如何从Array创建Set并删除JavaScript中的原始项

Question

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'
];
*/

Answer 1

Use Array#filter in combination 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); 

Answer 2

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); 

Answer 3

You should not use indexOf inside a loop, because it has linear cost, and the total cost becomes 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.

Answer 4

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); 

 .as-console-wrapper{top:0;max-height:100%!important;} 

Answer 5

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) 

Answer 6

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)