How to create a Set from Array and remove original items in JavaScript - Stack Overflow

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I have an Array with duplicate values.

I want to create a Set to get the distinct values of that array and remove or create a new Array that will have the same data MINUS the elements required to create the Set.

This is not just a matter of remove the duplicates, but remove a SINGLE entry of a each distinct value in the original array

Something like that works, but I wonder if there is a more direct approach:

let originalValues = [
  'a',
  'a',
  'a',
  'b',
  'b',
  'c',
  'c',
  'd'
];

let distinct = new Set(originalValues);
/*
distinct -> { 'a', 'b', 'c', 'd' }
*/

// Perhaps originalValues.extract(distinct) ??
for (let val of distinct.values()) {
  const index = originalValues.indexOf(val);
  originalValues.splice(index, 1);
}

/* 
originalValues -> [
  'a',
  'a',
  'b',
  'c'
];
*/

I have an Array with duplicate values.

I want to create a Set to get the distinct values of that array and remove or create a new Array that will have the same data MINUS the elements required to create the Set.

This is not just a matter of remove the duplicates, but remove a SINGLE entry of a each distinct value in the original array

Something like that works, but I wonder if there is a more direct approach:

let originalValues = [
  'a',
  'a',
  'a',
  'b',
  'b',
  'c',
  'c',
  'd'
];

let distinct = new Set(originalValues);
/*
distinct -> { 'a', 'b', 'c', 'd' }
*/

// Perhaps originalValues.extract(distinct) ??
for (let val of distinct.values()) {
  const index = originalValues.indexOf(val);
  originalValues.splice(index, 1);
}

/* 
originalValues -> [
  'a',
  'a',
  'b',
  'c'
];
*/

• javascript
• arrays
• ecmascript-6
• set

Share Improve this question Follow edited Jan 8, 2017 at 14:56 Lucas Ricoy asked Jan 8, 2017 at 14:45 Lucas RicoyLucas Ricoy 16311 silver badge88 bronze badges 8

• 1 If originalValues starts with 'a', 'a', 'a', should it end with 'a', 'a' or just 'a'? – Ry- ♦ Commented Jan 8, 2017 at 14:46
• It should just remove one, so 'a', 'a' – Lucas Ricoy Commented Jan 8, 2017 at 14:50
• 1 Reopening because this is about removing elements that appear in the set, not about removing duplicates. – Oriol Commented Jan 8, 2017 at 14:54
• 1 originalValues.splice(index, 1); ? – dandavis Commented Jan 8, 2017 at 14:55
• 1 in that code, watch out for splicing (w/remove) an index of -1, that can ruin your day! – dandavis Commented Jan 8, 2017 at 14:57

 |  Show 3 more ments

6 Answers 6

Sorted by: Reset to default 5

Use Array#filter in bination with the Set:

const originalValues = ['a', 'a', 'a', 'b', 'b', 'c',  'c', 'd'];

const remainingValues = originalValues.filter(function(val) {
  if (this.has(val)) { // if the Set has the value
    this.delete(val); // remove it from the Set
    return false; // filter it out
  }

  return true;
}, new Set(originalValues));



console.log(remainingValues);

You could use closure over a Set and check for existence.

let originalValues = ['a', 'a', 'a', 'b', 'b', 'c', 'c', 'd'],
    result = originalValues.filter((s => a => s.has(a) || !s.add(a))(new Set));

console.log(result);

You should not use indexOf inside a loop, because it has linear cost, and the total cost bees quadratic. What I would do is use a map to count the occurrences of each item in your array, and then convert back to an array subtracting one occurrence.

let originalValues = ['a', 'a', 'a', 'b', 'b', 'c', 'c', 'd'];
let freq = new Map(); // frequency table
for (let item of originalValues)
  if (freq.has(item)) freq.set(item, freq.get(item)+1);
  else freq.set(item, 1);
var arr = [];
for (let [item,count] of freq)
  for (let i=1; i<count; ++i)
    arr.push(item);
console.log(arr);

If all items are strings you can use a plain object instead of a map.

You can create a simple Array.prototype.reduce loop with a hash table to count the number of occurrences and populate the result only if it occurs more than once.

See demo below:

var originalValues=['a','a','a','a','b','b','b','c','c','d'];

var result = originalValues.reduce(function(hash) {
  return function(p,c) {
    hash[c] = (hash[c] || 0) + 1;
    if(hash[c] > 1)
      p.push(c);
    return p;  
  };     
}(Object.create(null)), []);

console.log(result);

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Instead of using Set for this you could just use reduce() and create new array with unique values and also update original array with splice().

let oV = ["a", "a", "a", "a", "b", "b", "c", "c", "d"]

var o = {}
var distinct = oV.reduce(function(r, e) {
  if (!o[e]) o[e] = 1 && r.push(e) && oV.splice(oV.indexOf(e), 1)
  return r;
}, [])

console.log(distinct)
console.log(oV)

As an alternate approach, you can use following algorithm that will remove only 1st entry of a duplicate element. If not duplicate, it will not remove anything.

const originalValues = ['a', 'a', 'a', 'b', 'b', 'c', 'c', 'd'];

var r = originalValues.reduce(function(p, c, i, a) {
  var lIndex = a.lastIndexOf(c);
  var index = a.indexOf(c)
  if (lIndex === index || index !== i)
    p.push(c);
  return p
}, [])

console.log(r)

---

If duplicates are not case, then you can directly remove first iteration directly

const originalValues = ['a', 'a', 'a', 'b', 'b', 'c', 'c', 'd'];

var r = originalValues.filter(function(el, i) {
  return originalValues.indexOf(el) !== i
})

console.log(r)

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