else ac[a]++;

  return ac;

},{})

//Using reduce on all the keys of object means all numbers.

let res = Object.keys(obj).reduce((ac,a) => {

  //check if count of current number 'a' is `1` in the above object then add it into array

  if(obj[a] === 1) ac.push(+a)

  return ac;

},)

console.log(res)

edited Mar 20 at 15:53

answered Mar 20 at 7:18

Maheer Ali

7,335619

edited Mar 20 at 15:53

answered Mar 20 at 7:18

Maheer Ali

7,335619

answered Mar 20 at 7:18

Maheer Ali

7,335619

answered Mar 20 at 7:18

Maheer Ali

7,335619

nice appreciate that, that 2nd one looks crazy. I'm assuming the time complexity for would be less than ideal compared to other results?

– mph85
Mar 20 at 7:24

@mph85 Yes its a little complex because it doesnot go through the array again and again instead it just store all the result obj and then filter it.Its better regarding performance

– Maheer Ali
Mar 20 at 7:26

could you remind me why we need the spread operator? what happens if we don't have it? @Maheer Ali

– mph85
Mar 20 at 7:28

Is it because if we don't have it, it'll just log an object?

– mph85
Mar 20 at 7:29

@mph85 No if we will not have it. We will have a Set() inside array. We use it convert Set() to Array

– Maheer Ali
Mar 20 at 7:30

|
show 1 more comment

nice appreciate that, that 2nd one looks crazy. I'm assuming the time complexity for would be less than ideal compared to other results?

– mph85
Mar 20 at 7:24

@mph85 Yes its a little complex because it doesnot go through the array again and again instead it just store all the result obj and then filter it.Its better regarding performance

– Maheer Ali
Mar 20 at 7:26

could you remind me why we need the spread operator? what happens if we don't have it? @Maheer Ali

– mph85
Mar 20 at 7:28

Is it because if we don't have it, it'll just log an object?

– mph85
Mar 20 at 7:29

@mph85 No if we will not have it. We will have a Set() inside array. We use it convert Set() to Array

– Maheer Ali
Mar 20 at 7:30

nice appreciate that, that 2nd one looks crazy. I'm assuming the time complexity for would be less than ideal compared to other results?

– mph85
Mar 20 at 7:24

@mph85 Yes its a little complex because it doesnot go through the array again and again instead it just store all the result obj and then filter it.Its better regarding performance

– Maheer Ali
Mar 20 at 7:26

could you remind me why we need the spread operator? what happens if we don't have it? @Maheer Ali

– mph85
Mar 20 at 7:28

Is it because if we don't have it, it'll just log an object?

– mph85
Mar 20 at 7:29

@mph85 No if we will not have it. We will have a Set() inside array. We use it convert Set() to Array

– Maheer Ali
Mar 20 at 7:30

|
show 1 more comment

You could sort the array before and filter the array by checking only one side for duplicates or both sides.

var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],

    result1,

    result2;



array.sort((a, b) => a - b);



result1 = array.filter((v, i, a) => a[i - 1] !== v);

result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);



console.log(...result1);

console.log(...result2)

answered Mar 20 at 7:34

Nina Scholz

193k15106177

thank for that, for the result1 what is going on with the a[i - 1] !== v?

– mph85
Mar 20 at 22:01

a is the array, i is the actual index, and v is the actual value. it takes the value from the index before of the actual index and looks if the values are not equal.

– Nina Scholz
Mar 20 at 22:03

ah ok, and sorry, it takes the value from the index before of the actual index? Not sure what you mean by that

– mph85
Mar 20 at 22:10

for example if you have the value 9 of the array as v, then the value before is 8.

– Nina Scholz
Mar 20 at 22:13

ah ok, thank you for that

– mph85
Mar 20 at 22:15

add a comment |

You could sort the array before and filter the array by checking only one side for duplicates or both sides.

var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],

    result1,

    result2;



array.sort((a, b) => a - b);



result1 = array.filter((v, i, a) => a[i - 1] !== v);

result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);



console.log(...result1);

console.log(...result2)

answered Mar 20 at 7:34

Nina Scholz

193k15106177

thank for that, for the result1 what is going on with the a[i - 1] !== v?

– mph85
Mar 20 at 22:01

a is the array, i is the actual index, and v is the actual value. it takes the value from the index before of the actual index and looks if the values are not equal.

– Nina Scholz
Mar 20 at 22:03

ah ok, and sorry, it takes the value from the index before of the actual index? Not sure what you mean by that

– mph85
Mar 20 at 22:10

for example if you have the value 9 of the array as v, then the value before is 8.

– Nina Scholz
Mar 20 at 22:13

ah ok, thank you for that

– mph85
Mar 20 at 22:15

add a comment |

You could sort the array before and filter the array by checking only one side for duplicates or both sides.

var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],

    result1,

    result2;



array.sort((a, b) => a - b);



result1 = array.filter((v, i, a) => a[i - 1] !== v);

result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);



console.log(...result1);

console.log(...result2)

answered Mar 20 at 7:34

Nina Scholz

193k15106177

You could sort the array before and filter the array by checking only one side for duplicates or both sides.

var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],

    result1,

    result2;



array.sort((a, b) => a - b);



result1 = array.filter((v, i, a) => a[i - 1] !== v);

result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);



console.log(...result1);

console.log(...result2)

var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],

    result1,

    result2;



array.sort((a, b) => a - b);



result1 = array.filter((v, i, a) => a[i - 1] !== v);

result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);



console.log(...result1);

console.log(...result2)

var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],

    result1,

    result2;



array.sort((a, b) => a - b);



result1 = array.filter((v, i, a) => a[i - 1] !== v);

result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);



console.log(...result1);

console.log(...result2)

answered Mar 20 at 7:34

Nina Scholz

193k15106177

answered Mar 20 at 7:34

Nina Scholz

193k15106177

answered Mar 20 at 7:34

Nina Scholz

193k15106177

answered Mar 20 at 7:34

Nina Scholz

193k15106177

thank for that, for the result1 what is going on with the a[i - 1] !== v?

– mph85
Mar 20 at 22:01

a is the array, i is the actual index, and v is the actual value. it takes the value from the index before of the actual index and looks if the values are not equal.

– Nina Scholz
Mar 20 at 22:03

ah ok, and sorry, it takes the value from the index before of the actual index? Not sure what you mean by that

– mph85
Mar 20 at 22:10

for example if you have the value 9 of the array as v, then the value before is 8.

– Nina Scholz
Mar 20 at 22:13

ah ok, thank you for that

– mph85
Mar 20 at 22:15

add a comment |

thank for that, for the result1 what is going on with the a[i - 1] !== v?

– mph85
Mar 20 at 22:01

a is the array, i is the actual index, and v is the actual value. it takes the value from the index before of the actual index and looks if the values are not equal.

– Nina Scholz
Mar 20 at 22:03

ah ok, and sorry, it takes the value from the index before of the actual index? Not sure what you mean by that

– mph85
Mar 20 at 22:10

for example if you have the value 9 of the array as v, then the value before is 8.

– Nina Scholz
Mar 20 at 22:13

ah ok, thank you for that

– mph85
Mar 20 at 22:15

thank for that, for the result1 what is going on with the a[i - 1] !== v?

– mph85
Mar 20 at 22:01

a is the array, i is the actual index, and v is the actual value. it takes the value from the index before of the actual index and looks if the values are not equal.

– Nina Scholz
Mar 20 at 22:03

ah ok, and sorry, it takes the value from the index before of the actual index? Not sure what you mean by that

– mph85
Mar 20 at 22:10

for example if you have the value 9 of the array as v, then the value before is 8.

– Nina Scholz
Mar 20 at 22:13

ah ok, thank you for that

– mph85
Mar 20 at 22:15

add a comment |

You can create both arrays in One Go

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let unique = new Set();

let repeated = Array.from(arr.reduce((acc, curr) => {

	acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);

	return acc;

}, new Set()));



console.log(Array.from(unique))

console.log(repeated)

answered Mar 20 at 7:35

AZ_

656210

1

Nice one mate +1

– Maheer Ali
Mar 20 at 7:47

I wonder if the ternary plus && could be unclear though (x?y:z&&w)? It's not obvious to me how JS's order of operations would handle that, and I wonder if you could get across the same logic and reasoning by using if/else?

– Florrie
Mar 20 at 14:56

add a comment |

You can create both arrays in One Go

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let unique = new Set();

let repeated = Array.from(arr.reduce((acc, curr) => {

	acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);

	return acc;

}, new Set()));



console.log(Array.from(unique))

console.log(repeated)

answered Mar 20 at 7:35

AZ_

656210

1

Nice one mate +1

– Maheer Ali
Mar 20 at 7:47

I wonder if the ternary plus && could be unclear though (x?y:z&&w)? It's not obvious to me how JS's order of operations would handle that, and I wonder if you could get across the same logic and reasoning by using if/else?

– Florrie
Mar 20 at 14:56

add a comment |

You can create both arrays in One Go

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let unique = new Set();

let repeated = Array.from(arr.reduce((acc, curr) => {

	acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);

	return acc;

}, new Set()));



console.log(Array.from(unique))

console.log(repeated)

answered Mar 20 at 7:35

AZ_

656210

You can create both arrays in One Go

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let unique = new Set();

let repeated = Array.from(arr.reduce((acc, curr) => {

	acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);

	return acc;

}, new Set()));



console.log(Array.from(unique))

console.log(repeated)

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let unique = new Set();

let repeated = Array.from(arr.reduce((acc, curr) => {

	acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);

	return acc;

}, new Set()));



console.log(Array.from(unique))

console.log(repeated)

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let unique = new Set();

let repeated = Array.from(arr.reduce((acc, curr) => {

	acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);

	return acc;

}, new Set()));



console.log(Array.from(unique))

console.log(repeated)

answered Mar 20 at 7:35

AZ_

656210

answered Mar 20 at 7:35

AZ_

656210

answered Mar 20 at 7:35

AZ_

656210

answered Mar 20 at 7:35

AZ_

656210

1

Nice one mate +1

– Maheer Ali
Mar 20 at 7:47

I wonder if the ternary plus && could be unclear though (x?y:z&&w)? It's not obvious to me how JS's order of operations would handle that, and I wonder if you could get across the same logic and reasoning by using if/else?

– Florrie
Mar 20 at 14:56

add a comment |

1

Nice one mate +1

– Maheer Ali
Mar 20 at 7:47

I wonder if the ternary plus && could be unclear though (x?y:z&&w)? It's not obvious to me how JS's order of operations would handle that, and I wonder if you could get across the same logic and reasoning by using if/else?

– Florrie
Mar 20 at 14:56

Nice one mate +1

– Maheer Ali
Mar 20 at 7:47

I wonder if the ternary plus && could be unclear though (x?y:z&&w)? It's not obvious to me how JS's order of operations would handle that, and I wonder if you could get across the same logic and reasoning by using if/else?

– Florrie
Mar 20 at 14:56

add a comment |

Then using Object.keys():

Remove all duplicates Object.keys(hash)

Remove all duplicates but filtering with Array.prototype.filter() to get the numbers with only one occurrence

Code:

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), {});



// [1, 2, 3, 4, 5, 8, 9, 10];

const finalResultOne = Object.keys(hash);



// [1, 9, 10];

const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);



console.log('finalResultOne:', ...finalResultOne);

console.log('finalResultTwo:', ...finalResultTwo);

edited Mar 21 at 3:23

answered Mar 20 at 7:47

Yosvel Quintero

11.9k42531

add a comment |

Then using Object.keys():

Remove all duplicates Object.keys(hash)

Remove all duplicates but filtering with Array.prototype.filter() to get the numbers with only one occurrence

Code:

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), {});



// [1, 2, 3, 4, 5, 8, 9, 10];

const finalResultOne = Object.keys(hash);



// [1, 9, 10];

const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);



console.log('finalResultOne:', ...finalResultOne);

console.log('finalResultTwo:', ...finalResultTwo);

edited Mar 21 at 3:23

answered Mar 20 at 7:47

Yosvel Quintero

11.9k42531

add a comment |

Then using Object.keys():

Remove all duplicates Object.keys(hash)

Remove all duplicates but filtering with Array.prototype.filter() to get the numbers with only one occurrence

Code:

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), {});



// [1, 2, 3, 4, 5, 8, 9, 10];

const finalResultOne = Object.keys(hash);



// [1, 9, 10];

const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);



console.log('finalResultOne:', ...finalResultOne);

console.log('finalResultTwo:', ...finalResultTwo);

edited Mar 21 at 3:23

answered Mar 20 at 7:47

Yosvel Quintero

11.9k42531

Then using Object.keys():

Remove all duplicates Object.keys(hash)

Remove all duplicates but filtering with Array.prototype.filter() to get the numbers with only one occurrence

Code:

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), {});



// [1, 2, 3, 4, 5, 8, 9, 10];

const finalResultOne = Object.keys(hash);



// [1, 9, 10];

const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);



console.log('finalResultOne:', ...finalResultOne);

console.log('finalResultTwo:', ...finalResultTwo);

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), {});



// [1, 2, 3, 4, 5, 8, 9, 10];

const finalResultOne = Object.keys(hash);



// [1, 9, 10];

const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);



console.log('finalResultOne:', ...finalResultOne);

console.log('finalResultTwo:', ...finalResultTwo);

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), {});



// [1, 2, 3, 4, 5, 8, 9, 10];

const finalResultOne = Object.keys(hash);



// [1, 9, 10];

const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);



console.log('finalResultOne:', ...finalResultOne);

console.log('finalResultTwo:', ...finalResultTwo);

edited Mar 21 at 3:23

answered Mar 20 at 7:47

Yosvel Quintero

11.9k42531

edited Mar 21 at 3:23

answered Mar 20 at 7:47

Yosvel Quintero

11.9k42531

answered Mar 20 at 7:47

Yosvel Quintero

11.9k42531

answered Mar 20 at 7:47

Yosvel Quintero

11.9k42531

add a comment |

You can use closure and Map

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const build = ar => {

  const mapObj = ar.reduce((acc, e) => {

    acc.has(e) ? acc.set(e, true) : acc.set(e, false)

    return acc

  }, new Map())

  

  return function(hasDup = true) {

    if(hasDup) return [...mapObj.keys()]

    else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)

  }

}



const getArr = build(arr)



console.log(getArr())

console.log(getArr(false))

answered Mar 20 at 7:30

bird

997620

add a comment |

You can use closure and Map

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const build = ar => {

  const mapObj = ar.reduce((acc, e) => {

    acc.has(e) ? acc.set(e, true) : acc.set(e, false)

    return acc

  }, new Map())

  

  return function(hasDup = true) {

    if(hasDup) return [...mapObj.keys()]

    else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)

  }

}



const getArr = build(arr)



console.log(getArr())

console.log(getArr(false))

answered Mar 20 at 7:30

bird

997620

add a comment |

You can use closure and Map

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const build = ar => {

  const mapObj = ar.reduce((acc, e) => {

    acc.has(e) ? acc.set(e, true) : acc.set(e, false)

    return acc

  }, new Map())

  

  return function(hasDup = true) {

    if(hasDup) return [...mapObj.keys()]

    else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)

  }

}



const getArr = build(arr)



console.log(getArr())

console.log(getArr(false))

answered Mar 20 at 7:30

bird

997620

You can use closure and Map

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const build = ar => {

  const mapObj = ar.reduce((acc, e) => {

    acc.has(e) ? acc.set(e, true) : acc.set(e, false)

    return acc

  }, new Map())

  

  return function(hasDup = true) {

    if(hasDup) return [...mapObj.keys()]

    else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)

  }

}



const getArr = build(arr)



console.log(getArr())

console.log(getArr(false))

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const build = ar => {

  const mapObj = ar.reduce((acc, e) => {

    acc.has(e) ? acc.set(e, true) : acc.set(e, false)

    return acc

  }, new Map())

  

  return function(hasDup = true) {

    if(hasDup) return [...mapObj.keys()]

    else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)

  }

}



const getArr = build(arr)



console.log(getArr())

console.log(getArr(false))

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const build = ar => {

  const mapObj = ar.reduce((acc, e) => {

    acc.has(e) ? acc.set(e, true) : acc.set(e, false)

    return acc

  }, new Map())

  

  return function(hasDup = true) {

    if(hasDup) return [...mapObj.keys()]

    else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)

  }

}



const getArr = build(arr)



console.log(getArr())

console.log(getArr(false))

answered Mar 20 at 7:30

bird

997620

answered Mar 20 at 7:30

bird

997620

answered Mar 20 at 7:30

bird

997620

answered Mar 20 at 7:30

bird

997620

add a comment |

As many other have said, the first one is just [...new Set(arr)]

For the second, just filter out those that occur more than once:

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const count = (arr, e) => arr.filter(n => n == e).length



const unique = arr => arr.filter(e => count(arr, e) < 2)



console.log(unique(arr));

answered Mar 20 at 11:37

JollyJoker

21115

add a comment |

As many other have said, the first one is just [...new Set(arr)]

For the second, just filter out those that occur more than once:

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const count = (arr, e) => arr.filter(n => n == e).length



const unique = arr => arr.filter(e => count(arr, e) < 2)



console.log(unique(arr));

answered Mar 20 at 11:37

JollyJoker

21115

add a comment |

As many other have said, the first one is just [...new Set(arr)]

For the second, just filter out those that occur more than once:

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const count = (arr, e) => arr.filter(n => n == e).length



const unique = arr => arr.filter(e => count(arr, e) < 2)



console.log(unique(arr));

answered Mar 20 at 11:37

JollyJoker

21115

As many other have said, the first one is just [...new Set(arr)]

For the second, just filter out those that occur more than once:

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const count = (arr, e) => arr.filter(n => n == e).length



const unique = arr => arr.filter(e => count(arr, e) < 2)



console.log(unique(arr));

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const count = (arr, e) => arr.filter(n => n == e).length



const unique = arr => arr.filter(e => count(arr, e) < 2)



console.log(unique(arr));

const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];



const count = (arr, e) => arr.filter(n => n == e).length



const unique = arr => arr.filter(e => count(arr, e) < 2)



console.log(unique(arr));

answered Mar 20 at 11:37

JollyJoker

21115

answered Mar 20 at 11:37

JollyJoker

21115

answered Mar 20 at 11:37

JollyJoker

21115

answered Mar 20 at 11:37

JollyJoker

21115

add a comment |

var arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

var map = {};

var finalResult = ;

for (var i = 0; i < arr.length; i++) {

  if (!map.hasOwnProperty(arr[i])) {

    map[arr[i]] = true;

    finalResult.push(arr[i]);

  }

}



//if you need it sorted otherwise it will be in order

finalResult.sort(function(a, b) {

  return a - b

});



console.log(finalResult);

edited 2 days ago

demo

2,38433984

answered Mar 20 at 13:54

Janspeed

1,3381315

add a comment |

var arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

var map = {};

var finalResult = ;

for (var i = 0; i < arr.length; i++) {

  if (!map.hasOwnProperty(arr[i])) {

    map[arr[i]] = true;

    finalResult.push(arr[i]);

  }

}



//if you need it sorted otherwise it will be in order

finalResult.sort(function(a, b) {

  return a - b

});



console.log(finalResult);

edited 2 days ago

demo

2,38433984

answered Mar 20 at 13:54

Janspeed

1,3381315

add a comment |

var arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

var map = {};

var finalResult = ;

for (var i = 0; i < arr.length; i++) {

  if (!map.hasOwnProperty(arr[i])) {

    map[arr[i]] = true;

    finalResult.push(arr[i]);

  }

}



//if you need it sorted otherwise it will be in order

finalResult.sort(function(a, b) {

  return a - b

});



console.log(finalResult);

edited 2 days ago

demo

2,38433984

answered Mar 20 at 13:54

Janspeed

1,3381315

var arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

var map = {};

var finalResult = ;

for (var i = 0; i < arr.length; i++) {

  if (!map.hasOwnProperty(arr[i])) {

    map[arr[i]] = true;

    finalResult.push(arr[i]);

  }

}



//if you need it sorted otherwise it will be in order

finalResult.sort(function(a, b) {

  return a - b

});



console.log(finalResult);

var arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

var map = {};

var finalResult = ;

for (var i = 0; i < arr.length; i++) {

  if (!map.hasOwnProperty(arr[i])) {

    map[arr[i]] = true;

    finalResult.push(arr[i]);

  }

}



//if you need it sorted otherwise it will be in order

finalResult.sort(function(a, b) {

  return a - b

});



console.log(finalResult);

var arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

var map = {};

var finalResult = ;

for (var i = 0; i < arr.length; i++) {

  if (!map.hasOwnProperty(arr[i])) {

    map[arr[i]] = true;

    finalResult.push(arr[i]);

  }

}



//if you need it sorted otherwise it will be in order

finalResult.sort(function(a, b) {

  return a - b

});



console.log(finalResult);

edited 2 days ago

demo

2,38433984

answered Mar 20 at 13:54

Janspeed

1,3381315

edited 2 days ago

demo

2,38433984

edited 2 days ago

demo

2,38433984

edited 2 days ago

demo

2,38433984

answered Mar 20 at 13:54

Janspeed

1,3381315

answered Mar 20 at 13:54

Janspeed

1,3381315

answered Mar 20 at 13:54

Janspeed

1,3381315

add a comment |

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