Remove Only One Duplicate from An Array

I'm trying to only remove one of the 2s from an array, but my code removes all of them. My code is as follows:

var arr = [2,7,9,5,2]

arr.filter(item => ((item !== 2)));

and:

var arr = [2,7,9,2,2,5,2]

arr.filter(item => ((item !== 2)));

Both remove all the 2s. I thought about removing duplicates, where it works if there's only one duplicate - e.g.:

Array.from(new Set([2,7,9,5,2]));

function uniq(a) {

 return Array.from(new Set(a)) 

}

But fails if there's multiple duplicates as it just removes them all, including any other duplicated numbers:

Array.from(new Set([2,7,9,9,2,2,5,2]));

function uniq(a) {

 return Array.from(new Set(a)) 

}

Does anyone know how to only remove one of the 2s? Thanks for any help here.

asked Nov 19 '18 at 10:32

user8758206

456110

1

create a simple loop, when found an index with value of 2: .splice(idx, 1) this index and break the loop

– Calvin Nunes
Nov 19 '18 at 10:34

1

Which one do you want to remove ? The first 2, or last, or ...

– Mihai Alexandru-Ionut
Nov 19 '18 at 10:34

1

Possible duplicate of How do I remove a particular element from an array in JavaScript?

– Haroldo_OK
Nov 19 '18 at 10:38

add a comment |

I'm trying to only remove one of the 2s from an array, but my code removes all of them. My code is as follows:

var arr = [2,7,9,5,2]

arr.filter(item => ((item !== 2)));

and:

var arr = [2,7,9,2,2,5,2]

arr.filter(item => ((item !== 2)));

Both remove all the 2s. I thought about removing duplicates, where it works if there's only one duplicate - e.g.:

Array.from(new Set([2,7,9,5,2]));

function uniq(a) {

 return Array.from(new Set(a)) 

}

But fails if there's multiple duplicates as it just removes them all, including any other duplicated numbers:

Array.from(new Set([2,7,9,9,2,2,5,2]));

function uniq(a) {

 return Array.from(new Set(a)) 

}

Does anyone know how to only remove one of the 2s? Thanks for any help here.

asked Nov 19 '18 at 10:32

user8758206

456110

1

create a simple loop, when found an index with value of 2: .splice(idx, 1) this index and break the loop

– Calvin Nunes
Nov 19 '18 at 10:34

1

Which one do you want to remove ? The first 2, or last, or ...

– Mihai Alexandru-Ionut
Nov 19 '18 at 10:34

1

Possible duplicate of How do I remove a particular element from an array in JavaScript?

– Haroldo_OK
Nov 19 '18 at 10:38

add a comment |

I'm trying to only remove one of the 2s from an array, but my code removes all of them. My code is as follows:

var arr = [2,7,9,5,2]

arr.filter(item => ((item !== 2)));

and:

var arr = [2,7,9,2,2,5,2]

arr.filter(item => ((item !== 2)));

Both remove all the 2s. I thought about removing duplicates, where it works if there's only one duplicate - e.g.:

Array.from(new Set([2,7,9,5,2]));

function uniq(a) {

 return Array.from(new Set(a)) 

}

But fails if there's multiple duplicates as it just removes them all, including any other duplicated numbers:

Array.from(new Set([2,7,9,9,2,2,5,2]));

function uniq(a) {

 return Array.from(new Set(a)) 

}

Does anyone know how to only remove one of the 2s? Thanks for any help here.

asked Nov 19 '18 at 10:32

user8758206

456110

I'm trying to only remove one of the 2s from an array, but my code removes all of them. My code is as follows:

var arr = [2,7,9,5,2]

arr.filter(item => ((item !== 2)));

and:

var arr = [2,7,9,2,2,5,2]

arr.filter(item => ((item !== 2)));

Both remove all the 2s. I thought about removing duplicates, where it works if there's only one duplicate - e.g.:

Array.from(new Set([2,7,9,5,2]));

function uniq(a) {

 return Array.from(new Set(a)) 

}

But fails if there's multiple duplicates as it just removes them all, including any other duplicated numbers:

Array.from(new Set([2,7,9,9,2,2,5,2]));

function uniq(a) {

 return Array.from(new Set(a)) 

}

Does anyone know how to only remove one of the 2s? Thanks for any help here.

javascript

asked Nov 19 '18 at 10:32

user8758206

456110

asked Nov 19 '18 at 10:32

user8758206

456110

asked Nov 19 '18 at 10:32

user8758206

456110

asked Nov 19 '18 at 10:32

user8758206

456110

asked Nov 19 '18 at 10:32

user8758206

456110

1

create a simple loop, when found an index with value of 2: .splice(idx, 1) this index and break the loop

– Calvin Nunes
Nov 19 '18 at 10:34

1

Which one do you want to remove ? The first 2, or last, or ...

– Mihai Alexandru-Ionut
Nov 19 '18 at 10:34

1

Possible duplicate of How do I remove a particular element from an array in JavaScript?

– Haroldo_OK
Nov 19 '18 at 10:38

add a comment |

1

create a simple loop, when found an index with value of 2: .splice(idx, 1) this index and break the loop

– Calvin Nunes
Nov 19 '18 at 10:34

1

Which one do you want to remove ? The first 2, or last, or ...

– Mihai Alexandru-Ionut
Nov 19 '18 at 10:34

1

Possible duplicate of How do I remove a particular element from an array in JavaScript?

– Haroldo_OK
Nov 19 '18 at 10:38

create a simple loop, when found an index with value of 2: .splice(idx, 1) this index and break the loop

– Calvin Nunes
Nov 19 '18 at 10:34

Which one do you want to remove ? The first 2, or last, or ...

– Mihai Alexandru-Ionut
Nov 19 '18 at 10:34

Possible duplicate of How do I remove a particular element from an array in JavaScript?

– Haroldo_OK
Nov 19 '18 at 10:38

add a comment |

6 Answers
6

active

oldest

votes

You could use indexOf method in combination with splice.

var arr = [2,7,9,5,2]

var idx = arr.indexOf(2)

if (idx >= 0) {

    arr.splice(idx, 1);

}

console.log(arr);

edited Nov 19 '18 at 10:49

Haroldo_OK

3,32521847

answered Nov 19 '18 at 10:39

Mihai Alexandru-Ionut

30.3k63971

Close, but it won't work correctly if the number isn't present on the array; checking if indexOf is non-negative would make it more robust.

– Haroldo_OK
Nov 19 '18 at 10:47

1

brilliant, thank you

– user8758206
Nov 19 '18 at 12:05

add a comment |

You could take a closure with a counter and remove only the first 2.

var array = [2, 7, 9, 2, 3, 2, 5, 2],

    result = array.filter((i => v => v !== 2 || --i)(1));

    

console.log(result);

For any other 2, you could adjust the start value for decrementing.

var array = [2, 7, 9, 2, 3, 2, 5, 2],

    result = array.filter((i => v => v !== 2 || --i)(2));

    

console.log(result);

edited Nov 19 '18 at 10:48

answered Nov 19 '18 at 10:42

Nina Scholz

182k1495164

interesting answer. What does the v and (1) and (2) represent? I'm new to JS and haven't seen this type of syntax before - can you link to anything I can read to learn this?

– user8758206
Nov 19 '18 at 16:44

1

one is to remove the first occurence of 2 and two for the second, and so on. it works, because the number gets decremented and if zero, the value is filterd out. the technique's name is closure with an IIFE (immediately-invoked function expression).

– Nina Scholz
Nov 19 '18 at 17:00

you're obviously a very advanced coder - thank you

– user8758206
Nov 19 '18 at 20:41

add a comment |

you can follow the following method

var arr= [2,3,4,2,4,5];

var unique = ;

$.each(arr, function(i, el){

    if($.inArray(el, unique) === -1) unique.push(el);

})

answered Nov 19 '18 at 10:42

Tijo John

372516

add a comment |

There are various ways to do that; one relatively simple way would be to use indexOf; see this other post: https://stackoverflow.com/a/5767357/679240

var array = [2, 7, 9, 5, 2];

console.log(array)

var index = array.indexOf(2);

if (index > -1) {

  array.splice(index, 1);

}

// array = [7, 9, 5, 2]

console.log(array);

edited Nov 19 '18 at 10:43

answered Nov 19 '18 at 10:38

Haroldo_OK

3,32521847

add a comment |

You can do:

const arr = [2, 7, 9, 2, 2, 5, 2];

const result = arr

  .reduce((a, c) => {

    a.temp[c] = ++a.temp[c] || 1;

    if (a.temp[c] !== 2) {

      a.array.push(c);

    }

    return a;

  }, {temp: {}, array: })

  .array;



console.log(result);

edited Nov 19 '18 at 10:46

answered Nov 19 '18 at 10:38

Yosvel Quintero

11.1k42429

add a comment |

Most simple way to filter all duplicates from array:

arr.filter((item, position) => arr.indexOf(item) === position)

This method skip element if another element with the same value already exist.

If you need to filter only first duplicate, you can use additional bool key:

arr.filter((item, position) => {

  if (!already && arr.indexOf(item) !== position) {

    already = true

    return false

  } else return true

})

But this method have overheaded. Smartest way is use for loop:

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

  if (arr.indexOf(arr[i]) !== i) {

    arr.splice(i,1);

    break;

  }

}

answered Nov 19 '18 at 11:21

Xeelley

314

add a comment |

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

You could use indexOf method in combination with splice.

var arr = [2,7,9,5,2]

var idx = arr.indexOf(2)

if (idx >= 0) {

    arr.splice(idx, 1);

}

console.log(arr);

edited Nov 19 '18 at 10:49

Haroldo_OK

3,32521847

answered Nov 19 '18 at 10:39

Mihai Alexandru-Ionut

30.3k63971

Close, but it won't work correctly if the number isn't present on the array; checking if indexOf is non-negative would make it more robust.

– Haroldo_OK
Nov 19 '18 at 10:47

1

brilliant, thank you

– user8758206
Nov 19 '18 at 12:05

add a comment |

You could use indexOf method in combination with splice.

var arr = [2,7,9,5,2]

var idx = arr.indexOf(2)

if (idx >= 0) {

    arr.splice(idx, 1);

}

console.log(arr);

edited Nov 19 '18 at 10:49

Haroldo_OK

3,32521847

answered Nov 19 '18 at 10:39

Mihai Alexandru-Ionut

30.3k63971

Close, but it won't work correctly if the number isn't present on the array; checking if indexOf is non-negative would make it more robust.

– Haroldo_OK
Nov 19 '18 at 10:47

1

brilliant, thank you

– user8758206
Nov 19 '18 at 12:05

add a comment |

You could use indexOf method in combination with splice.

var arr = [2,7,9,5,2]

var idx = arr.indexOf(2)

if (idx >= 0) {

    arr.splice(idx, 1);

}

console.log(arr);

edited Nov 19 '18 at 10:49

Haroldo_OK

3,32521847

answered Nov 19 '18 at 10:39

Mihai Alexandru-Ionut

30.3k63971

You could use indexOf method in combination with splice.

var arr = [2,7,9,5,2]

var idx = arr.indexOf(2)

if (idx >= 0) {

    arr.splice(idx, 1);

}

console.log(arr);

var arr = [2,7,9,5,2]

var idx = arr.indexOf(2)

if (idx >= 0) {

    arr.splice(idx, 1);

}

console.log(arr);

var arr = [2,7,9,5,2]

var idx = arr.indexOf(2)

if (idx >= 0) {

    arr.splice(idx, 1);

}

console.log(arr);

edited Nov 19 '18 at 10:49

Haroldo_OK

3,32521847

answered Nov 19 '18 at 10:39

Mihai Alexandru-Ionut

30.3k63971

edited Nov 19 '18 at 10:49

Haroldo_OK

3,32521847

edited Nov 19 '18 at 10:49

Haroldo_OK

3,32521847

edited Nov 19 '18 at 10:49

Haroldo_OK

3,32521847

answered Nov 19 '18 at 10:39

Mihai Alexandru-Ionut

30.3k63971

answered Nov 19 '18 at 10:39

Mihai Alexandru-Ionut

30.3k63971

answered Nov 19 '18 at 10:39

Mihai Alexandru-Ionut

30.3k63971

Close, but it won't work correctly if the number isn't present on the array; checking if indexOf is non-negative would make it more robust.

– Haroldo_OK
Nov 19 '18 at 10:47

1

brilliant, thank you

– user8758206
Nov 19 '18 at 12:05

add a comment |

Close, but it won't work correctly if the number isn't present on the array; checking if indexOf is non-negative would make it more robust.

– Haroldo_OK
Nov 19 '18 at 10:47

1

brilliant, thank you

– user8758206
Nov 19 '18 at 12:05

Close, but it won't work correctly if the number isn't present on the array; checking if indexOf is non-negative would make it more robust.

– Haroldo_OK
Nov 19 '18 at 10:47

brilliant, thank you

– user8758206
Nov 19 '18 at 12:05

add a comment |

You could take a closure with a counter and remove only the first 2.

var array = [2, 7, 9, 2, 3, 2, 5, 2],

    result = array.filter((i => v => v !== 2 || --i)(1));

    

console.log(result);

For any other 2, you could adjust the start value for decrementing.

var array = [2, 7, 9, 2, 3, 2, 5, 2],

    result = array.filter((i => v => v !== 2 || --i)(2));

    

console.log(result);

edited Nov 19 '18 at 10:48

answered Nov 19 '18 at 10:42

Nina Scholz

182k1495164

interesting answer. What does the v and (1) and (2) represent? I'm new to JS and haven't seen this type of syntax before - can you link to anything I can read to learn this?

– user8758206
Nov 19 '18 at 16:44

1

one is to remove the first occurence of 2 and two for the second, and so on. it works, because the number gets decremented and if zero, the value is filterd out. the technique's name is closure with an IIFE (immediately-invoked function expression).

– Nina Scholz
Nov 19 '18 at 17:00

you're obviously a very advanced coder - thank you

– user8758206
Nov 19 '18 at 20:41

add a comment |

You could take a closure with a counter and remove only the first 2.

var array = [2, 7, 9, 2, 3, 2, 5, 2],

    result = array.filter((i => v => v !== 2 || --i)(1));

    

console.log(result);

For any other 2, you could adjust the start value for decrementing.

var array = [2, 7, 9, 2, 3, 2, 5, 2],

    result = array.filter((i => v => v !== 2 || --i)(2));

    

console.log(result);

edited Nov 19 '18 at 10:48

answered Nov 19 '18 at 10:42

Nina Scholz

182k1495164

interesting answer. What does the v and (1) and (2) represent? I'm new to JS and haven't seen this type of syntax before - can you link to anything I can read to learn this?

– user8758206
Nov 19 '18 at 16:44

1

one is to remove the first occurence of 2 and two for the second, and so on. it works, because the number gets decremented and if zero, the value is filterd out. the technique's name is closure with an IIFE (immediately-invoked function expression).

– Nina Scholz
Nov 19 '18 at 17:00

you're obviously a very advanced coder - thank you

– user8758206
Nov 19 '18 at 20:41

add a comment |

You could take a closure with a counter and remove only the first 2.

var array = [2, 7, 9, 2, 3, 2, 5, 2],

    result = array.filter((i => v => v !== 2 || --i)(1));

    

console.log(result);

For any other 2, you could adjust the start value for decrementing.

var array = [2, 7, 9, 2, 3, 2, 5, 2],

    result = array.filter((i => v => v !== 2 || --i)(2));

    

console.log(result);

edited Nov 19 '18 at 10:48

answered Nov 19 '18 at 10:42

Nina Scholz

182k1495164

You could take a closure with a counter and remove only the first 2.

var array = [2, 7, 9, 2, 3, 2, 5, 2],

    result = array.filter((i => v => v !== 2 || --i)(1));

    

console.log(result);

For any other 2, you could adjust the start value for decrementing.

var array = [2, 7, 9, 2, 3, 2, 5, 2],

    result = array.filter((i => v => v !== 2 || --i)(2));

    

console.log(result);

var array = [2, 7, 9, 2, 3, 2, 5, 2],

    result = array.filter((i => v => v !== 2 || --i)(1));

    

console.log(result);

var array = [2, 7, 9, 2, 3, 2, 5, 2],

    result = array.filter((i => v => v !== 2 || --i)(1));

    

console.log(result);

var array = [2, 7, 9, 2, 3, 2, 5, 2],

    result = array.filter((i => v => v !== 2 || --i)(2));

    

console.log(result);

var array = [2, 7, 9, 2, 3, 2, 5, 2],

    result = array.filter((i => v => v !== 2 || --i)(2));

    

console.log(result);

edited Nov 19 '18 at 10:48

answered Nov 19 '18 at 10:42

Nina Scholz

182k1495164

edited Nov 19 '18 at 10:48

answered Nov 19 '18 at 10:42

Nina Scholz

182k1495164

answered Nov 19 '18 at 10:42

Nina Scholz

182k1495164

answered Nov 19 '18 at 10:42

Nina Scholz

182k1495164

interesting answer. What does the v and (1) and (2) represent? I'm new to JS and haven't seen this type of syntax before - can you link to anything I can read to learn this?

– user8758206
Nov 19 '18 at 16:44

1

one is to remove the first occurence of 2 and two for the second, and so on. it works, because the number gets decremented and if zero, the value is filterd out. the technique's name is closure with an IIFE (immediately-invoked function expression).

– Nina Scholz
Nov 19 '18 at 17:00

you're obviously a very advanced coder - thank you

– user8758206
Nov 19 '18 at 20:41

add a comment |

interesting answer. What does the v and (1) and (2) represent? I'm new to JS and haven't seen this type of syntax before - can you link to anything I can read to learn this?

– user8758206
Nov 19 '18 at 16:44

1

one is to remove the first occurence of 2 and two for the second, and so on. it works, because the number gets decremented and if zero, the value is filterd out. the technique's name is closure with an IIFE (immediately-invoked function expression).

– Nina Scholz
Nov 19 '18 at 17:00

you're obviously a very advanced coder - thank you

– user8758206
Nov 19 '18 at 20:41

interesting answer. What does the v and (1) and (2) represent? I'm new to JS and haven't seen this type of syntax before - can you link to anything I can read to learn this?

– user8758206
Nov 19 '18 at 16:44

one is to remove the first occurence of 2 and two for the second, and so on. it works, because the number gets decremented and if zero, the value is filterd out. the technique's name is closure with an IIFE (immediately-invoked function expression).

– Nina Scholz
Nov 19 '18 at 17:00

you're obviously a very advanced coder - thank you

– user8758206
Nov 19 '18 at 20:41

add a comment |

you can follow the following method

var arr= [2,3,4,2,4,5];

var unique = ;

$.each(arr, function(i, el){

    if($.inArray(el, unique) === -1) unique.push(el);

})

answered Nov 19 '18 at 10:42

Tijo John

372516

add a comment |

you can follow the following method

var arr= [2,3,4,2,4,5];

var unique = ;

$.each(arr, function(i, el){

    if($.inArray(el, unique) === -1) unique.push(el);

})

answered Nov 19 '18 at 10:42

Tijo John

372516

add a comment |

you can follow the following method

var arr= [2,3,4,2,4,5];

var unique = ;

$.each(arr, function(i, el){

    if($.inArray(el, unique) === -1) unique.push(el);

})

answered Nov 19 '18 at 10:42

Tijo John

372516

you can follow the following method

var arr= [2,3,4,2,4,5];

var unique = ;

$.each(arr, function(i, el){

    if($.inArray(el, unique) === -1) unique.push(el);

})

answered Nov 19 '18 at 10:42

Tijo John

372516

answered Nov 19 '18 at 10:42

Tijo John

372516

answered Nov 19 '18 at 10:42

Tijo John

372516

answered Nov 19 '18 at 10:42

Tijo John

372516

add a comment |

There are various ways to do that; one relatively simple way would be to use indexOf; see this other post: https://stackoverflow.com/a/5767357/679240

var array = [2, 7, 9, 5, 2];

console.log(array)

var index = array.indexOf(2);

if (index > -1) {

  array.splice(index, 1);

}

// array = [7, 9, 5, 2]

console.log(array);

edited Nov 19 '18 at 10:43

answered Nov 19 '18 at 10:38

Haroldo_OK

3,32521847

add a comment |

There are various ways to do that; one relatively simple way would be to use indexOf; see this other post: https://stackoverflow.com/a/5767357/679240

var array = [2, 7, 9, 5, 2];

console.log(array)

var index = array.indexOf(2);

if (index > -1) {

  array.splice(index, 1);

}

// array = [7, 9, 5, 2]

console.log(array);

edited Nov 19 '18 at 10:43

answered Nov 19 '18 at 10:38

Haroldo_OK

3,32521847

add a comment |

There are various ways to do that; one relatively simple way would be to use indexOf; see this other post: https://stackoverflow.com/a/5767357/679240

var array = [2, 7, 9, 5, 2];

console.log(array)

var index = array.indexOf(2);

if (index > -1) {

  array.splice(index, 1);

}

// array = [7, 9, 5, 2]

console.log(array);

edited Nov 19 '18 at 10:43

answered Nov 19 '18 at 10:38

Haroldo_OK

3,32521847

There are various ways to do that; one relatively simple way would be to use indexOf; see this other post: https://stackoverflow.com/a/5767357/679240

var array = [2, 7, 9, 5, 2];

console.log(array)

var index = array.indexOf(2);

if (index > -1) {

  array.splice(index, 1);

}

// array = [7, 9, 5, 2]

console.log(array);

var array = [2, 7, 9, 5, 2];

console.log(array)

var index = array.indexOf(2);

if (index > -1) {

  array.splice(index, 1);

}

// array = [7, 9, 5, 2]

console.log(array);

var array = [2, 7, 9, 5, 2];

console.log(array)

var index = array.indexOf(2);

if (index > -1) {

  array.splice(index, 1);

}

// array = [7, 9, 5, 2]

console.log(array);

edited Nov 19 '18 at 10:43

answered Nov 19 '18 at 10:38

Haroldo_OK

3,32521847

edited Nov 19 '18 at 10:43

answered Nov 19 '18 at 10:38

Haroldo_OK

3,32521847

answered Nov 19 '18 at 10:38

Haroldo_OK

3,32521847

answered Nov 19 '18 at 10:38

Haroldo_OK

3,32521847

add a comment |

You can do:

const arr = [2, 7, 9, 2, 2, 5, 2];

const result = arr

  .reduce((a, c) => {

    a.temp[c] = ++a.temp[c] || 1;

    if (a.temp[c] !== 2) {

      a.array.push(c);

    }

    return a;

  }, {temp: {}, array: })

  .array;



console.log(result);

edited Nov 19 '18 at 10:46

answered Nov 19 '18 at 10:38

Yosvel Quintero

11.1k42429

add a comment |

You can do:

const arr = [2, 7, 9, 2, 2, 5, 2];

const result = arr

  .reduce((a, c) => {

    a.temp[c] = ++a.temp[c] || 1;

    if (a.temp[c] !== 2) {

      a.array.push(c);

    }

    return a;

  }, {temp: {}, array: })

  .array;



console.log(result);

edited Nov 19 '18 at 10:46

answered Nov 19 '18 at 10:38

Yosvel Quintero

11.1k42429

add a comment |

You can do:

const arr = [2, 7, 9, 2, 2, 5, 2];

const result = arr

  .reduce((a, c) => {

    a.temp[c] = ++a.temp[c] || 1;

    if (a.temp[c] !== 2) {

      a.array.push(c);

    }

    return a;

  }, {temp: {}, array: })

  .array;



console.log(result);

edited Nov 19 '18 at 10:46

answered Nov 19 '18 at 10:38

Yosvel Quintero

11.1k42429

You can do:

const arr = [2, 7, 9, 2, 2, 5, 2];

const result = arr

  .reduce((a, c) => {

    a.temp[c] = ++a.temp[c] || 1;

    if (a.temp[c] !== 2) {

      a.array.push(c);

    }

    return a;

  }, {temp: {}, array: })

  .array;



console.log(result);

const arr = [2, 7, 9, 2, 2, 5, 2];

const result = arr

  .reduce((a, c) => {

    a.temp[c] = ++a.temp[c] || 1;

    if (a.temp[c] !== 2) {

      a.array.push(c);

    }

    return a;

  }, {temp: {}, array: })

  .array;



console.log(result);

const arr = [2, 7, 9, 2, 2, 5, 2];

const result = arr

  .reduce((a, c) => {

    a.temp[c] = ++a.temp[c] || 1;

    if (a.temp[c] !== 2) {

      a.array.push(c);

    }

    return a;

  }, {temp: {}, array: })

  .array;



console.log(result);

edited Nov 19 '18 at 10:46

answered Nov 19 '18 at 10:38

Yosvel Quintero

11.1k42429

edited Nov 19 '18 at 10:46

answered Nov 19 '18 at 10:38

Yosvel Quintero

11.1k42429

answered Nov 19 '18 at 10:38

Yosvel Quintero

11.1k42429

answered Nov 19 '18 at 10:38

Yosvel Quintero

11.1k42429

add a comment |

Most simple way to filter all duplicates from array:

arr.filter((item, position) => arr.indexOf(item) === position)

This method skip element if another element with the same value already exist.

If you need to filter only first duplicate, you can use additional bool key:

arr.filter((item, position) => {

  if (!already && arr.indexOf(item) !== position) {

    already = true

    return false

  } else return true

})

But this method have overheaded. Smartest way is use for loop:

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

  if (arr.indexOf(arr[i]) !== i) {

    arr.splice(i,1);

    break;

  }

}

answered Nov 19 '18 at 11:21

Xeelley

314

add a comment |

Most simple way to filter all duplicates from array:

arr.filter((item, position) => arr.indexOf(item) === position)

This method skip element if another element with the same value already exist.

If you need to filter only first duplicate, you can use additional bool key:

arr.filter((item, position) => {

  if (!already && arr.indexOf(item) !== position) {

    already = true

    return false

  } else return true

})

But this method have overheaded. Smartest way is use for loop:

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

  if (arr.indexOf(arr[i]) !== i) {

    arr.splice(i,1);

    break;

  }

}

answered Nov 19 '18 at 11:21

Xeelley

314

add a comment |

Most simple way to filter all duplicates from array:

arr.filter((item, position) => arr.indexOf(item) === position)

This method skip element if another element with the same value already exist.

If you need to filter only first duplicate, you can use additional bool key:

arr.filter((item, position) => {

  if (!already && arr.indexOf(item) !== position) {

    already = true

    return false

  } else return true

})

But this method have overheaded. Smartest way is use for loop:

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

  if (arr.indexOf(arr[i]) !== i) {

    arr.splice(i,1);

    break;

  }

}

answered Nov 19 '18 at 11:21

Xeelley

314

Most simple way to filter all duplicates from array:

arr.filter((item, position) => arr.indexOf(item) === position)

This method skip element if another element with the same value already exist.

If you need to filter only first duplicate, you can use additional bool key:

arr.filter((item, position) => {

  if (!already && arr.indexOf(item) !== position) {

    already = true

    return false

  } else return true

})

But this method have overheaded. Smartest way is use for loop:

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

  if (arr.indexOf(arr[i]) !== i) {

    arr.splice(i,1);

    break;

  }

}

answered Nov 19 '18 at 11:21

Xeelley

314

answered Nov 19 '18 at 11:21

Xeelley

314

answered Nov 19 '18 at 11:21

Xeelley

314

answered Nov 19 '18 at 11:21

Xeelley

314

add a comment |

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