Get distinct duplicates from array

Question

I have an array [1,1,1,1,2,3,4,5,5,6,7,8,8,8]

How can I get an array of the distinct duplicates [1,5,8] - each duplicate in the result only once, regardless of how many times it appears in the original array

my code:

var types = availControls.map(item => item.type);
var sorted_types = types.slice().sort();

availControlTypes = [];
for (var i = 0; i < sorted_types.length - 1, i++) {
   if (sorted_types[i + 1] == sorted_types[i])
   availControlTypes.push(sorted_types[i]);
}

This gets me the duplicates, but not unique.

OK, where is your code and what is the complicated part that you need help with? — Dalorzo
– Dalorzo, Commented Oct 28, 2019 at 20:46
.map and arrow functions are ES6 features... Are you sure you're using pre-ES5 code? — FZs
– FZs, Commented Oct 28, 2019 at 20:48
not sure, I tried using ...Set and it didnt' work... but I took that out of the question — full-stack
– full-stack, Commented Oct 28, 2019 at 20:49
the accepted answer for this question is what you need stackoverflow.com/questions/49215358/… — Mike Ezzati
– Mike Ezzati, Commented Oct 28, 2019 at 20:55

Mike Ezzati · Accepted Answer · 2019-10-28 20:59:15Z

1

This will do it

var input = [1, 1, 1, 1, 2, 3, 4, 5, 5, 6, 7, 8, 8, 8];

let filterDuplicates = arr => [...new Set(arr.filter((item, index) => arr.indexOf(item) != index))]


console.log(filterDuplicates(input))

answered Oct 28, 2019 at 20:59

Mike Ezzati

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Comments

Ori Drori · Accepted Answer · 2019-10-28 20:51:41Z

You need a for loop, with an object that will hold the number of times the number appeared in the array. When the count is already 1, we can add the item to the result. We should continue to increment the counter, so we won't add more than a single duplicate of the same number (although we can stop at 2).

function fn(arr) {
  var counts = {};
  var result = [];
  var n;
  
  for(var i = 0; i < arr.length; i++) {
    n = arr[i]; // get the current number
    
    if(counts[n] === 1) result.push(n); // if counts is exactly 1, we should add the number to results
    
    counts[n] = (counts[n] || 0) +1; // increment the counter
  }
  
  return result;
}

var arr = [1, 1, 1, 1, 2, 3, 4, 5, 5, 6, 7, 8, 8, 8];

var result = fn(arr);

console.log(result)

Hero Qu · Accepted Answer · 2019-10-28 21:00:11Z

0

const dupes = arr =>
  Array.from(
    arr.reduce((acc, item) => {
      acc.set(item, (acc.get(item) || 0) + 1);
      return acc;
    }, new Map())
  )
    .filter(x => x[1] > 1)
    .map(x => x[0]);

const arr = [1, 1, 1, 1, 2, 3, 4, 5, 5, 6, 7, 8, 8, 8];

console.log(dupes(arr));

// [ 1, 5, 8 ]

answered Oct 28, 2019 at 21:00

Hero Qu

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Comments

junvar · Accepted Answer · 2019-10-28 21:01:24Z

0

ES6 1 liner. This is very similar to how we find unique values, except instead of filtering for 1st occurrences, we're filtering for 2nd occurrences.

let nthOccurrences = (a, n = 1) => a.filter((v, i) => a.filter((vv, ii) => vv === v && ii <= i).length === n);

let x = [1, 1, 1, 1, 2, 3, 4, 5, 5, 6, 7, 8, 8, 8];

let uniques = nthOccurrences(x, 1);
let uniqueDuplicates = nthOccurrences(x, 2);
let uniqueTriplets = nthOccurrences(x, 3); // unique values with 3 or more occurrences

console.log(JSON.stringify(uniques));
console.log(JSON.stringify(uniqueDuplicates));
console.log(JSON.stringify(uniqueTriplets));

edited Oct 28, 2019 at 21:01

answered Oct 28, 2019 at 20:54

junvar

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